Match a string that contains a newline using sed - linux

I have a string like this one:
#
pap
which basically translates to a \t#\n\tpap and I want to replace it with:
#
pap
python
which translates to \t#\n\tpap\n\tpython.
Tried this with sed in a lot of ways but it's not working maybe because sed uses new lines in a different way. I tried with:
sed -i "s/\t#\n\tpap/\t#\tpython\n\tpap/" /etc/freeradius/sites-available/default
...and many different other ways with no result. Any idea how can I do my replace in this situation?

try this line with gawk:
awk -v RS="\0" -v ORS="" '{gsub(/\t#\n\tpap/,"yourNEwString")}7' file
if you want to let sed handle new lines, you have to read the whole file first:
sed ':a;N;$!ba;s/\t#\n\tpap/NewString/g' file

This might work for you (GNU sed):
sed '/^\t#$/{n;/^\tpap$/{p;s//\tpython/}}' file
If a line contains only \t# print it, then if the next line contains only \tpap print it too, then replace that line with \tpython and print that.

A GNU sed solution that doesn't require reading the entire file at once:
sed '/^\t#$/ {n;/^\tpap$/a\\tpython'$'\n''}' file
/^\t#$/ matches comment-only lines (matching \t# exactly), in which case (only) the entire {...} expression is executed:
n loads and prints the next line.
/^\tpap/ matches that next line against \tpap exactly.
in case of a match, a\\tpython will then output \n\tpython before the following line is read - note that the spliced-in newline ($'\n') is required to signal the end of the text passed to the a command (you can alternatively use multiple -e options).
(As an aside: with BSD sed (OS X), it gets cumbersome, because
Control chars. such as \n and \t aren't directly supported and must be spliced in as ANSI C-quoted literals.
Leading whitespace is invariably stripped from the text argument to the a command, so a substitution approach must be used: s//&\'$'\n\t'python'/ replaces the pap line with itself plus the line to append:
sed '/^'$'\t''#$/ {n; /^'$'\t''pap$/ s//&\'$'\n\t'python'/;}' file
)
An awk solution (POSIX-compliant) that also doesn't require reading the entire file at once:
awk '{print} /^\t#$/ {f=1;next} f && /^\tpap$/ {print "\tpython"} {f=0}' file
{print}: prints every input line
/^\t#$/ {f=1;next}: sets flag f (for 'found') to 1 if a comment-only line (matching \t# exactly) is found and moves on to the next line.
f && /^\tpap$/ {print "\tpython"}: if a line is preceded by a comment line and matches \tpap exactly, outputs extra line \tpython.
{f=0}: resets the flag that indicates a comment-only line.

A couple of pure bash solutions:
Concise, but somewhat fragile, using parameter expansion:
in=$'\t#\n\tpap\n' # input string
echo "${in/$'\t#\n\tpap\n'/$'\t#\n\tpap\n\tpython\n'}"
Parameter expansion only supports patterns (wildcard expressions) as search strings, which limits the matching abilities:
Here the assumption is made that pap is followed by \n, whereas no assumption is made about what precedes \t#, potentially resulting in false positives.
If the assumption could be made that \t#\n\tpap is always enclosed in \n, echo "${in/$'\n\t#\n\tpap\n'/$'\n\t#\n\tpap\n\tpython\n'}" would work robustly; otherwise, see below.
Robust, but verbose, using the =~ operator for regex matching:
The =~ operator supports extended regular expressions on the right-hand side and thus allows more flexible and robust matching:
in=$'\t#\n\tpap' # input string
# Search string and string to append after.
search=$'\t#\n\tpap'
append=$'\n\tpython'
out=$in # Initialize output string to input string.
if [[ $in =~ ^(.*$'\n')?("$search")($'\n'.*)?$ ]]; then # perform regex matching
out=${out/$search/$search$append} # replace match with match + appendage
fi
echo "$out"

You can just translate the character \n to another one, then apply sed, then apply the reverse translation. If tr is used, it must be a 1-byte character, for instance \v (vertical tabulation, nowadays almost unused).
cat FILE|tr '\n' '\v'|sed 's/\t#\v\tpap/&\v\tpython/'|tr '\v' '\n'|sponge FILE
or, without sponge:
cat FILE|tr '\n' '\v'|sed 's/\t#\v\tpap/&\v\tpython/'|tr '\v' '\n' >FILE.bak && mv FILE.bak FILE

Related

How to extract and replace columns with a multi-character delimiter?

I got a file with ^$ as delimiter, the text is like :
tony^$36^$developer^$20210310^$CA
I want to replace the datetime.
I tried awk -F '\^\$' '{print $4}' file.txt | sed -i '/20210310/20221210/' , but it returns nothing. Then I tried the awk part, it returns nothing, I guess it still treat the line as a whole and the delimiter doesn't work. Wondering why and how to solve it?
A simple solution would be:
sed 's/\^\$/\n/g; s/20210310/20221210/g' -i file.txt
which will modify the file to separate each section to a new line.
If you need a different delimiter, change the \n in the command to maybe space or , .. up to you.
And it will also replace the date in the file.
If you want to see the changes, and really modify the file, remove the -i from the command.
When I run your awk command, I get these warnings:
awk: warning: escape sequence `\^' treated as plain `^'
awk: warning: escape sequence `\$' treated as plain `$'
That explains why your output is blank: the field delimiter is interpreted as the regular expression '^$', which matches a completely blank line (only). As a result, each non-blank line of input is without any field separators, and therefore has only a single field. $4 can be non-empty only if there are at least four fields.
You can fix that by escaping the backslashes:
awk -F '\\^\\$' '{print $4}' file.txt
If all you want to do is print the modified datecodes py themselves, then that should get you going. However, the question ...
How to extract and replace columns with a multi-character delimiter?
... sounds like you may want actually to replace the datecode within each line, keeping the rest intact. In that case, it is a non-starter for the awk command to discard the other parts of the line. You have several options here, but two of the more likely would be
instead of sending field 4 out to sed for substitution, do the sub in the awk script, and then reconstitute the input line by printing all fields, with the expected delimiters. (This is left as an exercise.) OR
do the whole thing in sed:
sed -E 's/^((([^^]|\^[^$])*\^\$){3})20210310(\^\$.*)/\120221210\4/' file.txt
If you wanted to modify file.txt in-place then you could add the -i flag (which, on the other hand, is not useful in your original command, where sed's input is coming from a pipe rather than a file).
The -E option engages the POSIX extended regex dialect, which allows the given regex to be more readable (the alternative would require a bunch more \ characters).
Overall, presuming that there are five or more fields delimited by literal '^$' strings, and the fourth contains exactly "20210310", that matches the first three fields, including their trailing delimiters, and captures them all as group 1; matches the leading delimiter of the fifth field and all the remainder of the line and captures it as group 4; and substitutes replaces the whole line with group 1 followed by the new datecode followed by group 4.

