ETA:
Actually i want to check if the architecture contains string 64. Something like:
if %ARCH% == '*64*' ( echo ok )
How can i do that?
For this you can employ the help of a nice little tool called findstr
which searches for patterns in files or command output. Just pipe the value of the variable into findstr
:
echo %ARCH% | findstr 64
which will find any line in the output that contains 64 somewhere. In this case, there is only one line but that's fine.
However, this also causes the line to be output on screen, so change it into the following:
echo %ARCH% | findstr 64 > nul 2>&1
which will redirect any output (including error messages) into nothingness.
Now, findstr
sets a special value (the exit code, or error level) depending on whether the string was found or not. You can use that to test whether 64
in fact appeared in the output:
echo %ARCH% | findstr 64 > nul 2>&1
if errorlevel 1 (
echo 64 did NOT appear (error level greater 0 usually means failure)
) else (
echo 64 DID appear
)
To your original question:
The opening parenthesis must occur on the same line as the if
statement itself:
@echo off
set ARCH=%PROCESSOR_ARCHITECTURE%
if %ARCH% == x86 (
rem ...
)
does work.
Remember that cmd
's parser is line-based; therefore caution is advised with anything that needs to span multiple lines. Another option would be to escape the line break:
@echo off
set ARCH=%PROCESSOR_ARCHITECTURE%
if %ARCH% == x86 ^
(
rem ...
)
The ^
is the escape character in cmd
and in this use it's effectively a line-continuation operator so the parser accepts the opening parenthesis on the next line as part of the current statement (which usually would be terminated by a line break).
And yes, the use of single quotes for the literal string value in the comparison will get you into other problems as Carlos Gutiérrez noted, since essentially everything in cmd
is a string literal. So you would be comparing with the literal string
'x86'
instead of
x86
what is actually contained in the environment variable.