I am teaching myself commands and different ways to use grep. I know how to search for a string in a directory and its sub directories, but I am confused when it comes to searching for a split in the string.
for example: how could I search for all words (string size varies) that starts with a and ends with e. so that I could find ape or apple in text files?
EDIT UPDATE:
I am not sure of the grep version I am using, but I tried using:
" grep -nr "a[A-Za-z]*e" "
this produces the answer by including outputs like ape and apple but it also includes apes which is NOT wanted.
Simply:
grep '\ba\w*e\b'
or
grep --color '\ba\w*e\b'
or
grep -rn '\ba\w*e\b'
Some explanations
As this question is tagged linux, this answer use GNU grep: grep (GNU grep) 2.27.
The result of command man grep | grep -3 '\\b':
The Backslash Character and Special Expressions
The symbols \< and \> respectively match the empty string at the
beginning and end of a word. The symbol \b matches the empty string at
the edge of a word, and \B matches the empty string provided it's not
at the edge of a word. The symbol \w is a synonym for [_[:alnum:]] and
\W is a synonym for [^_[:alnum:]].
Let you show
\b mean edge of a word
\w mean [_[:alnum:]]
a and e are letters
you may already know* wich mean The preceding item will be matched zero or more times. (elsewhere in same man page: man grep | grep '^ *\*' ;)
... and finally... This could by written:
grep '\<a\w*e\>'
where
The symbols \< and > respectively match the empty string at the beginning and end of a word.
This could have near same effect, but description strictly correspond to title of this: grep: finding a string that starts and ends with a specific letter in directory
I suppose you could use:
find . -type f -name '*.txt' -exec cat {} \; | grep 'a[A-Za-z]\+e'
That should cat any .txt files in the current directory, recursively, and grep for "a... any characters... e"
The [A-Za-z] searches for a character of either case, the \+ says "any number of them".
I think that's what you're after?
Edit:
Word boundaries:
find . -type f -name '*.txt' -exec cat {} \+ | grep '\ba[A-Za-z]\+e\b'
As alluded to in various comments, it is possible to do this using POSIX standard grep -E, but it is not all that notationally convenient.
I used a script file grep-ape.sh containing:
grep -E -e '(^|[^[:alpha:]])a[[:alpha:]]+e($|[^[:alpha:]])' "$#"
The -E enables extended regular expressions. The -e is optional, but allows me to add extra options as 'file names' after the regular expression. The regular expression looks for either 'start of line' or a non-alpha character, followed by a, one or more additional alpha characters, an e and either 'end of line' or a non-alpha character.
Given the data file (called, unimaginatively, data):
I want to tape the apes that ate the grapes.
ape at the start.
Ending with ape
Situating ape in the middle
And an apple too.
But not apples, no way.
The tape ran out.
The apes ran out.
The grapes ran out.
They ate them.
I could run grep-ape.sh -n data (demonstrating the usefulness of the -e option, though GNU systems will permute options so you don't necessarily spot the problem), and got:
1:I want to tape the apes that ate the grapes.
2:ape at the start.
3:Ending with ape
4:Situating ape in the middle
5:And an apple too.
10:They ate them.
Using a non-POSIX option -o (supported by GNU and BSD versions of grep) to print only what is matched, I can get the output:
$ grep-ape.sh -n -o data
1: ate
2:ape
3: ape
4: ape
5: apple
10: ate
$
This shows that the regular expression is picking up the acceptable words, even on lines where there are words that would not be acceptable when not in the company of words that are acceptable.
Related
Is there a way to make grep output "words" from files that match the search expression?
If I want to find all the instances of, say, "th" in a number of files, I can do:
grep "th" *
but the output will be something like (bold is by me);
some-text-file : the cat sat on the mat
some-other-text-file : the quick brown fox
yet-another-text-file : i hope this explains it thoroughly
What I want it to output, using the same search, is:
the
the
the
this
thoroughly
Is this possible using grep? Or using another combination of tools?
Try grep -o:
grep -oh "\w*th\w*" *
Edit: matching from Phil's comment.
From the docs:
-h, --no-filename
Suppress the prefixing of file names on output. This is the default
when there is only one file (or only standard input) to search.
