Using sed command to find and replace code

Question

I have a block of php code (inside <? php ... ?>) in multiple files which I want to remove.

I tried to put the code into a variable:

var="<?php [some code in here] ?>"

Then run this:

find . -type f -name '*.php' -exec sed "s|$var| |" {} \;

Or

find . -type f -name '*.php' -exec sed -i 's/'"$var"'.*/'"$str"'/g' {} \;

(where str=" ")

Is there a way to do this?

You got success and looking for another way(s)? or you couldn't get success with above ones? BTW both working just fine — αғsнιη
– αғsнιη, Commented May 1, 2018 at 5:15
Somehow none of the above works for me. :( None of them could replace the string. — Vu Hoang
– Vu Hoang, Commented May 1, 2018 at 5:21
Is this on one line, or multiple? Are there special characters in the php element (relative to bash; $, !, ", ')? — cunninghamp3
– cunninghamp3, Commented May 1, 2018 at 9:20
You'll also run in trouble if your php code contains / or . or [ or * or something else that breaks your sed script. We could better help you if you tell us the actual code to be removed. — Philippos
– Philippos, Commented May 3, 2018 at 12:00

agc · Accepted Answer · 2018-05-02 15:17:08Z

0

Try this, but remove the -i the first time to make sure it works:

var='echo rand(1, 10);'  # dummy code, change to whatever.
sed -i 's/<? php '"${var}"' ?>//' *.php

Notes:

answered May 1, 2018 at 5:43

agc

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