How to substitute variable contents in a Windows batch file

Question

I\'m writing a simple script to substitute text in an environment variable with other text. The trouble I get is with the substituted or substituted text being pulled from o

Answer 1

Recently I came accross the same situation..As said earlier, I used like below and worked...

set filearg=C:\data\dev\log\process
set env=C:\data\dev

REM I wanted \log\process as output

SETLOCAL enabledelayedexpansion
set str2=!filearg:%env%=!
echo Output : %str2%
echo.
endlocal

Output :

\log\process

It worked..!!

Answer 2

And... How about this?

@echo off
setlocal enabledelayedexpansion

set str=The fat cat
set f=fat
set t=thin

echo.
echo       f = [%f%]
echo       t = [%t%]

echo.
echo     str = [%str%]

set str=!str:%f%=%t%!

echo str:f=t = [%str%]

Nifty eh?

Answer 3

Use CALL. Put the following in a batch script and run it:

set a=The fat cat
set b=fat
set c=thin

REM To replace "%b%" with "%c%" in "%a%", we can do:
call set a=%%a:%b%^=%c%%%
echo %a%
pause

As stated here, we use the fact that:

CALL internal_cmd

...

internal_cmd Run an internal command, first expanding any variables in the argument.

In our case internal_cmd is initially set a=%%a:%b%^=%c%%%.

After expansion internal_cmd becomes set a=%a:fat=thin%.

Thus, in our case running

call set a=%%a:%b%^=%c%%%

is equivalent to running:

set a=%a:fat=thin%.

Answer 4

Please try the following:

Copy and paste the code into Notepad and save it as a batch file.

   @echo off
   setlocal enabledelayedexpansion

   set str=The fat cat
   set f=fat

   echo.
   echo          f = [%f%]

   echo.
   echo        str = [%str%]

   set str=!str:%f%=thin!

   echo str:f=thin = [%str%]

I hope you're convinced!

Answer 5

The problem with:

echo %a:%c%=thin%

is that it tries to expand two variables: a: and =thin with a c constant string between them.

Try this:

echo echo ^%a:%c%=thin^% | cmd

The first command outputs:

echo %a:fat=thin%

which is piped into a second command shell for evaluation.

Answer 6

:: Use perl on %var% =~ s/$old/$new/

set var=The fat cat
set old=fat
set new=thin

ECHO before=%var%

for /f "delims=" %%m in ('echo %var% ^|perl -pe "s/%old%/%new%/" ') do set var=%%m

ECHO after=%var% 

Output:

before=The fat cat
after=The thin cat