How to check command line parameter in “.bat” file?

Question

My OS is Windows Vista. I need to have a \".bat\" file where I need to check if user enters any command-line parameter or not. If does then if the parameter equals to

Answer 1

Actually, all the other answers have flaws. The most reliable way is:

IF "%~1"=="-b" (GOTO SPECIFIC) ELSE (GOTO UNKNOWN)

Detailed Explanation:

Using "%1"=="-b" will flat out crash if passing argument with spaces and quotes. This is the least reliable method.

IF "%1"=="-b" (GOTO SPECIFIC) ELSE (GOTO UNKNOWN)

C:\> run.bat "a b"

b""=="-b" was unexpected at this time.

Using [%1]==[-b] is better because it will not crash with spaces and quotes, but it will not match if the argument is surrounded by quotes.

IF [%1]==[-b] (GOTO SPECIFIC) ELSE (GOTO UNKNOWN)

C:\> run.bat "-b"

(does not match, and jumps to UNKNOWN instead of SPECIFIC)

Using "%~1"=="-b" is the most reliable. %~1 will strip off surrounding quotes if they exist. So it works with and without quotes, and also with no args.

IF "%~1"=="-b" (GOTO SPECIFIC) ELSE (GOTO UNKNOWN)

C:\> run.bat
C:\> run.bat -b
C:\> run.bat "-b"
C:\> run.bat "a b"

(all of the above tests work correctly)