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
Cross distribution safe answer (including windows minGW?)
grep -h "[[:alpha:]]*th[[:alpha:]]*" 'filename' | tr ' ' '\n' | grep -h "[[:alpha:]]*th[[:alpha:]]*"
If you're using older versions of grep (like 2.4.2) which do not include the -o option, then use the above. Else use the simpler to maintain version below.
Linux cross distribution safe answer
grep -oh "[[:alpha:]]*th[[:alpha:]]*" 'filename'
To summarize: -oh outputs the regular expression matches to the file content (and not its filename), just like how you would expect a regular expression to work in vim/etc... What word or regular expression you would be searching for then, is up to you! As long as you remain with POSIX and not perl syntax (refer below)
More from the manual for grep
-o Print each match, but only the match, not the entire line.
-h Never print filename headers (i.e. filenames) with output lines.
-w The expression is searched for as a word (as if surrounded by
`[[:<:]]' and `[[:>:]]';
The reason why the original answer does not work for everyone
The usage of \w varies from platform to platform, as it's an extended "perl" syntax. As such, those grep installations that are limited to work with POSIX character classes use [[:alpha:]] and not its perl equivalent of \w. See the Wikipedia page on regular expression for more
Ultimately, the POSIX answer above will be a lot more reliable regardless of platform (being the original) for grep
As for support of grep without -o option, the first grep outputs the relevant lines, the tr splits the spaces to new lines, the final grep filters only for the respective lines.
(PS: I know most platforms by now would have been patched for \w.... but there are always those that lag behind)
Credit for the "-o" workaround from #AdamRosenfield answer
It's more simple than you think. Try this:
egrep -wo 'th.[a-z]*' filename.txt #### (Case Sensitive)
egrep -iwo 'th.[a-z]*' filename.txt ### (Case Insensitive)
Where,
egrep: Grep will work with extended regular expression.
w : Matches only word/words instead of substring.
o : Display only matched pattern instead of whole line.
i : If u want to ignore case sensitivity.
You could translate spaces to newlines and then grep, e.g.:
cat * | tr ' ' '\n' | grep th
Just awk, no need combination of tools.
# awk '{for(i=1;i<=NF;i++){if($i~/^th/){print $i}}}' file
the
the
the
this
thoroughly
grep command for only matching and perl
grep -o -P 'th.*? ' filename
I was unsatisfied with awk's hard to remember syntax but I liked the idea of using one utility to do this.
It seems like ack (or ack-grep if you use Ubuntu) can do this easily:
# ack-grep -ho "\bth.*?\b" *
the
the
the
this
thoroughly
If you omit the -h flag you get:
# ack-grep -o "\bth.*?\b" *
some-other-text-file
1:the
some-text-file
1:the
the
yet-another-text-file
1:this
thoroughly
As a bonus, you can use the --output flag to do this for more complex searches with just about the easiest syntax I've found:
# echo "bug: 1, id: 5, time: 12/27/2010" > test-file
# ack-grep -ho "bug: (\d*), id: (\d*), time: (.*)" --output '$1, $2, $3' test-file
1, 5, 12/27/2010
cat *-text-file | grep -Eio "th[a-z]+"
You can also try pcregrep. There is also a -w option in grep, but in some cases it doesn't work as expected.
From Wikipedia:
cat fruitlist.txt
apple
apples
pineapple
apple-
apple-fruit
fruit-apple
grep -w apple fruitlist.txt
apple
apple-
apple-fruit
fruit-apple
I had a similar problem, looking for grep/pattern regex and the "matched pattern found" as output.
At the end I used egrep (same regex on grep -e or -G didn't give me the same result of egrep) with the option -o
so, I think that could be something similar to (I'm NOT a regex Master) :
egrep -o "the*|this{1}|thoroughly{1}" filename
To search all the words with start with "icon-" the following command works perfect. I am using Ack here which is similar to grep but with better options and nice formatting.
ack -oh --type=html "\w*icon-\w*" | sort | uniq
You could pipe your grep output into Perl like this:
grep "th" * | perl -n -e'while(/(\w*th\w*)/g) {print "$1\n"}'
grep --color -o -E "Begin.{0,}?End" file.txt
? - Match as few as possible until the End
Tested on macos terminal
$ grep -w
Excerpt from grep man page:
-w: Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character.
ripgrep
Here are the example using ripgrep:
rg -o "(\w+)?th(\w+)?"
