Linux extract text between specific strings - regex

Question

I have multiple files with different job names. The job name is specified as follows.
#SBATCH --job-name=01_job1 #Set the job name
I want to use sed/awk/grep to automatically get the name, that is to say, what follows '--job-name=' and precedes the comment '#Set the job name'. For the example above, I want to get 01_job1. The job name could be longer for several files, and there are multiple = signs in following lines in the file.
I have tried using grep -oP "job-name=\s+\K\w+" file and get an empty output. I suspect that this doesn't work because there is no space between 'name=' and '01_job1', so they must be understood as a single word.
I also unsuccessfully tried using awk '{for (I=1;I<NF;I++) if ($I == "name=") print $(I+1)}' file, attempting to find the characters after 'name='.
Lastly, I also unsuccessfully tried sed -e 's/name=\(.*\)#Set/\1/' file to find the characters between 'name=' and the beginning of the comment '#Set'. I receive the whole file as my output when I attempt this.
I appreciate any guidance. Thank you!!

Answer 1

You need to match the whole string with sed and capture just what you need to get, and use -n option with the p flag:
sed -n 's/.*name=\([^[:space:]]*\).*/\1/p'
See the online demo:
#!/bin/bash
s='#SBATCH --job-name=01_job1 #Set the job name'
sed -n 's/.*name=\([^[:space:]]*\).*/\1/p' <<< "$s"
# => 01_job1
Details:
-n - suppresses default line output
.* - any text
name= - a literal name= string
\([^[:space:]]*\) - Group 1 (\1): any zero or more chars other than whitespace
.* - any text
p - print the result of the successful substitution.

Answer 2

Simlar to the answer of Gilles Quenot
grep -oP -- '--job-name=\K.*(?= *# *Set the job name)'
This adds a look-ahead to ensure that the string is followed by #Set the job name

Answer 3

1st solution: In GNU awk with your shown samples please try following awk code.
awk -v RS=' --job-name=\\S+' 'RT && split(RT,arr,"="){print arr[2]}' Input_file
OR a non-one liner form of above GNU awk code would be:
awk -v RS=' --job-name=\\S+' '
RT && split(RT,arr,"="){
print arr[2]
}
' Input_file
2nd solution: Using any awk please try following code.
awk -F'[[:space:]]+|--job-name=' '{print $3}' Input_file
3rd solution: Using GNU grep please try following code with your shown samples and using non-greedy .*? approach here in regex.
grep -oP '^.*?--job-name=\K\S+' Input_file

Answer 4

Use this, you was close, just correctness of your grep -oP attempt (the main issue if you are trying to match a space after = character):
$ grep -oP -- '--job-name=\K\S+' file
01_job1
The regular expression matches as follows:
Node
Explanation
job-name=
'job-name='
\K
resets the start of the match (what is Kept) as a shorter alternative to using a look-behind assertion: perlmonks look arounds and Support of K in regex
\S+
non-whitespace (all but \n, \r, \t, \f, and " ") (1 or more times (matching the most amount possible))

Answer 5

You can use a lookbehind and lookahead with GNU grep to get exactly what you describe:
grep -oP '(?<=--job-name=)\S+(?=\s+#Set the job name)' file
Or with awk:
awk '/^#SBATCH[[:space:]]+--job-name=/ &&
/#Set the job name$/ {
sub(/^[^=]*=/,"")
sub(/#[^#]*$/,"")
print
}' file
Or perl:
perl -lnE 'say $1 if /(?<=--job-name=)(\S+)(?=\s+#Set the job name)/' file
Any prints:
01_job1