Linux sed command to add text between first 2 | [duplicate]

I'm trying to use sed to clean up lines of URLs to extract just the domain.
So from:
http://www.suepearson.co.uk/product/174/71/3816/
I want:
http://www.suepearson.co.uk/
(either with or without the trailing slash, it doesn't matter)
I have tried:
sed 's|\(http:\/\/.*?\/\).*|\1|'
and (escaping the non-greedy quantifier)
sed 's|\(http:\/\/.*\?\/\).*|\1|'
but I can not seem to get the non-greedy quantifier (?) to work, so it always ends up matching the whole string.
Neither basic nor extended Posix/GNU regex recognizes the non-greedy quantifier; you need a later regex. Fortunately, Perl regex for this context is pretty easy to get:
perl -pe 's|(http://.*?/).*|\1|'
In this specific case, you can get the job done without using a non-greedy regex.
Try this non-greedy regex [^/]* instead of .*?:
sed 's|\(http://[^/]*/\).*|\1|g'
With sed, I usually implement non-greedy search by searching for anything except the separator until the separator :
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*\)/.*;\1;p'
Output:
http://www.suon.co.uk
this is:
don't output -n
search, match pattern, replace and print s/<pattern>/<replace>/p
use ; search command separator instead of / to make it easier to type so s;<pattern>;<replace>;p
remember match between brackets \( ... \), later accessible with \1,\2...
match http://
followed by anything in brackets [], [ab/] would mean either a or b or /
first ^ in [] means not, so followed by anything but the thing in the []
so [^/] means anything except / character
* is to repeat previous group so [^/]* means characters except /.
so far sed -n 's;\(http://[^/]*\) means search and remember http://followed by any characters except / and remember what you've found
we want to search untill the end of domain so stop on the next / so add another / at the end: sed -n 's;\(http://[^/]*\)/' but we want to match the rest of the line after the domain so add .*
now the match remembered in group 1 (\1) is the domain so replace matched line with stuff saved in group \1 and print: sed -n 's;\(http://[^/]*\)/.*;\1;p'
If you want to include backslash after the domain as well, then add one more backslash in the group to remember:
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*/\).*;\1;p'
output:
http://www.suon.co.uk/
Simulating lazy (un-greedy) quantifier in sed
And all other regex flavors!
Finding first occurrence of an expression:
POSIX ERE (using -r option)
Regex:
(EXPRESSION).*|.
Sed:
sed -r ‍'s/(EXPRESSION).*|./\1/g' # Global `g` modifier should be on
Example (finding first sequence of digits) Live demo:
$ sed -r 's/([0-9]+).*|./\1/g' <<< 'foo 12 bar 34'
12
How does it work?
This regex benefits from an alternation |. At each position engine tries to pick the longest match (this is a POSIX standard which is followed by couple of other engines as well) which means it goes with . until a match is found for ([0-9]+).*. But order is important too.
Since global flag is set, engine tries to continue matching character by character up to the end of input string or our target. As soon as the first and only capturing group of left side of alternation is matched (EXPRESSION) rest of line is consumed immediately as well .*. We now hold our value in the first capturing group.
POSIX BRE
Regex:
\(\(\(EXPRESSION\).*\)*.\)*
Sed:
sed 's/\(\(\(EXPRESSION\).*\)*.\)*/\3/'
Example (finding first sequence of digits):
$ sed 's/\(\(\([0-9]\{1,\}\).*\)*.\)*/\3/' <<< 'foo 12 bar 34'
12
This one is like ERE version but with no alternation involved. That's all. At each single position engine tries to match a digit.
If it is found, other following digits are consumed and captured and the rest of line is matched immediately otherwise since * means
more or zero it skips over second capturing group \(\([0-9]\{1,\}\).*\)* and arrives at a dot . to match a single character and this process continues.
Finding first occurrence of a delimited expression:
This approach will match the very first occurrence of a string that is delimited. We can call it a block of string.
sed 's/\(END-DELIMITER-EXPRESSION\).*/\1/; \
s/\(\(START-DELIMITER-EXPRESSION.*\)*.\)*/\1/g'
Input string:
foobar start block #1 end barfoo start block #2 end
-EDE: end
-SDE: start
$ sed 's/\(end\).*/\1/; s/\(\(start.*\)*.\)*/\1/g'
Output:
start block #1 end
First regex \(end\).* matches and captures first end delimiter end and substitues all match with recent captured characters which
is the end delimiter. At this stage our output is: foobar start block #1 end.
Then the result is passed to second regex \(\(start.*\)*.\)* that is same as POSIX BRE version above. It matches a single character
if start delimiter start is not matched otherwise it matches and captures the start delimiter and matches the rest of characters.
Directly answering your question
Using approach #2 (delimited expression) you should select two appropriate expressions:
EDE: [^:/]\/
SDE: http:
Usage:
$ sed 's/\([^:/]\/\).*/\1/g; s/\(\(http:.*\)*.\)*/\1/' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
Output:
http://www.suepearson.co.uk/
Note: this will not work with identical delimiters.
sed does not support "non greedy" operator.
You have to use "[]" operator to exclude "/" from match.
sed 's,\(http://[^/]*\)/.*,\1,'
P.S. there is no need to backslash "/".
sed - non greedy matching by Christoph Sieghart
The trick to get non greedy matching in sed is to match all characters excluding the one that terminates the match. I know, a no-brainer, but I wasted precious minutes on it and shell scripts should be, after all, quick and easy. So in case somebody else might need it:
Greedy matching
% echo "<b>foo</b>bar" | sed 's/<.*>//g'
bar
Non greedy matching
% echo "<b>foo</b>bar" | sed 's/<[^>]*>//g'
foobar
Non-greedy solution for more than a single character
This thread is really old but I assume people still needs it.
Lets say you want to kill everything till the very first occurrence of HELLO. You cannot say [^HELLO]...
So a nice solution involves two steps, assuming that you can spare a unique word that you are not expecting in the input, say top_sekrit.
In this case we can:
s/HELLO/top_sekrit/ #will only replace the very first occurrence
s/.*top_sekrit// #kill everything till end of the first HELLO
Of course, with a simpler input you could use a smaller word, or maybe even a single character.
HTH!
This can be done using cut:
echo "http://www.suepearson.co.uk/product/174/71/3816/" | cut -d'/' -f1-3
another way, not using regex, is to use fields/delimiter method eg
string="http://www.suepearson.co.uk/product/174/71/3816/"
echo $string | awk -F"/" '{print $1,$2,$3}' OFS="/"
sed certainly has its place but this not not one of them !
As Dee has pointed out: Just use cut. It is far simpler and much more safe in this case. Here's an example where we extract various components from the URL using Bash syntax:
url="http://www.suepearson.co.uk/product/174/71/3816/"
protocol=$(echo "$url" | cut -d':' -f1)
host=$(echo "$url" | cut -d'/' -f3)
urlhost=$(echo "$url" | cut -d'/' -f1-3)
urlpath=$(echo "$url" | cut -d'/' -f4-)
gives you:
protocol = "http"
host = "www.suepearson.co.uk"
urlhost = "http://www.suepearson.co.uk"
urlpath = "product/174/71/3816/"
As you can see this is a lot more flexible approach.
(all credit to Dee)
sed 's|(http:\/\/[^\/]+\/).*|\1|'
There is still hope to solve this using pure (GNU) sed. Despite this is not a generic solution in some cases you can use "loops" to eliminate all the unnecessary parts of the string like this:
sed -r -e ":loop" -e 's|(http://.+)/.*|\1|' -e "t loop"
-r: Use extended regex (for + and unescaped parenthesis)
":loop": Define a new label named "loop"
-e: add commands to sed
"t loop": Jump back to label "loop" if there was a successful substitution
The only problem here is it will also cut the last separator character ('/'), but if you really need it you can still simply put it back after the "loop" finished, just append this additional command at the end of the previous command line:
-e "s,$,/,"
sed -E interprets regular expressions as extended (modern) regular expressions
Update: -E on MacOS X, -r in GNU sed.
