I have a text file which have lots of lines
I have a line in it which is:
MyCar on
how can I turn my car off?
You could use sed:
sed -i 's/MyCar on/MyCar off/' path/to/file
sed 's/MyCar on/MyCar off/' >filename
more on sed
You can do this with shell only. This example uses an unnecessary case statement for this particular example, but I included it to show how you could incorporate multiple replacements. Although the code is larger than a sed 1-liner it is typically much faster since it uses only shell builtins (as much as 20x for small files).
REPLACEOLD="old"
WITHNEW="new"
FILE="tmpfile"
OUTPUT=""
while read LINE || [ "$LINE" ]; do
case "$LINE" in
*${REPLACEOLD}*)OUTPUT="${OUTPUT}${LINE//$REPLACEOLD/$WITHNEW}
";;
*)OUTPUT="${OUTPUT}${LINE}
";;
esac
done < "${FILE}"
printf "${OUTPUT}" > "${FILE}"
for the simple case one could omit the case statement:
while read LINE || [ "$LINE" ]; do
OUTPUT="${OUTPUT}${LINE//$REPLACEOLD/$WITHNEW}
"; done < "${FILE}"
printf "${OUTPUT}" > "${FILE}"
Note: the ...|| [ "$LINE" ]... bit is to prevent losing the last line of a file that doesn't end in a new line
(now you know at least one reasone why your text editor keeps adding those)
try this command when inside the file
:%s/old test/new text/g
Using sed with variables;
host=$(hostname)
se1=$(cat /opt/splunkforwarder/etc/system/local/server.conf | grep serverName)
sed -i "s/${se1}/serverName = ${host}/g" /opt/splunkforwarder/etc/system/local/server.conf`
Related
I have the following bash script:
while IFS= read -r line; do
line=$(echo $line | sed "s/\'/\'\'/")
[[ $line =~ ^\<ID\>(.*) ]] && printf "${BASH_REMATCH[1]}"
done < <(dos2unix < file)
EDITED version of script without dos2unix:
while IFS= read -r line && line=${line%$'\r'}; do
[[ $line =~ ^\<ID\>(.*) ]] && printf "${BASH_REMATCH[1]}"
done < file
I want to substitute every apostrophe in "file" with 2 apostrophes BEFORE I loop through it. How can I do this? I'd be grateful for any suggestions concerning any of the 2 versions.
IMPORTANT
Im NOT allowed to modify the original file!!
This is a job for sed alone:
sed 's/\r$//;s/\'/\'\'/g;s/^<ID>\(.*\)/\1/p;d' < file
The steps are:
sed accepts multiple commands separated with newlines, semicolons or given as multiple -e options.
sed 's/\r$//; removes the CR at end of each line like dos2unix.
The g flag added to s/\'/\'\'/ means replace all occurrences in the line; default is to replace just one.
The s/^<ID>\(.*\)/\1/ does the equivalent of that bash regex match and the p flag at the end makes sed print the matching lines now, because
The d command removes the line so it won't get printed by default (you could do that with the -n option instead).
On a side-note, my zsh does not accept \' in ', so I'd probably write it
sed -n -e 's/\r$//' -e "s/'/''/g" -e 's/^<ID>\(.*\)/\1/p'
It should be equivalent, just switching the quote style, separate options and the -n instead of final d.
While this is not a "solution" (your question is not clear on what is not working in your code), you certainly should avoid calling sed for each individual line. It is not "wrong" in the sense of producing an incorrect result, but it is so much slower that it should be avoided. There are ways do it that are both faster and simpler to code.
