I want to replace all occurrences of "foo" with "bar(" in all files that contain "foo". I have tried
perl -pie 's/foo/bar\(/g' `grep -ril foo .`
but that just hangs and nothing happens. I have tried varying the number of escape backslashes in front of the opening parenthesis, but to no success. I'm working in bash 4.1.5.
The replacement works fine if I remove the opening parenthesis. Does anyone know how to escape the opening parenthesis?
What you posted exits immediately as Perl tries to open s/foo/bar\(/g as a source file since the e is treated the argument of -i.
$ perl -pie 's/foo/bar\(/g' `grep -ril foo .`
Can't open perl script "s/foo/bar\(/g": No such file or directory
I'm guessing you ran the following instead:
perl -i -pe's/foo/bar\(/g' `grep -ril foo .`
This will hang when grep finds nothing. When no arguments are given, the -i is effectively ignored. The program will read from STDIN and write to STDOUT. So, when grep returns nothing, this program will block waiting for input from STDIN.
Solution:
grep -ril foo . | xargs perl -i -pe's/foo/bar\(/ig'
Related
I have approximately 1500 2GB files in a folder and would like to extract lines from them based on a regex. I tried:
find . -regex "filename pattern" -exec grep -P "pattern1\t|pattern2\t|pattern3\t|...|patternN\t" {} +
which works perfectly, but is pretty slow. I then read about running grep with GNU parallel, but couldn't figure out how to properly use it. Here's what I tried:
find . -regex "filename pattern" | parallel grep -P "pattern1\t|pattern2\t|pattern3\t|...|patternN\t" {}
along with a few variations of this command. However, I get in return:
/bin/bash: pattern1t: command not found
/bin/bash: pattern3t: command not found
/bin/bash: pattern2t: command not found
...
It seems the problem lies with the \t I use to ensure I match an entire string in a column of a TSV file. The grep command without parallel works perfectly with this regex.
How can I use escape characters in the grep regex with parallel?
As #Mark Setchell pointed out, I missed the "--quote" argument! This solution works:
find . -regex "filename pattern" -print0 | parallel -0 --quote grep -P "pattern1\t|pattern2\t|pattern3\t|...|patternN\t"
string='binddn:cn=SxX.UXxxxM-E2A,OU=CA,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com'
The working peice of code in Red Hat
dn=($(grep -oi 'cn=[^():]*dc=com' <<< "$string"))
I modified the code for AIX and modified code is
dn=($(grep -xi 'cn=[^():]*dc=com' "$string"))
The code is working perfect in RedHat server, the output in redhat is
dn[0]="cn=SxX.UXxxxM-E2A,OU=CA,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com"
The error in AIX is
grep: can't open binddn:cn=SxX.UXxxxM-E2A,OU=CA,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com
Edited:
Another example:
string = "userbasedn:DC=i,DC=companyname,DC=com?subtree?(&(objectcategory=person)(uidNumber=*)(|(memberOf:1.2.840.113556.1.4.1941:=cn=example1,OU=GROUPS,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com)(memberOf:1.2.840.11.1.4.1941:=cn=example2,OU=GROUPS,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com)))
groupbasedn:DC=i,DC=companyname,DC=com?subtree?(&(objectcategory=group)(gidNumber=*))"
expected output
dn[0]=cn=example1,OU=GROUPS,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com
dn[1]=cn=example2,OU=GROUPS,OU=AI INFRASTRUCTURE,DC=i,DC=companyname,DC=com
If you can use awk, try this:
echo "$string" | awk -F"cn=" 'NF>1{$0=tolower($0);for (i=2;i<=NF;i++) {split($i,a,"dc=com)");print FS a[1]"dc=com"}}'
cn=example1,ou=groups,ou=ai infrastructure,dc=i,dc=companyname,dc=com
cn=example2,ou=groups,ou=ai infrastructure,dc=i,dc=companyname,dc=com
The second argument to grep is a file name, not a string. AIX correctly reports that it cannot find a file with that name. You would get the same error on Red Hat if you tried the same command.
Unfortunately, the -x option doesn't do what you hope; it checks whether the entire line of input matches the regex. Again, you will find exactly the same behavior on Red Hat.