Because you specifically stated you're trying to use sed (instead of perl, cut, etc.), try grouping. This circumvents the non-greedy identifier potentially not being recognized. The first group is the protocol (i.e. 'http://', 'https://', 'tcp://', etc). The second group is the domain:
echo "http://www.suon.co.uk/product/1/7/3/" | sed "s|^\(.*//\)\([^/]*\).*$|\1\2|"
If you're not familiar with grouping, start here.
I realize this is an old entry, but someone may find it useful.
As the full domain name may not exceed a total length of 253 characters replace .* with .\{1, 255\}
This is how to robustly do non-greedy matching of multi-character strings using sed. Lets say you want to change every foo...bar to <foo...bar> so for example this input:
$ cat file
ABC foo DEF bar GHI foo KLM bar NOP foo QRS bar TUV
should become this output:
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
To do that you convert foo and bar to individual characters and then use the negation of those characters between them:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/g; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
In the above:
s/#/#A/g; s/{/#B/g; s/}/#C/g is converting { and } to placeholder strings that cannot exist in the input so those chars then are available to convert foo and bar to.
s/foo/{/g; s/bar/}/g is converting foo and bar to { and } respectively
s/{[^{}]*}/<&>/g is performing the op we want - converting foo...bar to <foo...bar>
s/}/bar/g; s/{/foo/g is converting { and } back to foo and bar.
s/#C/}/g; s/#B/{/g; s/#A/#/g is converting the placeholder strings back to their original characters.
Note that the above does not rely on any particular string not being present in the input as it manufactures such strings in the first step, nor does it care which occurrence of any particular regexp you want to match since you can use {[^{}]*} as many times as necessary in the expression to isolate the actual match you want and/or with seds numeric match operator, e.g. to only replace the 2nd occurrence:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/2; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC foo DEF bar GHI <foo KLM bar> NOP foo QRS bar TUV
Have not yet seen this answer, so here's how you can do this with vi or vim:
vi -c '%s/\(http:\/\/.\{-}\/\).*/\1/ge | wq' file &>/dev/null
This runs the vi :%s substitution globally (the trailing g), refrains from raising an error if the pattern is not found (e), then saves the resulting changes to disk and quits. The &>/dev/null prevents the GUI from briefly flashing on screen, which can be annoying.
I like using vi sometimes for super complicated regexes, because (1) perl is dead dying, (2) vim has a very advanced regex engine, and (3) I'm already intimately familiar with vi regexes in my day-to-day usage editing documents.
Since PCRE is also tagged here, we could use GNU grep by using non-lazy match in regex .*? which will match first nearest match opposite of .*(which is really greedy and goes till last occurrence of match).
grep -oP '^http[s]?:\/\/.*?/' Input_file
Explanation: using grep's oP options here where -P is responsible for enabling PCRE regex here. In main program of grep mentioning regex which is matching starting http/https followed by :// till next occurrence of / since we have used .*? it will look for first / after (http/https://). It will print matched part only in line.
echo "/home/one/two/three/myfile.txt" | sed 's|\(.*\)/.*|\1|'
don bother, i got it on another forum :)
sed 's|\(http:\/\/www\.[a-z.0-9]*\/\).*|\1| works too
Here is something you can do with a two step approach and awk:
A=http://www.suepearson.co.uk/product/174/71/3816/
echo $A|awk '
{
var=gensub(///,"||",3,$0) ;
sub(/\|\|.*/,"",var);
print var
}'
Output:
http://www.suepearson.co.uk
Hope that helps!
Another sed version:
sed 's|/[:alnum:].*||' file.txt
It matches / followed by an alphanumeric character (so not another forward slash) as well as the rest of characters till the end of the line. Afterwards it replaces it with nothing (ie. deletes it.)
#Daniel H (concerning your comment on andcoz' answer, although long time ago): deleting trailing zeros works with
s,([[:digit:]]\.[[:digit:]]*[1-9])[0]*$,\1,g
it's about clearly defining the matching conditions ...
You should also think about the case where there is no matching delims. Do you want to output the line or not. My examples here do not output anything if there is no match.
You need prefix up to 3rd /, so select two times string of any length not containing / and following / and then string of any length not containing / and then match / following any string and then print selection. This idea works with any single char delims.
echo http://www.suepearson.co.uk/product/174/71/3816/ | \
sed -nr 's,(([^/]*/){2}[^/]*)/.*,\1,p'
Using sed commands you can do fast prefix dropping or delim selection, like:
echo 'aaa #cee: { "foo":" #cee: " }' | \
sed -r 't x;s/ #cee: /\n/;D;:x'
This is lot faster than eating char at a time.
Jump to label if successful match previously. Add \n at / before 1st delim. Remove up to first \n. If \n was added, jump to end and print.
If there is start and end delims, it is just easy to remove end delims until you reach the nth-2 element you want and then do D trick, remove after end delim, jump to delete if no match, remove before start delim and and print. This only works if start/end delims occur in pairs.
echo 'foobar start block #1 end barfoo start block #2 end bazfoo start block #3 end goo start block #4 end faa' | \
sed -r 't x;s/end//;s/end/\n/;D;:x;s/(end).*/\1/;T y;s/.*(start)/\1/;p;:y;d'
If you have access to gnu grep, then can utilize perl regex:
grep -Po '^https?://([^/]+)(?=)' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
http://www.suepearson.co.uk
Alternatively, to get everything after the domain use
grep -Po '^https?://([^/]+)\K.*' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
/product/174/71/3816/
The following solution works for matching / working with multiply present (chained; tandem; compound) HTML or other tags. For example, I wanted to edit HTML code to remove <span> tags, that appeared in tandem.
Issue: regular sed regex expressions greedily matched over all the tags from the first to the last.
Solution: non-greedy pattern matching (per discussions elsewhere in this thread; e.g. https://stackoverflow.com/a/46719361/1904943).
Example:
echo '<span>Will</span>This <span>remove</span>will <span>this.</span>remain.' | \
sed 's/<span>[^>]*>//g' ; echo
This will remain.
Explanation:
s/<span> : find <span>
[^>] : followed by anything that is not >
*> : until you find >
//g : replace any such strings present with nothing.
Addendum
I was trying to clean up URLs, but I was running into difficulty matching / excluding a word - href - using the approach above. I briefly looked at negative lookarounds (Regular expression to match a line that doesn't contain a word) but that approach seemed overly complex and did not provide a satisfactory solution.
I decided to replace href with ` (backtick), do the regex substitutions, then replace ` with href.
Example (formatted here for readability):
printf '\n
<a aaa h href="apple">apple</a>
<a bbb "c=ccc" href="banana">banana</a>
<a class="gtm-content-click"
data-vars-link-text="nope"
data-vars-click-url="https://blablabla"
data-vars-event-category="story"
data-vars-sub-category="story"
data-vars-item="in_content_link"
data-vars-link-text
href="https:example.com">Example.com</a>\n\n' |
sed 's/href/`/g ;
s/<a[^`]*`/\n<a href/g'
apple
banana
Example.com
Explanation: basically as above. Here,
s/href/` : replace href with ` (backtick)
s/<a : find start of URL
[^`] : followed by anything that is not ` (backtick)
*` : until you find a `
/<a href/g : replace each of those found with <a href
Unfortunately, as mentioned, this it is not supported in sed.
To overcome this, I suggest to use the next best thing(actually better even), to use vim sed-like capabilities.
define in .bash-profile
vimdo() { vim $2 --not-a-term -c "$1" -es +"w >> /dev/stdout" -cq! ; }
That will create headless vim to execute a command.
Now you can do for example:
echo $PATH | vimdo "%s_\c:[a-zA-Z0-9\\/]\{-}python[a-zA-Z0-9\\/]\{-}:__g" -
to filter out python in $PATH.
Use - to have input from pipe in vimdo.
While most of the syntax is the same. Vim features more advanced features, and using \{-} is standard for non-greedy match. see help regexp.