Do it this way :
while IFS= read -r line; do
[[ $line =~ ^\<ID\>(.*) ]] && printf "${BASH_REMATCH[1]}"
done < <(dos2unix < file | sed "s/\'/\'\'/")
I am trying to parse each line of a file and look for a particular string. The script seems to be doing its intended job, however, in parallel it tries to execute the if command on line 6:
#!/bin/bash
for line in $(cat $1)
do
echo $line | grep -e "Oct/2015"
if($?==0); then
echo "current line is: $line"
fi
done
and I get the following (my script is readlines.sh)
./readlines.sh: line 6: 0==0: command not found
First: As Mr. Llama says, you need more spaces. Right now your script tries to look for a file named something like /usr/bin/0==0 to run. Instead:
[ "$?" -eq 0 ] # POSIX-compliant numeric comparison
[ "$?" = 0 ] # POSIX-compliant string comparison
(( $? == 0 )) # bash-extended numeric comparison
Second: Don't test $? at all in this case. In fact, you don't even have good cause to use grep; the following is both more efficient (because it uses only functionality built into bash and requires no invocation of external commands) and more readable:
if [[ $line = *"Oct/2015"* ]]; then
echo "Current line is: $line"
fi
If you really do need to use grep, write it like so:
if echo "$line" | grep -q "Oct/2015"; then
echo "Current line is: $line"
fi
That way if operates directly on the pipeline's exit status, rather than running a second command testing $? and operating on that command's exit status.
#Charles Duffy has a good answer which I have up-voted as correct (and it is), but here's a detailed, line by line breakdown of your script and the correct thing to do for each part of it.
for line in $(cat $1)
As I noted in my comment elsewhere this should be done as a while read construct instead of a for cat construct.
This construct will wordsplit each line making spaces in the file separate "lines" in the output.
All empty lines will be skipped.
In addition when you cat $1 the variable should be quoted. If it is not quoted spaces and other less-usual characters appearing in the file name will cause the cat to fail and the loop will not process the file.
The complete line would read:
while IFS= read -r line
An illustrative example of the tradeoffs can be found here. The linked test script follows. I tried to include an indication of why IFS= and -r are important.
#!/bin/bash
mkdir -p /tmp/testcase
pushd /tmp/testcase >/dev/null
printf '%s\n' '' two 'three three' '' ' five with leading spaces' 'c:\some\dos\path' '' > testfile
printf '\nwc -l testfile:\n'
wc -l testfile
printf '\n\nfor line in $(cat) ... \n\n'
let n=1
for line in $(cat testfile) ; do
echo line $n: "$line"
let n++
done
printf '\n\nfor line in "$(cat)" ... \n\n'
let n=1
for line in "$(cat testfile)" ; do
echo line $n: "$line"
let n++
done
let n=1
printf '\n\nwhile read ... \n\n'
while read line ; do
echo line $n: "$line"
let n++
done < testfile
printf '\n\nwhile IFS= read ... \n\n'
let n=1
while IFS= read line ; do
echo line $n: "$line"
let n++
done < testfile
printf '\n\nwhile IFS= read -r ... \n\n'
let n=1
while IFS= read -r line ; do
echo line $n: "$line"
let n++
done < testfile
rm -- testfile
popd >/dev/null
rmdir /tmp/testcase
Note that this is a bash-heavy example. Other shells do not tend to support -r for read, for example, nor is let portable. On to the next line of your script.
do
As a matter of style I prefer do on the same line as the for or while declaration, but there's no convention on this.
echo $line | grep -e "Oct/2015"
The variable $line should be quoted here. In general, meaning always unless you specifically know better, you should double-quote all expansion--and that means subshells as well as variables. This insulates you from most unexpected shell weirdness.
You decclared your shell as bash which means you will have there "Here string" operator <<< available to you. When available it can be used to avoid the pipe; each element of a pipeline executes in a subshell, which incurs extra overhead and can lead to unexpected behavior if you try to modify variables. This would be written as
grep -e "Oct/2015" <<<"$line"
Note that I have quoted the line expansion.
You have called grep with -e, which is not incorrect but is needless since your pattern does not begin with -. In addition you have full-quoted a string in shell but you don't attempt to expand a variable or use other shell interpolation inside of it. When you don't expect and don't want the contents of a quoted string to be treated as special by the shell you should single quote them. Furthermore, your use of grep is inefficient: because your pattern is a fixed string and not a regular expression you could have used fgrep or grep -F, which does string contains rather than regular expression matching (and is far faster because of this). So this could be
grep -F 'Oct/2015' <<<"$line"
Without altering the behavior.
if($?==0); then
This is the source of your original problem. In shell scripts commands are separated by whitespace; when you say if($?==0) the $? expands, probably to 0, and bash will try to execute a command called if(0==0) which is a legal command name. What you wanted to do was invoke the if command and give it some parameters, which requires more whitespace. I believe others have covered this sufficiently.
