How to grep a string in a file and output the digits next to the string

Question

I am trying to print out a number next to a string that I search (using shell script).

For ex., my file contains a line

arg[30] = -Xmx4096m

I search the file with string Xmx and I need the output of next digits attached to the string and nothing else is needed in the output. How can I achieve this?

Answer 1

If your version of grep supports PCRE mode, you can use grep -Po 'Xmx\K\d+'

Ex.

echo 'F arg[30] = -Xmx4096m' | grep -Po 'Xmx\K\d+'
4096

Answer 2

sed -n 's/.*Xmx\([0-9]\+\).*/\1/p' filename

Explanation:

• sed -n makes sed only print output when told (by p)
• s/<p1>/<p2>/p search for <p1> and replace with <p2>. then print line
• .*Xmx\([0-9]\+\).* any string with Xmx followed by digits. the digits are "captured".
• \1 the first captured group.

In human words: search for lines containing Xmx followed by digits. Take note of the digits. Discard everything but keep the digits. Print the digits.