Find out whether an environment variable contains a substring

Question

I need to find out if a certain environment variable (let\'s say Foo) contains a substring (let\'s say BAR) in a windows batch file. Is there any way to do this using only

Answer 1

I wrote this function for a nice script integration. Code looks better and it's also easier to remember. This function is based on Joey's answer on this page. I know it's not the fastest code but it's seems to work very well for what I need to do.

Just copy the function's code at the end of your script and you can use it like in this example here:

Example:

set "Main_String=This is just a test"
set "Search_String= just "

call :FindString Main_String Search_String

if "%_FindString%" == "true" (
    echo String Found
) else (
    echo String Not Found
)

Note that you don't need to add % to your variables when giving them to this function, it will automatically take care of this. (This is a method that I found which it letting me use spaces in my function's arguments/variables without the need of using undesirables quotes in them.)

Function:

:FindString

rem Example:
rem 
rem set "Main_String=This is just a test"
rem set "Search_String= just "
rem 
rem call :FindString Main_String Search_String
rem 
rem if "%_FindString%" == "true" echo Found
rem if "%_FindString%" == "false" echo Not Found

SETLOCAL

for /f "delims=" %%A in ('echo %%%1%%') do set str1=%%A
for /f "delims=" %%A in ('echo %%%2%%') do set str2=%%A

echo.%str1%|findstr /C:"%str2%" >nul 2>&1
if not errorlevel 1 (
   set "_Result=true"
) else (
   set "_Result=false"
)

ENDLOCAL & SET _FindString=%_Result%
Goto :eof

Answer 2

Of course, just use good old findstr:

echo.%Foo%|findstr /C:"BAR" >nul 2>&1 && echo Found || echo Not found.

Instead of echoing you can also branch elsewhere there, but I think if you need multiple statements based on that the following is easier:

echo.%Foo%|findstr /C:"BAR" >nul 2>&1
if not errorlevel 1 (
   echo Found
) else (
    echo Not found.
)

Edit: Take note of jeb's solution as well which is more succinct, although it needs an additional mental step to figure out what it does when reading.

Answer 3

The findstr solution works, it's a little bit slow and in my opinion with findstr you break a butterfly on a wheel.

A simple string replace should also work

if "%foo%"=="%foo:bar=%" (
    echo Not Found
) ELSE (
    echo found
)

Or with inverse logic

if NOT "%foo%"=="%foo:bar=%" echo FOUND

If both sides of the comparision are not equal, then there must be the text inside the variable, so the search text is removed.

A small sample how the line will be expanded

set foo=John goes to the bar.
if NOT "John goes to the bar."=="John goes to the ." echo FOUND

Answer 4

@mythofechelon: The %var:str=% part removes str from var. So if var contains str on the left side of the equation, it will be removed on the right side - thus the equation will result in "false" if str was found in var or "true" if str was not present in var.