Windows Batch Files: if else

Question

I\'m doing a simple batch file that requires one argument (you can provide more, but I ignore them).

For testing, this is what I have so far.

if not          


        

Answer 1

Another related tip is to use "%~1" instead of "%1". Type "CALL /?" at the command line in Windows to get more details.

Answer 2

if not %1 == "" (

must be

if not "%1" == "" (

If an argument isn't given, it's completely empty, not even "" (which represents an empty string in most programming languages). So we use the surrounding quotes to detect an empty argument.

Answer 3

An alternative would be to set a variable, and check whether it is defined:

SET ARG=%1
IF DEFINED ARG (echo "It is defined: %1") ELSE (echo "%%1 is not defined")

Unfortunately, using %1 directly with DEFINED doesn't work.

Answer 4

you have to do like this...

if not "A%1" == "A"

if the input argument %1 is null, your code will have problem.

Answer 5

You have to do the following:

if "%1" == "" (
    echo The variable is empty
) ELSE (
    echo The variable contains %1
)

Answer 6

Surround your %1 with something.

Eg:

if not "%1" == ""

Another one I've seen fairly often:

if not {%1} == {}

And so on...

The problem, as you can likely guess, is that the %1 is literally replaced with emptiness. It is not 'an empty string' it is actually a blank spot in your source file at that point.

Then after the replacement, the interpreter tries to parse the if statement and gets confused.