You should never need to test the value of $? in a shell script. The if command exists for branching behavior based on the return code of whatever command you pass to it, so you can inline your grep call and have if check its return code directly, thus:
if grep -F 'Oct/2015` <<<"$line" ; then
Note the generous whitespace around the ; delimiter. I do this because in shell whitespace is usually required and can only sometiems be omitted. Rather than try to remember when you can do which I recommend an extra one space padding around everything. It's never wrong and can make other mistakes easier to notice.
As others have noted this grep will print matched lines to stdout, which is probably not something you want. If you are using GNU grep, which is standard on Linux, you will have the -q switch available to you. This will suppress the output from grep
if grep -q -F 'Oct/2015' <<<"$line" ; then
If you are trying to be strictly standards compliant or are in any environment with a grep that doesn't know -q the standard way to achieve this effect is to redirect stdout to /dev/null/
if printf "$line" | grep -F 'Oct/2015' >/dev/null ; then
In this example I also removed the here string bashism just to show a portable version of this line.
echo "current line is: $line"
There is nothing wrong with this line of your script, except that although echo is standard implementations vary to such an extent that it's not possible to absolutely rely on its behavior. You can use printf anywhere you would use echo and you can be fairly confident of what it will print. Even printf has some caveats: Some uncommon escape sequences are not evenly supported. See mascheck for details.
printf 'current line is: %s\n' "$line"
Note the explicit newline at the end; printf doesn't add one automatically.
fi
No comment on this line.
done
In the case where you did as I recommended and replaced the for line with a while read construct this line would change to:
done < "$1"
This directs the contents of the file in the $1 variable to the stdin of the while loop, which in turn passes the data to read.
In the interests of clarity I recommend copying the value from $1 into another variable first. That way when you read this line the purpose is more clear.
I hope no one takes great offense at the stylistic choices made above, which I have attempted to note; there are many ways to do this (but not a great many correct) ways.
Be sure to always run interesting snippets through the excellent shellcheck and explain shell when you run into difficulties like this in the future.
And finally, here's everything put together:
#!/bin/bash
input_file="$1"
while IFS= read -r line ; do
if grep -q -F 'Oct/2015' <<<"$line" ; then
printf 'current line is %s\n' "$line"
fi
done < "$input_file"
If you like one-liners, you may use AND operator (&&), for example:
echo "$line" | grep -e "Oct/2015" && echo "current line is: $line"
or:
grep -qe "Oct/2015" <<<"$line" && echo "current line is: $line"
Spacing is important in shell scripting.
Also, double-parens is for numerical comparison, not single-parens.
if (( $? == 0 )); then
I have a text file with a list of files with the structure ABC123456A or ABC123456AA. What I would like to do is check whether the files ABC123456ZZP also exists. i.e I want to substitute the letter(s) after ABC123456 with ZZP
Can I do this using sed?
Like this?
X=ABC123456 ; echo ABC123456AA | sed -e "s,\(${X}\).*,\1ZZP,"
You could use sed as wilx suggests but I think a better option would be bash.
while read file; do
base=${file:0:9}
[[ -f ${base}ZZP ]] && echo "${base}ZZP exists!"
done < file
This will loop over each line in file
then base is set to the first 9 characters of the line (excluding whitespace)
then check to see if a file exists with ZZP on the end of base and print a message if it does.
Look:
$ str="ABC123456AA"
$ echo "${str%[[:alpha:]][[:alpha:]]*}"
ABC123456
so do this:
while IFS= read -r tgt; do
tgt="${tgt%[[:alpha:]][[:alpha:]]*}ZZP"
[[ -f "$tgt" ]] && printf "%s exists!\n" "$tgt"
done < file
It will still fail for file names that contain newlines so let us know if you have that situation but unlike the other posted solutions it will work for file names with other than 9 key characters, file names containing spaces, commas, backslashes, globbing characters, etc., etc. and it is efficient.