Getting Exact Pattern Match with grep and sed

I'm solving a bunch of text strings using grep and sed in which I only want the stdout to print the data after package: and ends at the folder name without the ending /.
For example:
data/dataapp/com.android.chrome-DeX_54==
System/app/Keychain
vendor/app/NlpService
This is the sample...
package:data/app/com.android.chrome-DeX_54==/base.apk=com.android.chrome
package:data/dataapp/ExactCalculator/ExactCalculator.apk=com.android.calculator2
package:data/hw_init/cust/app/Email/Email.apk=com.android.email
package:system/app/KeyChain/KeyChain.apk=com.android.keychain
package:system/delapp/WallpaperBackup/WallpaperBackup.apk=com.android.wallpaperbackup
package:system/framework/framework-res.apk=android
package:system/priv-app/CIT/CIT.apk=com.ontim.cit
package:vendor/app/NlpService/NlpService.apk=com.mediatek.nlpservice
I'm not getting the exact output I want so any help would be appreciated.
P.S: I'm learning grep and sed just for fun.
Would you please try:
grep -Po '(?<=package:).+(?=/[^/]*$)' input.txt
Results:
data/app/com.android.chrome-DeX_54==
data/dataapp/ExactCalculator
data/hw_init/cust/app/Email
system/app/KeyChain
system/delapp/WallpaperBackup
system/framework
system/priv-app/CIT
vendor/app/NlpService
The -P option enables a Perl compatible regex.
The -o option tells grep to print only the matched substring(s).
The pattern (?<=package:) is a positive lookbehind assertion and
the matched substring is not included in the output of grep -o.
The pattern (?=/[^/]*$) is a positive lookahead assertion as well.
The sed alternative will be:
sed 's#\(^package:\)\(.\+\)\(/[^/]*$\)#\2#' input.txt
or
sed -E 's#(^package:)(.+)(/[^/]*$)#\2#' input.txt
The latter will be more legible.
You'll see the positive lookarounds can be substituted with the back reference of sed just by discarding the unnecessary groups.
Hope this helps.
This might work for you (GNU sed):
sed -n 's#^package:\(.*\)/.*#\1#p' file
As this may be a filtering operation use the -n option to explicitly print results. The regexp starts with ^ in the substitution command which anchors package: to the start of the line and the uses .* to greedily consume the remainder of the line. However, the next character it tries to match is a / and so the regexp engine backtracks to find it and then the following .* again swallows the remainder of the line. The quoted parens \(...\) capture this part of the regexp and it is represented in the RHS of the substitute command by \1 known as a back reference. The p flag at the end of the substitute command explicitly prints the amended line in its current state.
N.B. That with the substitute command, the programmer can choose its delimiter. In documentation the command will usually be written s/LHS/RHS/flags where the delimiter is / but can be any character as in the above solution # was chosen to reduce the need for quoting the / character, LHS = regexp on the left hand side, RHS = replacement and flags = additional operations such as g meaning substitute globally throughout the line/file and p meaning print the line in its current state following a successful substitution (there are others see sed documentation.