It'll match all words matching th.
I want to find the number of 8 letter words that do not contain the letter "e" in a number of text files (*.txt). In the process I ran into two issues: my lack of understanding in quantifiers and how to exclude characters.
I'm quite new to the Unix terminal, but this is what I have tried:
cat *.txt | grep -Eo "\w+" | grep -i ".*[^e].*"
I need to include the cat command because it otherwise includes the names of the text files in the pipe. The second pipe is to have all the words in a list, and it works, but the last pipe was meant to find all the words that do not have the letter "e" in them, but doesn't seem to work. (I thought "." for no or any number of any character, followed by a character that is not an "e", and followed by another "." for no or any number of any character.)
cat *.txt | grep -Eo "\w+" | grep -wi "[a-z][a-z][a-z][a-z][a-z][a-z][a-z][a-z]"
This command works to find the words that contain 8 characters, but it is quite ineffective, because I have to repeat "[a-z]" 8 times. I thought it could also be "[a-z]{8}", but that doesn't seem to work.
cat *.txt | grep -Eo "\w+" | grep -wi "[a-z][a-z][a-z][a-z][a-z][a-z][a-z][a-z]" | grep -i ".*[^e].*"
So finally, this would be my best guess, however, the third pipe is ineffective and the last pipe doesn't work.
You may use this grep:
grep -hEiwo '[a-df-z]{8}' *.txt
Here:
[a-df-z]{8}: Matches all letters except e
-h: Don't print filename in output
-i: Ignore case search
-o: Print matches only
-w: Match complete words
In case you are ok with GNU awk and assuming that you want to print only the exact words and could be multiple matches in a line if this is the case one could try following.
awk -v IGNORECASE="1" '{for(i=1;i<=NF;i++){if($i~/^[a-df-z]{8}$/){print $i}}}' *.txt
OR without the use of IGNORCASE one could try:
awk '{for(i=1;i<=NF;i++){if(tolower($i)~/^[a-df-z]{8}$/){print $i}}}' *.txt
NOTE: Considering that you want exact matches of 8 letters only in lines. 8 letter words followed by a punctuation mark will be excluded.
Here is a crazy thought with GNU awk:
awk 'BEGIN{FPAT="\\<\\w{8}\\>"}{c+=NF}END{print c}' file
Or if you want to make it work only on a select set of characters:
awk 'BEGIN{FPAT="\\<[a-df-z]{8}\\>"}{c+=NF}END{print c}' file
What this does is, it defines the fields, to be a set of 8 characters (\w as a word-constituent or [a-df-z] as a selected set) which is enclosed by word-boundaries (\< and \>). This is done with FPAT (note the Gory details about escaping).
Sometimes you might also have words which contain diatrics, so you have to expand. Then this might be the best solution:
awk 'BEGIN{FPAT="\\<\\w{8}\\>"}{for(i=1;i<=NF;++i) if($i !~ /e/) c++}END{print c}' file
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.
Using grep
How many 12 letter - alphabet only lines are in testing.txt?
excerpt of testing.txt
tyler1
Tanktop_Paedo
xyz2#geocities.com
milt#uole.com
justincrump
cranges10
namer#uole.com
soulfunkbrotha
timetolearnz
hotbooby#geocities.com
Fire_Crazy
helloworldad
dingbat#geocities.com
from this excerpt, I want to get a result of 2. (helloworldad, and timetolearnz)
I want to check every line and grep only those that have 12 characters in each line. I can't think of a way to do this with grep though.
For the alphabet only, I think I can use
grep [A-Za-z] testing.txt
However, how do I make it so only the characters [A-Za-z] show up in those 12 characters?
You can do it with extended regex -E and by specifying that the match is exactly {12} characters from start ^ to finish $
$ grep -E "^[A-Za-z]{12}$" testing.txt
timetolearnz
helloworldad
Or if you want to get the count -c of the lines you can use
$ grep -cE "^[A-Za-z]{12}$" testing.txt
2
grep supports whole-line match and counting, e.g.:
grep -xc '[[:alpha:]]\{12\}' testing.txt
Output:
2
The [:alpha:] character class is another way of saying [A-Za-z]. See section 3.2 of the the info pages: info grep 'Regular Expressions' 'Character Classes and Bracket Expressions' for more on this subject. Or look it up in the pdf manual online.
