Does PHP ignore undefined variables

[mrgrammar]

I'm passing a variable from one page to the next via a form post. On the next page, the code checks to see if that variable equals a preset value. If not, the code doesn't execute.

$jehosophat = $_POST['variablepassed'];
if ($blahblah == "jehosophat")
{
print "execute code";
}

What about when $blahblah isn't defined such as by a person who accesses the page without coming through the form? The code won't execute, but I also don't want the page to return an error and I want to send the user somewhere else.

In Code Fusion, I protected against this by using:

if (isDefined($blahblah))
code to execute
end if

Using PHP on our Linux server, I tried using

$jehosophat = $_POST['variablepassed'];
if (defined($jehosophat))
{
if ($blahblah == "jehosophat"){print "execute code";}
else{other code;}
}
else
{
print "Get Out!";
}

but it never qualified $jehosophat as defined even when it was. What am I doing wrong and is there a better code to use?

Brian

[bad76]

Hi,
you get a look at isset() and empty() function.

I think is that you are searching for...

(And do the test for $POST['variablepassed']; too...
if(isset($POST['variablepassed']))$jehosophat=$_POST['variablepassed'];
if(isset($jehosophat))
{
...
)

[tbone]

Try this

$jehosophat = $_POST['variablepassed'];

if(!$jehosophat){
print "Get Out!";
}else{

if ($blahblah == "jehosophat"){
print "execute code";
}else{
other code;
}

}

[laserlight]

um tbone, the problem with that is that if notices are reported, and $_POST['variablepassed'] is not defined, a notice will be reported.