According to the AIX grep manual page it supports the -o option just fine, though.
The Bash "here string" syntax <<<"string" is not available if you don't have Bash, but it is easy to rephrase portably:
printf '%s\n' "$string" |
grep -oi 'cn=[^():]*dc=com'
If you don't have grep -o, try with sed:
printf '%s\n' "$string" |
sed -n 's/.*\(cn=[^():]*dc=com\).*/\1/p'
This is not exactly the same, because the first .* is greedy. If you expect more than one match on a line, you will need a slightly more complex regex.
The file that I want to grep contains many lines.
I want to grep lines which contain only 1 digit: "0" or "1".
I used this command:
exec grep -e "^\[0-1\]{1}$" file
But I got:
child process exited abnormally
What's wrong with RegExp of grep?
The most common issue when running grep as a Tcl subprocess is that it exits with a non-zero error code when it doesn't find anything at all. This always causes Tcl to throw an exception. The simplest workaround is perhaps this:
exec /bin/sh -c {grep -e '^[0-1]{1}$'; true} < file
Note that we are feeding in the file using a redirection here; this means that it is not necessary to strip the name of the file from the results.
In a sh script, I am trying to loop over all files that match the following pattern
abc.123 basically abc. followed by only numbers, number following . can be of any length.
Using
$ shopt -s extglob
$ ls abc.+([0-9])
does the job but on terminal and not through the script. How can I get only files that match the pattern?
if I understood you right, the pattern could be translated into regex:
^abc\.[0-9]+$
so you could
keep using ls and grep the output. for example:
ls *.*|xargs -n1|grep -E '^abc\.[0-9]+$'
or use find
find has an option -regex
If you're using sh and not bash, and presumably you also want to be POSIX compliant, you can use:
for f in ./*
do
echo "$f" | grep -Eq '^\./abc.[0-9]+$' && continue
echo "Something with $f here"
done
It will work fine with filenames with spaces, quotes and such, but may match some filenames with line feeds in them that it shouldn't.
If you tagged your question bash because you're using bash, then just use extglob like you described.
I have sed running with the following argument fine if I copy and paste this into an open shell:
cat test.txt | sed '/[,0-9]\{0,\}[0-9]\{1,\}[acd][0-9]\{1,\}[,0-9]\{0,\}/{N
s/[,0-9]\{0,\}[0-9]\{1,\}[acd][0-9]\{1,\}[,0-9]\{0,\}\n\-\-\-//}'
The problem is that when I try to move this into a KornShell (ksh) script, the ksh throws errors because of what I think is that new line character. Can anyone give me a hand with this? FYI: the regular expression is supposed to be a multiple line replacement.
Thank you!
This: \{0,\} can be replaced by this: *
This: \{1,\} can be replaced by this: \+
It's not necessary to escape hyphens.
The newline can be replaced by -e (or by a semicolon)
The cat can be replaced by using the filename as an argument to sed
The result:
sed -e '/[,0-9]*[0-9]\+[acd][0-9]\+[,0-9]*/{N' -e 's/[,0-9]*[0-9]\+[acd][0-9]\+[,0-9]*\n---//}' test.txt
or
sed '/[,0-9]*[0-9]\+[acd][0-9]\+[,0-9]*/{N;s/[,0-9]*[0-9]\+[acd][0-9]\+[,0-9]*\n---//}' test.txt
(untested)
can you try to put your regex in a file and call sed with the option -f ?
cat test.txt | sed -f file.sed
Can you try to replace the new line character with `echo -e \\r`
The Korn Shell - unlike the C Shell - has no problem with newlines in strings. The newline is very unlikely to be your problem, therefore. The same comments apply to Bourne and POSIX shells, and to Bash. I've copied your example and run it on Linux under both Bash and Korn shell without any problem.
If you use C Shell for your work, are you sure you're running 'ksh ./script' and not './script'?
Otherwise, there is some other problem - an unbalanced quote somewhere, perhaps.
Check out the '-v' and '-n' options as well as the '-x' option to the Korn Shell. That may tell you more about where the problem is.