Replace text in files by variable values using Bash

Question

I have php files with the flag like /* TEST */.

In the script, I have variables with this flag and the value to replace:

test="defined('_TEST') or die('Test test');"
testPattern="/* TEST */"

My code returns no errors, but flags aren't replaced:

find . -name "*.php" -exec sed -i -e 's/"$testPattern"/"$test"/g' {} \;

Poshi · Accepted Answer · 2019-02-27 09:45:35Z

2

Your sed string is in single quotes. Variable expansion does not occur there. Switch to double quotes:

find . -name "*.php" -exec sed -i -e "s/$testPattern/$test/g" {} \;

answered Feb 27, 2019 at 9:45

Poshi

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2 Comments

Irina Over a year ago

Thank you for the answer. In this case I get errors: sed: 1: "s//* TEST *//defined('_J ...": bad flag in substitute command: '/'

Poshi Over a year ago

You have to adjust the pattern to be properly escaped, otherwise, sed is interpreting it according its rules. In particular, you should escape the characters with special meaning for sed, the slash and the start of $testPattern.

gromi08 · Accepted Answer · 2019-02-27 10:03:11Z

0

Enclosing characters in single quotes preserves the literal value of each character within the quotes.

Try this one:

sed -i '' "s~$testPattern~$test~g" $(find . -name '*.php')

edited Feb 27, 2019 at 10:03

answered Feb 27, 2019 at 9:55

gromi08

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3 Comments

Irina Over a year ago

Thanks for the answer. Unfortunately it also doesn't replace.

gromi08 Over a year ago

Can you check the latest edit? on MacOS the -i flag requires a specified extension following the flag, in this case ''. (If a zero-length extension is given, no backup will be saved)

Irina Over a year ago

Yes. I've checked this the last edit. The same result.

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