Shell-Script – How to Delete Everything Until a Pattern and After Another Pattern from a Line

Doing this for single line strings is very simple:

sed 's/((([^)]*)))//g' file

If you need it to deal with multiline strings, it gets more complex. One approach would be to use tr to replace all newlines with the null character (\0), make the substitution and translate back again:

tr '\n' '\0' < file | sed 's/((([^)]*)))//g' | tr '\0' '\n'

Alternatively, you could just use perl:

perl -0pe 's/\(\(\([^)]+\)\)\)//g;' file

The -0 causes perl to read the entire file into memory (this might be a problem for huge files), the -p means "print each line" but because of the -0, the "line" is actually the entire file. The s/// is the same idea as for sed.

Using cut :

$ cut -sd'/' -f2 file.txt   ##This will print only the lines containing /
7fad416d-f2b3-4259-b98d-2449957a3123
8a8589bf-49e3-4cd7-af15-6753067355c6

The following suggestions assumes that / appears only once in a line :

Using grep :

$ grep -o '[^/]*$' file.txt  ##This will print the lines not having / too
7fad416d-f2b3-4259-b98d-2449957a3123
8a8589bf-49e3-4cd7-af15-6753067355c6

If you have / in all of the lines, you can use these too:

Using bash parameter expansion:

$ while read line; do echo "${line#*/}"; done <file.txt 
7fad416d-f2b3-4259-b98d-2449957a3123
8a8589bf-49e3-4cd7-af15-6753067355c6

Or python :

#!/usr/bin/env python2
with open('file.txt') as f:
    for line in f:
        print line.split('/')[1].rstrip()

Note that as far as your example is concerned all of the above suggestions are valid.