This question already has answers here:
Colorized grep -- viewing the entire file with highlighted matches
(24 answers)
Closed 7 years ago.
When using grep, it will highlight any text in a line with a match to your regular expression.
What if I want this behaviour, but have grep print out all lines as well? I came up empty after a quick look through the grep man page.
Use ack. Checkout its --passthru option here: ack. It has the added benefit of allowing full perl regular expressions.
$ ack --passthru 'pattern1' file_name
$ command_here | ack --passthru 'pattern1'
You can also do it using grep like this:
$ grep --color -E '^|pattern1|pattern2' file_name
$ command_here | grep --color -E '^|pattern1|pattern2'
This will match all lines and highlight the patterns. The ^ matches every start of line, but won't get printed/highlighted since it's not a character.
(Note that most of the setups will use --color by default. You may not need that flag).
You can make sure that all lines match but there is nothing to highlight on irrelevant matches
egrep --color 'apple|' test.txt
Notes:
egrep may be spelled also grep -E
--color is usually default in most distributions
some variants of grep will "optimize" the empty match, so you might want to use "apple|$" instead (see: https://stackoverflow.com/a/13979036/939457)
EDIT:
This works with OS X Mountain Lion's grep:
grep --color -E 'pattern1|pattern2|$'
This is better than '^|pattern1|pattern2' because the ^ part of the alternation matches at the beginning of the line whereas the $ matches at the end of the line. Some regular expression engines won't highlight pattern1 or pattern2 because ^ already matched and the engine is eager.
Something similar happens for 'pattern1|pattern2|' because the regex engine notices the empty alternation at the end of the pattern string matches the beginning of the subject string.
[1]: http://www.regular-expressions.info/engine.html
FIRST EDIT:
I ended up using perl:
perl -pe 's:pattern:\033[31;1m$&\033[30;0m:g'
This assumes you have an ANSI-compatible terminal.
ORIGINAL ANSWER:
If you're stuck with a strange grep, this might work:
grep -E --color=always -A500 -B500 'pattern1|pattern2' | grep -v '^--'
Adjust the numbers to get all the lines you want.
The second grep just removes extraneous -- lines inserted by the BSD-style grep on Mac OS X Mountain Lion, even when the context of consecutive matches overlap.
I thought GNU grep omitted the -- lines when context overlaps, but it's been awhile so maybe I remember wrong.
You can use my highlight script from https://github.com/kepkin/dev-shell-essentials
It's better than grep cause you can highlight each match with it's own color.
$ command_here | highlight green "input" | highlight red "output"
Since you want matches highlighted, this is probably for human consumption (as opposed to piping to another program for instance), so a nice solution would be to use:
less -p <your-pattern> <your-file>
And if you don't care about case sensitivity:
less -i -p <your-pattern> <your-file>
This also has the advantage of having pages, which is nice when having to go through a long output
You can do it using only grep by:
reading the file line by line
matching a pattern in each line and highlighting pattern by grep
if there is no match, echo the line as is
which gives you the following:
while read line ; do (echo $line | grep PATTERN) || echo $line ; done < inputfile
If you want to print "all" lines, there is a simple working solution:
grep "test" -A 9999999 -B 9999999
A => After
B => Before
If you are doing this because you want more context in your search, you can do this:
cat BIG_FILE.txt | less
Doing a search in less should highlight your search terms.
Or pipe the output to your favorite editor. One example:
cat BIG_FILE.txt | vim -
Then search/highlight/replace.
If you are looking for a pattern in a directory recursively, you can either first save it to file.
ls -1R ./ | list-of-files.txt
And then grep that, or pipe it to the grep search
ls -1R | grep --color -rE '[A-Z]|'
This will look of listing all files, but colour the ones with uppercase letters. If you remove the last | you will only see the matches.
I use this to find images named badly with upper case for example, but normal grep does not show the path for each file just once per directory so this way I can see context.
Maybe this is an XY problem, and what you are really trying to do is to highlight occurrences of words as they appear in your shell. If so, you may be able to use your terminal emulator for this. For instance, in Konsole, start Find (ctrl+shift+F) and type your word. The word will then be highlighted whenever it occurs in new or existing output until you cancel the function.