We can replace strings in a batch file using the following command
set str=\"jump over the chair\"
set str=%str:chair=table%
These lines wo
This works fine
@echo off
set word=table
set str=jump over the chair
set rpl=%str:chair=%%word%
echo %rpl%
You can use !, but you must have the ENABLEDELAYEDEXPANSION switch set.
setlocal ENABLEDELAYEDEXPANSION
set word=table
set str="jump over the chair"
set str=%str:chair=!word!%
You can use the following little trick:
set word=table
set str="jump over the chair"
call set str=%%str:chair=%word%%%
echo %str%
The call
there causes another layer of variable expansion, making it necessary to quote the original %
signs but it all works out in the end.
I was able to use Joey's Answer to create a function:
Use it as:
@echo off
SETLOCAL ENABLEDELAYEDEXPANSION
SET "MYTEXT=jump over the chair"
echo !MYTEXT!
call:ReplaceText "!MYTEXT!" chair table RESULT
echo !RESULT!
GOTO:EOF
And these Functions to the bottom of your Batch File.
:FUNCTIONS
@REM FUNCTIONS AREA
GOTO:EOF
EXIT /B
:ReplaceText
::Replace Text In String
::USE:
:: CALL:ReplaceText "!OrginalText!" OldWordToReplace NewWordToUse Result
::Example
::SET "MYTEXT=jump over the chair"
:: echo !MYTEXT!
:: call:ReplaceText "!MYTEXT!" chair table RESULT
:: echo !RESULT!
::
:: Remember to use the "! on the input text, but NOT on the Output text.
:: The Following is Wrong: "!MYTEXT!" !chair! !table! !RESULT!
:: ^^Because it has a ! around the chair table and RESULT
:: Remember to add quotes "" around the MYTEXT Variable when calling.
:: If you don't add quotes, it won't treat it as a single string
::
set "OrginalText=%~1"
set "OldWord=%~2"
set "NewWord=%~3"
call set OrginalText=%%OrginalText:!OldWord!=!NewWord!%%
SET %4=!OrginalText!
GOTO:EOF
And remember you MUST add "SETLOCAL ENABLEDELAYEDEXPANSION" to the top of your batch file or else none of this will work properly.
SETLOCAL ENABLEDELAYEDEXPANSION
@REM # Remember to add this to the top of your batch file.