URL with operators (=, -) not working within if-statement

Question

I'm having an issue with an URL in an if-statement using batch.

@ECHO OFF

SET /P input="Insert Link "

if %input%==l (echo true) else (echo false)

cmd /k

I want to determine if the user input is a link or a single character (in this case l). I get that the operators within the URL might cause the problem. But shouldn't the if-statement just check if %input% is l and everything else would trigger the else? Using SET %input%=l leads to the true case being triggered.

Any ideas on how to make this work in a simple way? Am I missing something regarding syntax?

If you "want to determine if the user input is"... you need to use the correct set option: Set /P "input=PromptString". The correct If command would be If "%input%" == "l" (Echo true) Else Echo false or If /I "%input%" == "l" (Echo true) Else Echo false (for L or l). — Compo
– Compo, Commented Oct 18, 2022 at 15:18
Of course. I was using SET input=https://www.youtube.com/watch?v=6zswl5YrvVw for testing purposes. I'm going to edit the OP so my question and the given code are matching. — fassimcgee
– fassimcgee, Commented Oct 19, 2022 at 8:21

fassimcgee · Accepted Answer · 2022-10-19 11:39:18Z

0

Putting %input% and l within the if-statement in quotes solved the problem.

@ECHO OFF

SET /P input="Insert Link "

if "%input%"=="l" (echo true) else echo false

cmd /k

answered Oct 19, 2022 at 11:39

fassimcgee

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SyntaxError · Accepted Answer · 2022-10-18 14:06:10Z

-1

The way the Command Line works, %ThisIsAVariable% will be replaced with whatever ThisIsAVariable contains and then be interpreted as such. Hence, running your example prompts the following error:

=6zswl5YrvVw==l was unexpected at this time.

The simplest way to solve this is to wrap your %input% with ""

e.g.

@ECHO OFF

SET input=https://www.youtube.com/watch?v=6zswl5YrvVw

if "%input%"==l (echo true) else (echo false)

cmd /k

That would prompt false

answered Oct 18, 2022 at 14:06

SyntaxError

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3 Comments

SomethingDark Over a year ago

You'll want to put quotes around the l since quotes themselves are included in the comparison. You'll also want to wrap input=https://www.youtube.com/watch?v=6zswl5YrvVw in quotes (so set "input=https://www.youtube.com/watch?v=6zswl5YrvVw").

fassimcgee Over a year ago

The quotes did the trick. if "%input%"=="l" (echo true) else (echo false) it is. Thank you for your help and the explanation!

Compo Over a year ago

Please do not mark an answer as a solution to a problem when the answer does not solve that problem @fassimcgee. The above is so very wrong, and by accepting it you're potentially sending future readers down the wrong path. In the above code, only line 1 is acceptable.

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