Nested If Statement with Variables declared outside

Question

I have this block of code which is throwing error : "The syntax is incorrect".

SET "VAR1=UP"
SET "VAR2=UP"

IF ("%VAR1%"=="UP")
(
  IF "(%VAR2%"=="UP")
  ( ECHO "VALUES MATCH")
 )

But this keeps throwing error as mentioned above. Is the comparison syntax ( %VAR1%"=="UP") incorrect or there is any issue with Variable declaration?

The function of the code will be: There will be four such If blocks, each will compare the four possible values of the two variables and throw specific display message .

Would really be obliged if anyone can help me in this.

That is not the correct syntax for the IF command. Open up a cmd prompt and type IF /? to read the help file for the IF command. — Squashman
– Squashman, Commented Aug 28, 2017 at 14:31

Magoo · Accepted Answer · 2017-08-28 14:34:44Z

2

IF ("%VAR1%"=="UP")

The string IF ("%VAR1%" will never equal the string "UP"), regardless of the value of var1 - these parenthesis for part of the comparison, you can't group them as a "Boolean".

IF "%VAR1%"=="UP"

is the correct comparison syntax, BUT

Which starts the grouping of the statements to cascade if the comparison is evaluated as true absolutely MUST be on the same physical line as the if.

The same goes for else - the syntax ) else ( must appear just exactly so.

answered Aug 28, 2017 at 14:34

Magoo

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1 Comment

Pratik Datta Over a year ago

This has worked.Thanks a lot for your help.Much obliged.

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