[SOLVED] How do I fetch lines before/after the grep result in bash?

Issue

I want a way to search in a given text. For that, I use grep:

grep -i "my_regex"

That works. But given the data like this:

This is the test data
This is the error data as follows
. . . 
. . . .
. . . . . . 
. . . . . . . . .
Error data ends

Once I found the word error (using grep -i error data), I wish to find the 10 lines that follow the word error. So my output should be:

. . . 
. . . .
. . . . . . 
. . . . . . . . .
Error data ends

Are there any way to do it?

Solution

You can use the -B and -A to print lines before and after the match.

grep -i -B 10 'error' data

Will print the 10 lines before the match, including the matching line itself.

Answered By - Jon Lin

Answer Checked By - David Goodson (WPSolving Volunteer)