[SOLVED] bash - get previous 4 digits of a hex pattern search

Issue

I'm trying to do a hex search for a pattern.

I have a file and I search for a pattern on the file with...

xxd -g 2 -c 32 -u file | grep "0045 5804 0001 0000"

This returns the lines that contain that pattern.

FFFF FFFF FFFF 4556 4E54 0000 0116 0100 08B9 0045 5804 0001 0000 2008 0000 0001

But I want it is return the 4 digits before that pattern.

In this case the 08B9

How could I do it?

Solution

Don't use grep, use sed, e.g. using any sed:

$ xxd whataver | sed -n 's/.*\(....\) 0045 5804 0001 0000.*/\1/p'
08B9

Answered By - Ed Morton

Answer Checked By - David Goodson (WPSolving Volunteer)