Add # (pound) sign at begining of the line matching.(search string : file path)

Question

file:

sh  xyz/abc/www/def/gef/rqs/
sh xyz/toh/aaa/def/gef/rqs/
sh xyz/123/qqq/def/gef/rqs/

I want changes like this

sh  xyz/abc/www/def/gef/rqs/
#sh xyz/toh/aaa/def/gef/rqs/
sh xyz/123/qqq/def/gef/rqs/

edited: file paths to match my problem statement.

Based on what criteria, exactly?
– steeldriver
May 1, 2019 at 18:02 — steeldriver, May 1, 2019 at 18:02
The line with xyz/toh/aaa should be commented
– Tohid Tamboli
May 1, 2019 at 18:04 — Tohid Tamboli, May 1, 2019 at 18:04

steeldriver · Accepted Answer · 2019-05-01 18:08:32Z

1

You can use

sed -i '\%xyz/toh/aaa% s/^/#/' file

Explanation:

\% set the regex delimiter to % (so as not to conflict with the path separator /)
match xyz/toh/aaa
replace the start of line ^ by `#

answered May 1, 2019 at 18:08

steeldriver

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Thanks you made my day, also i want to add, sed -i '\%'$file_path/$var'% s/^/#/' file if you need to use variables in it.
– Tohid Tamboli
May 1, 2019 at 18:13

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score 0 · Accepted Answer · 2019-05-03 11:18:00Z

0

And why not using vim?

vi -c 'g/xyz\/toh\/aaa/s/^/#/ | x' file

or

vi -c '%s!.* xyz/toh/aaa!#&! | x' file

answered May 3, 2019 at 11:04

user216043

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2 Answers 2