Since you said now that you only need the first 9 characters of each line and you were happy with piping every line to sed, here's another solution you might like:
cut -c1-9 file |
while IFS= read -r tgt; do
[[ -f "${tgt}ZZP" ]] && printf "%sZZP exists!\n" "$tgt"
done
It'd be MUCH more efficient and more robust than the sed solution, and similar in both contexts to the other shell solutions.
I am trying to find a clever way to figure out if the file passed to sed has been altered successfully or not.
Basically, I want to know if the file has been changed or not without having to look at the file modification date.
The reason why I need this is because I need to do some extra stuff if sed has successfully replaced a pattern.
I currently have:
grep -q $pattern $filename
if [ $? -eq 0 ]
then
sed -i s:$pattern:$new_pattern: $filename
# DO SOME OTHER STUFF HERE
else
# DO SOME OTHER STUFF HERE
fi
The above code is a bit expensive and I would love to be able to use some hacks here.
A bit late to the party but for the benefit of others, I found the 'w' flag to be exactly what I was looking for.
sed -i "s/$pattern/$new_pattern/w changelog.txt" "$filename"
if [ -s changelog.txt ]; then
# CHANGES MADE, DO SOME STUFF HERE
else
# NO CHANGES MADE, DO SOME OTHER STUFF HERE
fi
changelog.txt will contain each change (ie the changed text) on it's own line. If there were no changes, changelog.txt will be zero bytes.
A really helpful sed resource (and where I found this info) is http://www.grymoire.com/Unix/Sed.html.
I believe you may find these GNU sed extensions useful
t label
If a s/// has done a successful substitution since the last input line
was read and since the last t or T command, then branch to label; if
label is omitted, branch to end of script.
and
q [exit-code]
Immediately quit the sed script without processing any more input, except
that if auto-print is not disabled the current pattern space will be printed.
The exit code argument is a GNU extension.
It seems like exactly what are you looking for.
This might work for you (GNU sed):
sed -i.bak '/'"$old_pattern"'/{s//'"$new_pattern"'/;h};${x;/./{x;q1};x}' file || echo changed
Explanation:
/'"$old_pattern"'/{s//'"$new_pattern"'/;h} if the pattern space (PS) contains the old pattern, replace it by the new pattern and copy the PS to the hold space (HS).
${x;/./{x;q1};x} on encountering the last line, swap to the HS and test it for the presence of any string. If a string is found in the HS (i.e. a substitution has taken place) swap back to the original PS and exit using the exit code of 1, otherwise swap back to the original PS and exit with the exit code of 0 (the default).
You can diff the original file with the sed output to see if it changed:
sed -i.bak s:$pattern:$new_pattern: "$filename"
if ! diff "$filename" "$filename.bak" &> /dev/null; then
echo "changed"
else
echo "not changed"
fi
rm "$filename.bak"
You could use awk instead:
awk '$0 ~ p { gsub(p, r); t=1} 1 END{ exit (!t) }' p="$pattern" r="$repl"
I'm ignoring the -i feature: you can use the shell do do redirections as necessary.
Sigh. Many comments below asking for basic tutorial on the shell. You can use the above command as follows:
if awk '$0 ~ p { gsub(p, r); t=1} 1 END{ exit (!t) }' \
p="$pattern" r="$repl" "$filename" > "${filename}.new"; then
cat "${filename}.new" > "${filename}"
# DO SOME OTHER STUFF HERE
else
# DO SOME OTHER STUFF HERE
fi
It is not clear to me if "DO SOME OTHER STUFF HERE" is the same in each case. Any similar code in the two blocks should be refactored accordingly.
