89

How do we compare strings which got space and special chars in batch file?

I am trying:

if %DevEnvDir% == "C:\Program Files (x86)\Microsoft Visual Studio 10.0\Common7\IDE\"(
echo VS2010
)

But it gives an error "Files was unexpected at this time."

I tried:

if "%DevEnvDir%" == "C:\Program Files (x86)\Microsoft Visual Studio 10.0\Common7\IDE\"(
echo VS2010
)

But it gives an error "The syntax of the command is incorrect."

Any ideas?

dushyantp
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  • Judging by the linked answer, you could do `if [%DevEnvDir%] == ["C:\..."]`. Link: https://stackoverflow.com/a/4953226/2428861 – Mladen B. Aug 07 '19 at 09:11

3 Answers3

108

Just put quotes around the Environment variable (as you have done) :
if "%DevEnvDir%" == "C:\Program Files (x86)\Microsoft Visual Studio 10.0\Common7\IDE\"
but it's the way you put opening bracket without a space that is confusing it.

Works for me... 

C:\if "%gtk_basepath%" == "C:\Program Files\GtkSharp\2.12\" (echo yes)
yes
AjV Jsy
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    thanks for the hint!! The mistake I did was that, there was no space between by "string-value" & (. Now works fine! thanks. – dushyantp Feb 19 '13 at 10:59
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    This seems to fail if the variable already contains quotes. For example: if "%1"=="" goto Help fails if the first argument to the batch file already contains double quotes around it. – MarioVilas Jun 06 '13 at 13:54
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    So what the usage of not equal? – terwxqian Jan 20 '21 at 03:05
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    if not "example" == "example" See https://docs.microsoft.com/en-us/windows-server/administration/windows-commands/if – AjV Jsy Jan 20 '21 at 08:56
  • "bracket" - do you mean parentheses? This is a bracket: `[` – johny why Nov 23 '21 at 23:07
43

While @ajv-jsy's answer works most of the time, I had the same problem as @MarioVilas. If one of the strings to be compared contains a double quote ("), the variable expansion throws an error.

Example:

@echo off
SetLocal

set Lhs="
set Rhs="

if "%Lhs%" == "%Rhs%" echo Equal

Error:

echo was unexpected at this time.

Solution:

Enable delayed expansion and use ! instead of %.

@echo off
SetLocal EnableDelayedExpansion

set Lhs="
set Rhs="

if !Lhs! == !Rhs! echo Equal

:: Surrounding with double quotes also works but appears (is?) unnecessary.
if "!Lhs!" == "!Rhs!" echo Equal

I have not been able to break it so far using this technique. It works with empty strings and all the symbols I threw at it.

Test:

@echo off
SetLocal EnableDelayedExpansion

:: Test empty string
set Lhs=
set Rhs=
echo Lhs: !Lhs! & echo Rhs: !Rhs!
if !Lhs! == !Rhs! (echo Equal) else (echo Not Equal)
echo.

:: Test symbols
set Lhs= \ / : * ? " ' < > | %% ^^ ` ~ @ # $ [ ] & ( ) + - _ =
set Rhs= \ / : * ? " ' < > | %% ^^ ` ~ @ # $ [ ] & ( ) + - _ =
echo Lhs: !Lhs! & echo Rhs: !Rhs!
if !Lhs! == !Rhs! (echo Equal) else (echo Not Equal)
echo.
Community
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skataben
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  • try this >>> @echo off setlocal EnableDelayedExpansion set Lhs="hello" set Rhs="hello" if !Lhs! == "hello" (echo 1Equal) :: Surrounding with double quotes also works but appears (is?) unnecessary. if "!Lhs!" == "hello" echo Equal timeout 10 endlocal pause – Subham Tripathi Dec 09 '14 at 06:43
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    You can use `if /I` if you wish to do case insensitive string comparison – Carlos P Feb 11 '15 at 09:54
-5

The solution is DO NOT USE SPACES!

IF "%DevEnvDir%"=="C:\" (
Kiquenet
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Bernardo Ravazzoni
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    This is wrong. You can use spaces, as long as you quote them. The problem was the *missing* space at the end `if "%env%"=="xyz"(` – jeb Nov 14 '16 at 08:54