Linux extract text between specific strings

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.