In macos I just do it as follows:
changes=""
changes+=$(sed -i '' "s/$to_replace/$replacement/g w /dev/stdout" "$f")
if [ "$changes" != "" ]; then
echo "CHANGED!"
fi
I checked, and this is faster than md5, cksum and sha comparisons
I know it is a old question and using awk instead of sed is perhaps the best idea, but if one wants to stick with sed, an idea is to use the -w flag. The file argument to the w flag only contains the lines with a match. So, we only need to check that it is not empty.
perl -sple '$replaced++ if s/$from/$to/g;
END{if($replaced != 0){ print "[Info]: $replaced replacement done in $ARGV(from/to)($from/$to)"}
else {print "[Warning]: 0 replacement done in $ARGV(from/to)($from/$to)"}}' -- -from="FROM_STRING" -to="$DESIRED_STRING" </file/name>
Example:
The command will produce the following output, stating the number of changes made/file.
perl -sple '$replaced++ if s/$from/$to/g;
END{if($replaced != 0){ print "[Info]: $replaced replacement done in $ARGV(from/to)($from/$to)"}
else {print "[Warning]: 0 replacement done in $ARGV(from/to)($from/$to)"}}' -- -from="timeout" -to="TIMEOUT" *
[Info]: 5 replacement done in main.yml(from/to)(timeout/TIMEOUT)
[Info]: 1 replacement done in task/main.yml(from/to)(timeout/TIMEOUT)
[Info]: 4 replacement done in defaults/main.yml(from/to)(timeout/TIMEOUT)
[Warning]: 0 replacement done in vars/main.yml(from/to)(timeout/TIMEOUT)
Note: I have removed -i from the above command , so it will not update the files for the people who are just trying out the command. If you want to enable in-place replacements in the file add -i after perl in above command.
check if sed has changed MANY files
recursive replace of all files in one directory
produce a list of all modified files
workaround with two stages: match + replace
g='hello.*world'
s='s/hello.*world/bye world/g;'
d='./' # directory of input files
o='modified-files.txt'
grep -r -l -Z -E "$g" "$d" | tee "$o" | xargs -0 sed -i "$s"
the file paths in $o are zero-delimited
$ echo hi > abc.txt
$ sed "s/hi/bye/g; t; q1;" -i abc.txt && (echo "Changed") || (echo "Failed")
Changed
$ sed "s/hi/bye/g; t; q1;" -i abc.txt && (echo "Changed") || (echo "Failed")
Failed
https://askubuntu.com/questions/1036912/how-do-i-get-the-exit-status-when-using-the-sed-command/1036918#1036918
Don't use sed to tell if it has changed a file; instead, use grep to tell if it is going to change a file, then use sed to actually change the file. Notice the single line of sed usage at the very end of the Bash function below:
# Usage: `gs_replace_str "regex_search_pattern" "replacement_string" "file_path"`
gs_replace_str() {
REGEX_SEARCH="$1"
REPLACEMENT_STR="$2"
FILENAME="$3"
num_lines_matched=$(grep -c -E "$REGEX_SEARCH" "$FILENAME")
# Count number of matches, NOT lines (`grep -c` counts lines),
# in case there are multiple matches per line; see:
# https://superuser.com/questions/339522/counting-total-number-of-matches-with-grep-instead-of-just-how-many-lines-match/339523#339523
num_matches=$(grep -o -E "$REGEX_SEARCH" "$FILENAME" | wc -l)
# If num_matches > 0
if [ "$num_matches" -gt 0 ]; then
echo -e "\n${num_matches} matches found on ${num_lines_matched} lines in file"\
"\"${FILENAME}\":"
# Now show these exact matches with their corresponding line 'n'umbers in the file
grep -n --color=always -E "$REGEX_SEARCH" "$FILENAME"
# Now actually DO the string replacing on the files 'i'n place using the `sed`
# 's'tream 'ed'itor!
sed -i "s|${REGEX_SEARCH}|${REPLACEMENT_STR}|g" "$FILENAME"
fi
}
Place that in your ~/.bashrc file, for instance. Close and reopen your terminal and then use it.
Usage:
gs_replace_str "regex_search_pattern" "replacement_string" "file_path"
Example: replace do with bo so that "doing" becomes "boing" (I know, we should be fixing spelling errors not creating them :) ):
$ gs_replace_str "do" "bo" test_folder/test2.txt
9 matches found on 6 lines in file "test_folder/test2.txt":
1:hey how are you doing today
2:hey how are you doing today
3:hey how are you doing today
4:hey how are you doing today hey how are you doing today hey how are you doing today hey how are you doing today
5:hey how are you doing today
6:hey how are you doing today?