Sed: How to replace multiple times per line with wildcard? [duplicate]

I'm trying to use sed to clean up lines of URLs to extract just the domain.
So from:
http://www.suepearson.co.uk/product/174/71/3816/
I want:
http://www.suepearson.co.uk/
(either with or without the trailing slash, it doesn't matter)
I have tried:
sed 's|\(http:\/\/.*?\/\).*|\1|'
and (escaping the non-greedy quantifier)
sed 's|\(http:\/\/.*\?\/\).*|\1|'
but I can not seem to get the non-greedy quantifier (?) to work, so it always ends up matching the whole string.
Neither basic nor extended Posix/GNU regex recognizes the non-greedy quantifier; you need a later regex. Fortunately, Perl regex for this context is pretty easy to get:
perl -pe 's|(http://.*?/).*|\1|'
In this specific case, you can get the job done without using a non-greedy regex.
Try this non-greedy regex [^/]* instead of .*?:
sed 's|\(http://[^/]*/\).*|\1|g'
With sed, I usually implement non-greedy search by searching for anything except the separator until the separator :
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*\)/.*;\1;p'
Output:
http://www.suon.co.uk
this is:
don't output -n
search, match pattern, replace and print s/<pattern>/<replace>/p
use ; search command separator instead of / to make it easier to type so s;<pattern>;<replace>;p
remember match between brackets \( ... \), later accessible with \1,\2...
match http://
followed by anything in brackets [], [ab/] would mean either a or b or /
first ^ in [] means not, so followed by anything but the thing in the []
so [^/] means anything except / character
* is to repeat previous group so [^/]* means characters except /.
so far sed -n 's;\(http://[^/]*\) means search and remember http://followed by any characters except / and remember what you've found
we want to search untill the end of domain so stop on the next / so add another / at the end: sed -n 's;\(http://[^/]*\)/' but we want to match the rest of the line after the domain so add .*
now the match remembered in group 1 (\1) is the domain so replace matched line with stuff saved in group \1 and print: sed -n 's;\(http://[^/]*\)/.*;\1;p'
If you want to include backslash after the domain as well, then add one more backslash in the group to remember:
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*/\).*;\1;p'
output:
http://www.suon.co.uk/
Simulating lazy (un-greedy) quantifier in sed
And all other regex flavors!
Finding first occurrence of an expression:
POSIX ERE (using -r option)
Regex:
(EXPRESSION).*|.
Sed:
sed -r ‍'s/(EXPRESSION).*|./\1/g' # Global `g` modifier should be on
Example (finding first sequence of digits) Live demo:
$ sed -r 's/([0-9]+).*|./\1/g' <<< 'foo 12 bar 34'
12
How does it work?
This regex benefits from an alternation |. At each position engine tries to pick the longest match (this is a POSIX standard which is followed by couple of other engines as well) which means it goes with . until a match is found for ([0-9]+).*. But order is important too.
Since global flag is set, engine tries to continue matching character by character up to the end of input string or our target. As soon as the first and only capturing group of left side of alternation is matched (EXPRESSION) rest of line is consumed immediately as well .*. We now hold our value in the first capturing group.
POSIX BRE
Regex:
\(\(\(EXPRESSION\).*\)*.\)*
Sed:
sed 's/\(\(\(EXPRESSION\).*\)*.\)*/\3/'
Example (finding first sequence of digits):
$ sed 's/\(\(\([0-9]\{1,\}\).*\)*.\)*/\3/' <<< 'foo 12 bar 34'
12
This one is like ERE version but with no alternation involved. That's all. At each single position engine tries to match a digit.
If it is found, other following digits are consumed and captured and the rest of line is matched immediately otherwise since * means
more or zero it skips over second capturing group \(\([0-9]\{1,\}\).*\)* and arrives at a dot . to match a single character and this process continues.
Finding first occurrence of a delimited expression:
This approach will match the very first occurrence of a string that is delimited. We can call it a block of string.
sed 's/\(END-DELIMITER-EXPRESSION\).*/\1/; \
s/\(\(START-DELIMITER-EXPRESSION.*\)*.\)*/\1/g'
Input string:
foobar start block #1 end barfoo start block #2 end
-EDE: end
-SDE: start
$ sed 's/\(end\).*/\1/; s/\(\(start.*\)*.\)*/\1/g'
Output:
start block #1 end
First regex \(end\).* matches and captures first end delimiter end and substitues all match with recent captured characters which
is the end delimiter. At this stage our output is: foobar start block #1 end.
Then the result is passed to second regex \(\(start.*\)*.\)* that is same as POSIX BRE version above. It matches a single character
if start delimiter start is not matched otherwise it matches and captures the start delimiter and matches the rest of characters.
Directly answering your question
Using approach #2 (delimited expression) you should select two appropriate expressions:
EDE: [^:/]\/
SDE: http:
Usage:
$ sed 's/\([^:/]\/\).*/\1/g; s/\(\(http:.*\)*.\)*/\1/' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
Output:
http://www.suepearson.co.uk/
Note: this will not work with identical delimiters.
sed does not support "non greedy" operator.
You have to use "[]" operator to exclude "/" from match.
sed 's,\(http://[^/]*\)/.*,\1,'
P.S. there is no need to backslash "/".
sed - non greedy matching by Christoph Sieghart
The trick to get non greedy matching in sed is to match all characters excluding the one that terminates the match. I know, a no-brainer, but I wasted precious minutes on it and shell scripts should be, after all, quick and easy. So in case somebody else might need it:
Greedy matching
% echo "<b>foo</b>bar" | sed 's/<.*>//g'
bar
Non greedy matching
% echo "<b>foo</b>bar" | sed 's/<[^>]*>//g'
foobar
Non-greedy solution for more than a single character
This thread is really old but I assume people still needs it.
Lets say you want to kill everything till the very first occurrence of HELLO. You cannot say [^HELLO]...
So a nice solution involves two steps, assuming that you can spare a unique word that you are not expecting in the input, say top_sekrit.
In this case we can:
s/HELLO/top_sekrit/ #will only replace the very first occurrence
s/.*top_sekrit// #kill everything till end of the first HELLO
Of course, with a simpler input you could use a smaller word, or maybe even a single character.
HTH!
This can be done using cut:
echo "http://www.suepearson.co.uk/product/174/71/3816/" | cut -d'/' -f1-3
another way, not using regex, is to use fields/delimiter method eg
string="http://www.suepearson.co.uk/product/174/71/3816/"
echo $string | awk -F"/" '{print $1,$2,$3}' OFS="/"
sed certainly has its place but this not not one of them !
As Dee has pointed out: Just use cut. It is far simpler and much more safe in this case. Here's an example where we extract various components from the URL using Bash syntax:
url="http://www.suepearson.co.uk/product/174/71/3816/"
protocol=$(echo "$url" | cut -d':' -f1)
host=$(echo "$url" | cut -d'/' -f3)
urlhost=$(echo "$url" | cut -d'/' -f1-3)
urlpath=$(echo "$url" | cut -d'/' -f4-)
gives you:
protocol = "http"
host = "www.suepearson.co.uk"
urlhost = "http://www.suepearson.co.uk"
urlpath = "product/174/71/3816/"
As you can see this is a lot more flexible approach.
(all credit to Dee)
sed 's|(http:\/\/[^\/]+\/).*|\1|'
There is still hope to solve this using pure (GNU) sed. Despite this is not a generic solution in some cases you can use "loops" to eliminate all the unnecessary parts of the string like this:
sed -r -e ":loop" -e 's|(http://.+)/.*|\1|' -e "t loop"
-r: Use extended regex (for + and unescaped parenthesis)
":loop": Define a new label named "loop"
-e: add commands to sed
"t loop": Jump back to label "loop" if there was a successful substitution
The only problem here is it will also cut the last separator character ('/'), but if you really need it you can still simply put it back after the "loop" finished, just append this additional command at the end of the previous command line:
-e "s,$,/,"
sed -E interprets regular expressions as extended (modern) regular expressions
Update: -E on MacOS X, -r in GNU sed.
