[SOLVED] Using sed to remove all lines between two patters if text is found between them

Issue

I have a file

###
A
B
C
@@@
###
D
E
F
@@@
###
G
H
I
@@@

I want to remove all lines between ### and @@@ (inclusive) if it contains A

How can I achieve this ?

expected result

###
D
E
F
@@@
###
G
H
I
@@@

I have tried

sed "/###/,/@@@/d"

but that is removing everything.

If we can have multiple text patterns like remove everything between ### and @@@ if it contains either 'A' or 'D' would be even helpful.

Thanks.

Solution

You can use

sed '/###/{:a;N;/@@@/!ba;/A/d}' file

Details:

• /###/ - looks for a ### and once it is found,
• :a - set a label at the current location
• N - reads the next line appending it to pattern space
• /@@@/!ba - if there is @@@ stop processing block, else, go to a label
• /A/d - if A is inside the block matched, delete it.

See an online test:

#!/bin/bash
text=$(echo -e '###\nA\nB\nC\n@@@\n###\nD\nE\nF\n@@@\n###\nG\nH\nI\n@@@')
sed '/###/{:a;N;/@@@/!ba;/A/d}' <<< "$text"

Output:

###
D
E
F
@@@
###
G
H
I
@@@

Answered By - Wiktor Stribiżew

Answer Checked By - Willingham (WPSolving Volunteer)