Strange IF statement problem

[CyberFuture]

Something is not working right here.

echo "TESTER = $tester <br>";
    if ($tester == "mod")
     {
     echo "SECOND TEST :: TESTER = $tester <br>";
} 

The output is:

TESTER = 1
SECOND TEST :: TESTER = 1

As the output clearly doesn't equal "mod", so why is the secont test being echoed????!!!

Any ideas?

Thanks

[Huscy]

strange one indeed, try doing

if ($tester == "mod" )
{
echo "$tester";
} 
else
{
echo "$tester";
} 

just out of curiosity

before i post, is "mod" not some PHP reserved word?
(not sure, would be in pascal, just a thought)

[Jeff Lange]

mod is in a string, so it doesn't matter anyways. (But no I don't think it is reserved.)

anyways, when you are echoing a variable, lose the double quotes.

[CyberFuture]

I changed "mod" to "modd" and here's the new code:

echo "TESTER = $tester <br>";
if ($tester == "modd")
     {
     echo "SECOND TEST :: TESTER = $tester <br>";
      }
else
     { 
     echo "THIRD TEST :: TESTER = $tester <br>"; 
     } 

The output is:
TESTER = 1
SECOND TEST :: TESTER = 1

I'm scratching my head on this one

[Jeff Lange]

try

if ($tester==='modd') 

[CyberFuture]

when you are echoing a variable, lose the double quotes.

Why?? I was taught NEVER use single quotes on variables, because PHP treats variables in single quotes as literals. But not with double quotes.

[CyberFuture]

Thanks cyborg that did the trick.

[anode]

Why?? I was taught NEVER use single quotes on variables, because PHP treats variables in single quotes as literals. But not with double quotes.

He was referring to

echo "$tester"; 

where there is no need for any kind of quotes.

[Jeff Lange]

echo $tester;