Because you specifically stated you're trying to use sed (instead of perl, cut, etc.), try grouping. This circumvents the non-greedy identifier potentially not being recognized. The first group is the protocol (i.e. 'http://', 'https://', 'tcp://', etc). The second group is the domain:
echo "http://www.suon.co.uk/product/1/7/3/" | sed "s|^\(.*//\)\([^/]*\).*$|\1\2|"
If you're not familiar with grouping, start here.
I realize this is an old entry, but someone may find it useful.
As the full domain name may not exceed a total length of 253 characters replace .* with .\{1, 255\}
This is how to robustly do non-greedy matching of multi-character strings using sed. Lets say you want to change every foo...bar to <foo...bar> so for example this input:
$ cat file
ABC foo DEF bar GHI foo KLM bar NOP foo QRS bar TUV
should become this output:
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
To do that you convert foo and bar to individual characters and then use the negation of those characters between them:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/g; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
In the above:
s/#/#A/g; s/{/#B/g; s/}/#C/g is converting { and } to placeholder strings that cannot exist in the input so those chars then are available to convert foo and bar to.
s/foo/{/g; s/bar/}/g is converting foo and bar to { and } respectively
s/{[^{}]*}/<&>/g is performing the op we want - converting foo...bar to <foo...bar>
s/}/bar/g; s/{/foo/g is converting { and } back to foo and bar.
s/#C/}/g; s/#B/{/g; s/#A/#/g is converting the placeholder strings back to their original characters.
Note that the above does not rely on any particular string not being present in the input as it manufactures such strings in the first step, nor does it care which occurrence of any particular regexp you want to match since you can use {[^{}]*} as many times as necessary in the expression to isolate the actual match you want and/or with seds numeric match operator, e.g. to only replace the 2nd occurrence:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/2; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC foo DEF bar GHI <foo KLM bar> NOP foo QRS bar TUV
Have not yet seen this answer, so here's how you can do this with vi or vim:
vi -c '%s/\(http:\/\/.\{-}\/\).*/\1/ge | wq' file &>/dev/null
This runs the vi :%s substitution globally (the trailing g), refrains from raising an error if the pattern is not found (e), then saves the resulting changes to disk and quits. The &>/dev/null prevents the GUI from briefly flashing on screen, which can be annoying.
I like using vi sometimes for super complicated regexes, because (1) perl is dead dying, (2) vim has a very advanced regex engine, and (3) I'm already intimately familiar with vi regexes in my day-to-day usage editing documents.
Since PCRE is also tagged here, we could use GNU grep by using non-lazy match in regex .*? which will match first nearest match opposite of .*(which is really greedy and goes till last occurrence of match).
grep -oP '^http[s]?:\/\/.*?/' Input_file
Explanation: using grep's oP options here where -P is responsible for enabling PCRE regex here. In main program of grep mentioning regex which is matching starting http/https followed by :// till next occurrence of / since we have used .*? it will look for first / after (http/https://). It will print matched part only in line.
echo "/home/one/two/three/myfile.txt" | sed 's|\(.*\)/.*|\1|'
don bother, i got it on another forum :)
sed 's|\(http:\/\/www\.[a-z.0-9]*\/\).*|\1| works too
Here is something you can do with a two step approach and awk:
A=http://www.suepearson.co.uk/product/174/71/3816/
echo $A|awk '
{
var=gensub(///,"||",3,$0) ;
sub(/\|\|.*/,"",var);
print var
}'
Output:
http://www.suepearson.co.uk
Hope that helps!
Another sed version:
sed 's|/[:alnum:].*||' file.txt
It matches / followed by an alphanumeric character (so not another forward slash) as well as the rest of characters till the end of the line. Afterwards it replaces it with nothing (ie. deletes it.)
#Daniel H (concerning your comment on andcoz' answer, although long time ago): deleting trailing zeros works with
s,([[:digit:]]\.[[:digit:]]*[1-9])[0]*$,\1,g
it's about clearly defining the matching conditions ...
You should also think about the case where there is no matching delims. Do you want to output the line or not. My examples here do not output anything if there is no match.
You need prefix up to 3rd /, so select two times string of any length not containing / and following / and then string of any length not containing / and then match / following any string and then print selection. This idea works with any single char delims.
echo http://www.suepearson.co.uk/product/174/71/3816/ | \
sed -nr 's,(([^/]*/){2}[^/]*)/.*,\1,p'
Using sed commands you can do fast prefix dropping or delim selection, like:
echo 'aaa #cee: { "foo":" #cee: " }' | \
sed -r 't x;s/ #cee: /\n/;D;:x'
This is lot faster than eating char at a time.
Jump to label if successful match previously. Add \n at / before 1st delim. Remove up to first \n. If \n was added, jump to end and print.
If there is start and end delims, it is just easy to remove end delims until you reach the nth-2 element you want and then do D trick, remove after end delim, jump to delete if no match, remove before start delim and and print. This only works if start/end delims occur in pairs.
echo 'foobar start block #1 end barfoo start block #2 end bazfoo start block #3 end goo start block #4 end faa' | \
sed -r 't x;s/end//;s/end/\n/;D;:x;s/(end).*/\1/;T y;s/.*(start)/\1/;p;:y;d'
If you have access to gnu grep, then can utilize perl regex:
grep -Po '^https?://([^/]+)(?=)' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
http://www.suepearson.co.uk
Alternatively, to get everything after the domain use
grep -Po '^https?://([^/]+)\K.*' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
/product/174/71/3816/
The following solution works for matching / working with multiply present (chained; tandem; compound) HTML or other tags. For example, I wanted to edit HTML code to remove <span> tags, that appeared in tandem.
Issue: regular sed regex expressions greedily matched over all the tags from the first to the last.
Solution: non-greedy pattern matching (per discussions elsewhere in this thread; e.g. https://stackoverflow.com/a/46719361/1904943).
Example:
echo '<span>Will</span>This <span>remove</span>will <span>this.</span>remain.' | \
sed 's/<span>[^>]*>//g' ; echo
This will remain.
Explanation:
s/<span> : find <span>
[^>] : followed by anything that is not >
*> : until you find >
//g : replace any such strings present with nothing.
Addendum
I was trying to clean up URLs, but I was running into difficulty matching / excluding a word - href - using the approach above. I briefly looked at negative lookarounds (Regular expression to match a line that doesn't contain a word) but that approach seemed overly complex and did not provide a satisfactory solution.
I decided to replace href with ` (backtick), do the regex substitutions, then replace ` with href.
Example (formatted here for readability):
printf '\n
<a aaa h href="apple">apple</a>
<a bbb "c=ccc" href="banana">banana</a>
<a class="gtm-content-click"
data-vars-link-text="nope"
data-vars-click-url="https://blablabla"
data-vars-event-category="story"
data-vars-sub-category="story"
data-vars-item="in_content_link"
data-vars-link-text
href="https:example.com">Example.com</a>\n\n' |
sed 's/href/`/g ;
s/<a[^`]*`/\n<a href/g'
apple
banana
Example.com
Explanation: basically as above. Here,
s/href/` : replace href with ` (backtick)
s/<a : find start of URL
[^`] : followed by anything that is not ` (backtick)
*` : until you find a `
/<a href/g : replace each of those found with <a href
Unfortunately, as mentioned, this it is not supported in sed.
To overcome this, I suggest to use the next best thing(actually better even), to use vim sed-like capabilities.
define in .bash-profile
vimdo() { vim $2 --not-a-term -c "$1" -es +"w >> /dev/stdout" -cq! ; }
That will create headless vim to execute a command.
Now you can do for example:
echo $PATH | vimdo "%s_\c:[a-zA-Z0-9\\/]\{-}python[a-zA-Z0-9\\/]\{-}:__g" -
to filter out python in $PATH.
Use - to have input from pipe in vimdo.
While most of the syntax is the same. Vim features more advanced features, and using \{-} is standard for non-greedy match. see help regexp.