$SHLVL:3
Screenshot of the output:
References:
https://superuser.com/questions/339522/counting-total-number-of-matches-with-grep-instead-of-just-how-many-lines-match/339523#339523
https://unix.stackexchange.com/questions/112023/how-can-i-replace-a-string-in-a-files/580328#580328
Trying to debug an issue with a server and my only log file is a 20GB log file (with no timestamps even! Why do people use System.out.println() as logging? In production?!)
Using grep, I've found an area of the file that I'd like to take a look at, line 347340107.
Other than doing something like
head -<$LINENUM + 10> filename | tail -20
... which would require head to read through the first 347 million lines of the log file, is there a quick and easy command that would dump lines 347340100 - 347340200 (for example) to the console?
update I totally forgot that grep can print the context around a match ... this works well. Thanks!
I found two other solutions if you know the line number but nothing else (no grep possible):
Assuming you need lines 20 to 40,
sed -n '20,40p;41q' file_name
or
awk 'FNR>=20 && FNR<=40' file_name
When using sed it is more efficient to quit processing after having printed the last line than continue processing until the end of the file. This is especially important in the case of large files and printing lines at the beginning. In order to do so, the sed command above introduces the instruction 41q in order to stop processing after line 41 because in the example we are interested in lines 20-40 only. You will need to change the 41 to whatever the last line you are interested in is, plus one.
# print line number 52
sed -n '52p' # method 1
sed '52!d' # method 2
sed '52q;d' # method 3, efficient on large files
method 3 efficient on large files
fastest way to display specific lines
with GNU-grep you could just say
grep --context=10 ...
No there isn't, files are not line-addressable.
There is no constant-time way to find the start of line n in a text file. You must stream through the file and count newlines.
Use the simplest/fastest tool you have to do the job. To me, using head makes much more sense than grep, since the latter is way more complicated. I'm not saying "grep is slow", it really isn't, but I would be surprised if it's faster than head for this case. That'd be a bug in head, basically.
What about:
tail -n +347340107 filename | head -n 100
I didn't test it, but I think that would work.
I prefer just going into less and
typing 50% to goto halfway the file,
43210G to go to line 43210
:43210 to do the same
and stuff like that.
Even better: hit v to start editing (in vim, of course!), at that location. Now, note that vim has the same key bindings!
You can use the ex command, a standard Unix editor (part of Vim now), e.g.
display a single line (e.g. 2nd one):
ex +2p -scq file.txt
corresponding sed syntax: sed -n '2p' file.txt
range of lines (e.g. 2-5 lines):
ex +2,5p -scq file.txt
sed syntax: sed -n '2,5p' file.txt
from the given line till the end (e.g. 5th to the end of the file):
ex +5,p -scq file.txt
sed syntax: sed -n '2,$p' file.txt
multiple line ranges (e.g. 2-4 and 6-8 lines):
ex +2,4p +6,8p -scq file.txt
sed syntax: sed -n '2,4p;6,8p' file.txt
Above commands can be tested with the following test file:
seq 1 20 > file.txt
Explanation:
+ or -c followed by the command - execute the (vi/vim) command after file has been read,
-s - silent mode, also uses current terminal as a default output,
q followed by -c is the command to quit editor (add ! to do force quit, e.g. -scq!).
I'd first split the file into few smaller ones like this
$ split --lines=50000 /path/to/large/file /path/to/output/file/prefix
and then grep on the resulting files.
If your line number is 100 to read
head -100 filename | tail -1
Get ack
Ubuntu/Debian install:
$ sudo apt-get install ack-grep
Then run:
$ ack --lines=$START-$END filename
Example:
$ ack --lines=10-20 filename
From $ man ack:
--lines=NUM
Only print line NUM of each file. Multiple lines can be given with multiple --lines options or as a comma separated list (--lines=3,5,7). --lines=4-7 also works.
