[SOLVED] Difference between isset and != NULL

[SchweppesAle]

was hoping someone could explain the difference.  Does isset just require fewer operations?

 

edit:

also, I just found this empty() function.

 

empty — Tests whether a variable is defined and contains a non-empty and nonzero value.

 

Again, what's the difference 

[Daniel0]

The language construct isset() checks if a variable is set while a null comparison checks if some sort of value equals the special value "null".

 

empty() checks if the value of something can be considered empty (the manual tells what values those would be).

[SchweppesAle]

The language construct isset() checks if a variable is set while a null comparison checks if some sort of value equals the special value "null".

 

right, but what exactly does "set" mean.  I understand NULL

[Daniel0]

It means that it has been defined somewhere. For instance, if the following is a complete script, the variable $foo is not set, but the variable $var is:

<?php
$bar = 'foo';
var_dump(isset($foo), isset($bar));

[Alex]

The language construct isset() checks if a variable is set while a null comparison checks if some sort of value equals the special value "null".

 

right, but what exactly does "set" mean.  I understand NULL

Has a value other than NULL.

 

Ex:

$var = null; // is NOT set
while..
$var = ''; // is set

[Daniel0]

Right, but there is the important distinction that isset() works on undefined variables.

[SchweppesAle]

It means that it has been defined somewhere. For instance, if the following is a complete script, the variable $foo is not set, but the variable $var is:

<?php
$bar = 'foo';
var_dump(isset($foo), isset($bar));

 

right, so I'm assuming isset($foo) will return false and isset($bar) will return true.

 

in a conditional statement I would say

 

if(isset($bar))
{
echo var_dump($bar);
}

 

how is that any different from the following though?

if($bar != NULL)
{
echo var_dump($bar);
}

 

[Mchl]

In toher words checking if $var != null will throw E_NOTICE error while isset($var) will not (when $var is not set)

[SchweppesAle]

Right, but there is the important distinction that isset() works on undefined variables.

 

I guess that's true if I hadn't declared $bar at all. 

 

Still, I guess it comes down to performance if that's the case.  Should I avoid "!= NULL" condition statements and stick with isset() for now on?

[SchweppesAle]

Man i love google.  Thanks again, I think I understand now.

http://www.danielnorton.com/nerd/code/php/isdef

 

I was assuming that undefined == NULL.

 

 

[gizmola]

Right, but there is the important distinction that isset() works on undefined variables.

 

I guess that's true if I hadn't declared $bar at all. 

 

Still, I guess it comes down to performance if that's the case.  Should I avoid "!= NULL" condition statements and stick with isset() for now on?

 

Again, there are 2 different concepts:

 

"isset" -> does this symbol exist in the PHP symbol table. 

"NULL" -> does a matching symbol exist, however there is no associated value.