Replace whole line containing a string using Sed

I have a text file which has a particular line something like
sometext sometext sometext TEXT_TO_BE_REPLACED sometext sometext sometext
I need to replace the whole line above with
This line is removed by the admin.
The search keyword is TEXT_TO_BE_REPLACED
I need to write a shell script for this. How can I achieve this using sed?
You can use the change command to replace the entire line, and the -i flag to make the changes in-place. For example, using GNU sed:
sed -i '/TEXT_TO_BE_REPLACED/c\This line is removed by the admin.' /tmp/foo
You need to use wildcards (.*) before and after to replace the whole line:
sed 's/.*TEXT_TO_BE_REPLACED.*/This line is removed by the admin./'
The Answer above:
sed -i '/TEXT_TO_BE_REPLACED/c\This line is removed by the admin.' /tmp/foo
Works fine if the replacement string/line is not a variable.
The issue is that on Redhat 5 the \ after the c escapes the $. A double \\ did not work either (at least on Redhat 5).
Through hit and trial, I discovered that the \ after the c is redundant if your replacement string/line is only a single line. So I did not use \ after the c, used a variable as a single replacement line and it was joy.
The code would look something like:
sed -i "/TEXT_TO_BE_REPLACED/c $REPLACEMENT_TEXT_STRING" /tmp/foo
Note the use of double quotes instead of single quotes.
The accepted answer did not work for me for several reasons:
my version of sed does not like -i with a zero length extension
the syntax of the c\ command is weird and I couldn't get it to work
I didn't realize some of my issues are coming from unescaped slashes
So here is the solution I came up with which I think should work for most cases:
function escape_slashes {
sed 's/\//\\\//g'
}
function change_line {
local OLD_LINE_PATTERN=$1; shift
local NEW_LINE=$1; shift
local FILE=$1
local NEW=$(echo "${NEW_LINE}" | escape_slashes)
# FIX: No space after the option i.
sed -i.bak '/'"${OLD_LINE_PATTERN}"'/s/.*/'"${NEW}"'/' "${FILE}"
mv "${FILE}.bak" /tmp/
}
So the sample usage to fix the problem posed:
change_line "TEXT_TO_BE_REPLACED" "This line is removed by the admin." yourFile
All of the answers provided so far assume that you know something about the text to be replaced which makes sense, since that's what the OP asked. I'm providing an answer that assumes you know nothing about the text to be replaced and that there may be a separate line in the file with the same or similar content that you do not want to be replaced. Furthermore, I'm assuming you know the line number of the line to be replaced.
The following examples demonstrate the removing or changing of text by specific line numbers:
# replace line 17 with some replacement text and make changes in file (-i switch)
# the "-i" switch indicates that we want to change the file. Leave it out if you'd
# just like to see the potential changes output to the terminal window.
# "17s" indicates that we're searching line 17
# ".*" indicates that we want to change the text of the entire line
# "REPLACEMENT-TEXT" is the new text to put on that line
# "PATH-TO-FILE" tells us what file to operate on
sed -i '17s/.*/REPLACEMENT-TEXT/' PATH-TO-FILE
# replace specific text on line 3
sed -i '3s/TEXT-TO-REPLACE/REPLACEMENT-TEXT/'
for manipulation of config files
i came up with this solution inspired by skensell answer
configLine [searchPattern] [replaceLine] [filePath]
it will:
create the file if not exists
replace the whole line (all lines) where searchPattern matched
add replaceLine on the end of the file if pattern was not found
Function:
function configLine {
local OLD_LINE_PATTERN=$1; shift
local NEW_LINE=$1; shift
local FILE=$1
local NEW=$(echo "${NEW_LINE}" | sed 's/\//\\\//g')
touch "${FILE}"
sed -i '/'"${OLD_LINE_PATTERN}"'/{s/.*/'"${NEW}"'/;h};${x;/./{x;q100};x}' "${FILE}"
if [[ $? -ne 100 ]] && [[ ${NEW_LINE} != '' ]]
then
echo "${NEW_LINE}" >> "${FILE}"
fi
}
the crazy exit status magic comes from https://stackoverflow.com/a/12145797/1262663
In my makefile I use this:
#sed -i '/.*Revision:.*/c\'"`svn info -R main.cpp | awk '/^Rev/'`"'' README.md
PS: DO NOT forget that the -i changes actually the text in the file... so if the pattern you defined as "Revision" will change, you will also change the pattern to replace.
Example output:
Abc-Project written by John Doe
Revision: 1190
So if you set the pattern "Revision: 1190" it's obviously not the same as you defined them as "Revision:" only...
bash-4.1$ new_db_host="DB_HOSTNAME=good replaced with 122.334.567.90"
bash-4.1$
bash-4.1$ sed -i "/DB_HOST/c $new_db_host" test4sed
vim test4sed
'
'
'
DB_HOSTNAME=good replaced with 122.334.567.90
'
it works fine
To do this without relying on any GNUisms such as -i without a parameter or c without a linebreak:
sed '/TEXT_TO_BE_REPLACED/c\
This line is removed by the admin.
' infile > tmpfile && mv tmpfile infile
In this (POSIX compliant) form of the command
c\
text
text can consist of one or multiple lines, and linebreaks that should become part of the replacement have to be escaped:
c\
line1\
line2
s/x/y/
