Extract a value from a file

Question

I have a file with many lines, one is:

COMPOSER_HOME=/home/glen/.composer

I want to extract the string /home/glen/.composer from this file in my shell script. How can I?

I can get the whole line with grep but not sure how to remove the first part.

Answer 1

Here:

grep 'COMPOSER_HOME=' file| cut -d= -f2

cut cut's by delimiter = and the 2nd portion would be whatever is After the = eg: /home/glen/.composer , with -f1 you would get COMPOSER_HOME

Answer 2

Since you tagged linux , you have GNU grep which includes PCRE

grep -oP 'COMPOSER_HOME=\K.+' file

The \\K means match what comes before, then throw it out and operate on the rest of the line.

You can also use awk

awk -F "=" '$1 == "COMPOSER_HOME" {print $2}' file

Answer 3

Maybe this is enough

sed -nE 's/COMPOSER_HOME=(.*)/\1/p' your_file

It does not print any line unless you explicitly request it ( -n ), it matches the line starting with COMPOSER_HOME= and captures what follows (.*) (using () instead of \\(\\) , thanks to -E ), and puts in the replacement only what is captured. Then requests the printing of the line with the p flag of the s ubstitution command.