newbie: failing IF statement

[AlbertAguirre]

Why does this fail???

$setType="raise";
if ($setType=="lower"){
print "in if \n";
}else{
print "in else \n";
}

The output is consitantly:

in if

Why???

 

[AlbertAguirre]

OK. after hours of searching i found the "eq"

This works.

$setType="raise";
if ($setType eq "lower"){
print "in if \n";
}else{
print "in else \n";
}

in else


Sorry to trouble you!

 

[KevinADC]

hours of searching? Dang.

------------------------------------------
- Kevin, perl coder unexceptional!

 

[MillerH]

Heh.

Yes, next time go back to perldoc and look up the relevant source material:

perlop - Equality Operators

- Miller