where s/x/y/ is a new sed command after the pattern space has been replaced by the two lines
line1
line2
cat find_replace | while read pattern replacement ; do
sed -i "/${pattern}/c ${replacement}" file
done
find_replace file contains 2 columns, c1 with pattern to match, c2 with replacement, the sed loop replaces each line conatining one of the pattern of variable 1
To replace whole line containing a specified string with the content of that line
Text file:
Row: 0 last_time_contacted=0, display_name=Mozart, _id=100, phonebook_bucket_alt=2
Row: 1 last_time_contacted=0, display_name=Bach, _id=101, phonebook_bucket_alt=2
Single string:
$ sed 's/.* display_name=\([[:alpha:]]\+\).*/\1/'
output:
100
101
Multiple strings delimited by white-space:
$ sed 's/.* display_name=\([[:alpha:]]\+\).* _id=\([[:digit:]]\+\).*/\1 \2/'
output:
Mozart 100
Bach 101
Adjust regex to meet your needs
[:alpha] and [:digit:]
are Character Classes and Bracket Expressions
This worked for me:
sed -i <extension> 's/.*<Line to be replaced>.*/<New line to be added>/'
An example is:
sed -i .bak -e '7s/.*version.*/ version = "4.33.0"/'
-i: The extension for the backup file after the replacement. In this case, it is .bak.
-e: The sed script. In this case, it is '7s/.*version.*/ version = "4.33.0"/'. If you want to use a sed file use the -f flag
s: The line number in the file to be replaced. In this case, it is 7s which means line 7.
Note:
If you want to do a recursive find and replace with sed then you can grep to the beginning of the command:
grep -rl --exclude-dir=<directory-to-exclude> --include=\*<Files to include> "<Line to be replaced>" ./ | sed -i <extension> 's/.*<Line to be replaced>.*/<New line to be added>/'
The question asks for solutions using sed, but if that's not a hard requirement then there is another option which might be a wiser choice.
The accepted answer suggests sed -i and describes it as replacing the file in-place, but -i doesn't really do that and instead does the equivalent of sed pattern file > tmp; mv tmp file, preserving ownership and modes. This is not ideal in many circumstances. In general I do not recommend running sed -i non-interactively as part of an automatic process--it's like setting a bomb with a fuse of an unknown length. Sooner or later it will blow up on someone.
To actually edit a file "in place" and replace a line matching a pattern with some other content you would be well served to use an actual text editor. This is how it's done with ed, the standard text editor.
printf '%s\n' '/TEXT_TO_BE_REPLACED/' d i 'This line is removed by the admin' . w q | \
ed -s /tmp/foo > /dev/null
Note that this only replaces the first matching line, which is what the question implied was wanted. This is a material difference from most of the other answers.
That disadvantage aside, there are some advantages to using ed over sed:
You can replace the match with one or multiple lines without any extra effort.
The replacement text can be arbitrarily complex without needing any escaping to protect it.
Most importantly, the original file is opened, modified, and saved. A copy is not made.
How it works
How it works:
printf will use its first argument as a format string and print each of its other arguments using that format, effectively meaning that each argument to printf becomes a line of output, which is all sent to ed on stdin.
The first line is a regex pattern match which causes ed to move its notion of "the current line" forward to the first line that matches (if there is no match the current line is set to the last line of the file).
The next is the d command which instructs ed to delete the entire current line.
After that is the i command which puts ed into insert mode;
after that all subsequent lines entered are written to the current line (or additional lines if there are any embedded newlines). This means you can expand a variable (e.g. "$foo") containing multiple lines here and it will insert all of them.
Insert mode ends when ed sees a line consisting of .
The w command writes the content of the file to disk, and
the q command quits.
The ed command is given the -s switch, putting it into silent mode so it doesn't echo any information as it runs,
the file to be edited is given as an argument to ed,
and, finally, stdout is thrown away to prevent the line matching the regex from being printed.
Some Unix-like systems may (inappropriately) ship without an ed installed, but may still ship with an ex; if so you can simply use it instead. If have vim but no ex or ed you can use vim -e instead. If you have only standard vi but no ex or ed, complain to your sysadmin.
It is as similar to above one..
sed 's/[A-Za-z0-9]*TEXT_TO_BE_REPLACED.[A-Za-z0-9]*/This line is removed by the admin./'
Below command is working for me. Which is working with variables
sed -i "/\<$E\>/c $D" "$B"
I very often use regex to extract data from files I just used that to replace the literal quote \" with // nothing :-)
cat file.csv | egrep '^\"([0-9]{1,3}\.[0-9]{1,3}\.)' | sed s/\"//g | cut -d, -f1 > list.txt

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