The lines are always output in ascending order, no matter the order given on the command line.
sed will need to read the data too to count the lines.
The only way a shortcut would be possible would there to be context/order in the file to operate on. For example if there were log lines prepended with a fixed width time/date etc.
you could use the look unix utility to binary search through the files for particular dates/times
Use
x=`cat -n <file> | grep <match> | awk '{print $1}'`
Here you will get the line number where the match occurred.
Now you can use the following command to print 100 lines
awk -v var="$x" 'NR>=var && NR<=var+100{print}' <file>
or you can use "sed" as well
sed -n "${x},${x+100}p" <file>
With sed -e '1,N d; M q' you'll print lines N+1 through M. This is probably a bit better then grep -C as it doesn't try to match lines to a pattern.
Building on Sklivvz' answer, here's a nice function one can put in a .bash_aliases file. It is efficient on huge files when printing stuff from the front of the file.
function middle()
{
startidx=$1
len=$2
endidx=$(($startidx+$len))
filename=$3
awk "FNR>=${startidx} && FNR<=${endidx} { print NR\" \"\$0 }; FNR>${endidx} { print \"END HERE\"; exit }" $filename
}
To display a line from a <textfile> by its <line#>, just do this:
perl -wne 'print if $. == <line#>' <textfile>
If you want a more powerful way to show a range of lines with regular expressions -- I won't say why grep is a bad idea for doing this, it should be fairly obvious -- this simple expression will show you your range in a single pass which is what you want when dealing with ~20GB text files:
perl -wne 'print if m/<regex1>/ .. m/<regex2>/' <filename>
(tip: if your regex has / in it, use something like m!<regex>! instead)
This would print out <filename> starting with the line that matches <regex1> up until (and including) the line that matches <regex2>.
It doesn't take a wizard to see how a few tweaks can make it even more powerful.
Last thing: perl, since it is a mature language, has many hidden enhancements to favor speed and performance. With this in mind, it makes it the obvious choice for such an operation since it was originally developed for handling large log files, text, databases, etc.
print line 5
sed -n '5p' file.txt
sed '5q' file.txt
print everything else than line 5
`sed '5d' file.txt
and my creation using google
#!/bin/bash
#removeline.sh
#remove deleting it comes move line xD
usage() { # Function: Print a help message.
echo "Usage: $0 -l LINENUMBER -i INPUTFILE [ -o OUTPUTFILE ]"
echo "line is removed from INPUTFILE"
echo "line is appended to OUTPUTFILE"
}
exit_abnormal() { # Function: Exit with error.
usage
exit 1
}
while getopts l:i:o:b flag
do
case "${flag}" in
l) line=${OPTARG};;
i) input=${OPTARG};;
o) output=${OPTARG};;
esac
done
if [ -f tmp ]; then
echo "Temp file:tmp exist. delete it yourself :)"
exit
fi
if [ -f "$input" ]; then
re_isanum='^[0-9]+$'
if ! [[ $line =~ $re_isanum ]] ; then
echo "Error: LINENUMBER must be a positive, whole number."
exit 1
elif [ $line -eq "0" ]; then
echo "Error: LINENUMBER must be greater than zero."
exit_abnormal
fi
if [ ! -z $output ]; then
sed -n "${line}p" $input >> $output
fi
if [ ! -z $input ]; then
# remove this sed command and this comes move line to other file
sed "${line}d" $input > tmp && cp tmp $input
fi
fi
if [ -f tmp ]; then
rm tmp
fi
You could try this command:
egrep -n "*" <filename> | egrep "<line number>"
Easy with perl! If you want to get line 1, 3 and 5 from a file, say /etc/passwd:
perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd
I am surprised only one other answer (by Ramana Reddy) suggested to add line numbers to the output. The following searches for the required line number and colours the output.
file=FILE
lineno=LINENO
wb="107"; bf="30;1"; rb="101"; yb="103"
cat -n ${file} | { GREP_COLORS="se=${wb};${bf}:cx=${wb};${bf}:ms=${rb};${bf}:sl=${yb};${bf}" grep --color -C 10 "^[[:space:]]\\+${lineno}[[:space:]]"; }