If statement variable assignment in batch file ignored

Question

I am new to batch file coding and am trying to write a simple If-Else statement.

I have tried to use delayedexpansion as suggested in other questions but still don't seem to get the right value

@echo off
setlocal ENABLEDELAYEDEXPANSION

SET VAR=portrait
IF %VAR% == portrait do (
    SET /a height=1920;
    set /a width=1080;
) else do(
    set /a height=1080;
    set /a width=1920;
)

echo %height%

The code above gives the output as 1080, while I am expecting 1920 based on the logic I wrote. Can someone please help me understand?

Your example is not working. If you'll fix all syntactic errors it will work as expected. — montonero
– montonero, Commented May 30, 2019 at 7:14
I didn’t get any syntax error when I ran this. I do get an output. Are you getting an error? — s_om
– s_om, Commented May 30, 2019 at 7:15
Have you tried this exact example? Mind do and semi-colons. — montonero
– montonero, Commented May 30, 2019 at 7:18

Compo · Accepted Answer · 2019-05-30 09:48:37Z

You issue is mainly because there's no such thing as an If / Do or Else / Do statement. Also you should try to remember that Set /A is really for arithmetic, and you're not doing math, just setting values to names.

If "%VAR%" == "portrait" (
    Set "height=1920"
    Set "width=1080"
) Else (
    Set "height=1080"
    Set "width=1920"
)
Echo %height%

However, you can in this case, still use Set /A to shorten your commands a little; because Set /A allows for multiple names to integer values within the same statement.

If "%VAR%" == "portrait" (
   Set /A height=1920, width=1080
) Else (
   Set /A height=1080, width=1920
)
Echo %height%

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