If condition in batch files

Question

@echo off SET var1="Yes" SET var2="No" SET var3="Yes" if %var1%=="Yes"     echo Var1 set if %var2%=="Yes"     echo Var2 set if %var3%=="Yes"     echo Var3 set 

If I run the above script I get the following error. Can anyone pls help?

The syntax of the command is incorrect.

Thanks.

Answer 1

The echo needs to either be at the end of the if statement:

if %var1%=="Yes" echo Var1 set 

or of the following form:

if %var1%=="Yes" (     echo Var1 set ) 

I tend to use the former for very simple conditionals and the latter for multi-command ones and primitive while statements:

:while1     if %var1%=="Yes" (         :: Do something that potentially changes var1         goto :while1     ) 

What your particular piece of code is doing is trying to execute the command if %var1%=="Yes" which is not valid in and of itself.

Answer 2

You can't put a newline like that in the middle of the IF. So you could do this:

if %var1%=="Yes" echo Var1 set 

Or, if you do want your statements spread over multiple lines you can use brackets:

if %var1%=="Yes" (    echo Var1 set ) 

However, when you're using brackets be careful, because variable expansion might not behave as you expect. For example:

set myvar=orange  if 1==1 (    set myvar=apple    echo %myvar% ) 

Outputs:

orange 

This is because everything between the brackets is treated as a single statement and all variables are expanded before any of the command between the brackets are run. You can work around this using delayed expansion:

setlocal enabledelayedexpansion set myvar=orange  if 1==1 (    set myvar=apple    echo !myvar! )