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

Question

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?

From your description it seems you want the 10 lines proceeding the word `error`. — ThomasW, Mar 19 '15 at 05:33
Does this answer your question? [grep a file, but show several surrounding lines?](https://stackoverflow.com/questions/9081/grep-a-file-but-show-several-surrounding-lines) — Organic Advocate, Jan 30 '20 at 23:09

score 304 · Accepted Answer · answered Sep 16 '12 at 06:16

304

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 Sep 16 '12 at 06:16

Jon Lin

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1

Thanks this is working. But when I tried to store this execution in the variable like this `test=$(grep -i -B 10 'error' data)`, and print it using `echo $test`, I get the straight long lines as output. – sriram Sep 16 '12 at 09:35
1

Thanks I figured out I need to do like this `echo "$test"` rather than `echo $test` – sriram Sep 16 '12 at 09:37
38

`-C 10` will print out 10 lines before AND after in one fell swoop! – Joshua Pinter Jan 22 '18 at 15:19
is there a way to do this using a specific before point? say the length i have to grab prior is variable? – Erudaki Jul 17 '18 at 19:05

Charlotte Russell · Answer 2 · 2017-07-12T08:19:03.400

This prints 10 lines of trailing context after matching lines

grep -i "my_regex" -A 10

If you need to print 10 lines of leading context before matching lines,

grep -i "my_regex" -B 10

And if you need to print 10 lines of leading and trailing output context.

grep -i "my_regex" -C 10

Example

user@box:~$ cat out 
line 1
line 2
line 3
line 4
line 5 my_regex
line 6
line 7
line 8
line 9
user@box:~$

Normal grep

user@box:~$ grep my_regex out 
line 5 my_regex
user@box:~$

Grep exact matching lines and 2 lines after

user@box:~$ grep -A 2 my_regex out   
line 5 my_regex
line 6
line 7
user@box:~$

Grep exact matching lines and 2 lines before

user@box:~$ grep -B 2 my_regex out  
line 3
line 4
line 5 my_regex
user@box:~$

Grep exact matching lines and 2 lines before and after

user@box:~$ grep -C 2 my_regex out  
line 3
line 4
line 5 my_regex
line 6
line 7
user@box:~$

Reference: manpage grep

-A num
--after-context=num

    Print num lines of trailing context after matching lines.
-B num
--before-context=num

    Print num lines of leading context before matching lines.
-C num
-num
--context=num

    Print num lines of leading and trailing output context.

Nice, ive had to look this up a few times now, maybe I can remember it as -A(FTER) -B(EFORE) -C(ONTEXT) — Opentuned, Aug 30 '18 at 13:47

score 11 · Answer 3 · answered Sep 16 '12 at 06:17

11

The way to do this is near the top of the man page

grep -i -A 10 'error data'

answered Sep 16 '12 at 06:17

Ray Toal

79,229
13
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score 8 · Answer 4 · answered Sep 16 '12 at 06:17

8

Try this:

grep -i -A 10 "my_regex"

-A 10 means, print ten lines after match to "my_regex"

answered Sep 16 '12 at 06:17

Desislav Kamenov

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

4 Answers4