Replace a substring in batch script

Question

I have three variables: string, search, replace. I wish to substitute the %search% in %string% with %replace%.

This works but needs hard characters.

SET modified=%string:morning=evening%

This seems to be answer in the forums but does not work. It simply stores the entire line at %modified%

SET modified=!string:%search%=%replace%!

That syntax only works if you use delayed variable expansion. Use setlocal enabledelayedexpansion at the top of your script, or call your script using cmd.exe /V:ON. — Bill_Stewart
– Bill_Stewart, Commented Mar 11, 2013 at 2:57

niCk cAMel · Accepted Answer · 2018-10-05 13:06:44Z

1

The ! format is doing delayed expansion--the % variables get expanded immediately, but the ! variable gets expanded only when it's needed. I believe that only works in a batch file, so if you're experimenting directly at the command line you won't get the same behaviour as if you are running a batch file.

Make sure to enable delayed expansion in your batch file before using the ! notation, like this:

SETLOCAL ENABLEDELAYEDEXPANSION
SET string=This morning
SET search=morning
SET replace=evening
SET modified=!string:%search%=%replace%!
ECHO %modified%
ENDLOCAL

This will echo This evening.

edited Oct 5, 2018 at 13:06

niCk cAMel

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answered Mar 11, 2013 at 3:37

Nate Hekman

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1 Comment

srrvnn Over a year ago

Thanks. Enabling delayed expansion is what I wasn't doing.

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