This is probably easier that it seems...

[buzzt]

All I want to do is pass ALL the information from the array to the if statement without specifying {0],[1]... How do I do this?

$page = $_SERVER['PHP_SELF'];
$page = "

http://www.ANYSITE.com"

. $page;
$except = array('a.php', 'b.php');

if ($page != $except[0])
{
Whatever code
}
else
{
Other code
}

 

[sleipnir214]

I don't understand.

Since and if-block is not a function, it uses the same variable scope as the surrounding code. You don't have to pass it anything as all variables are available.

If you want to base the logical clause of the if-statement on an entire array, it's possible to do so, but you'll have to use meaningful PHP statements to operate on the array.



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[AnakinPt]

Looking for a dummy code i thing you are trying to know the name of the script to see if it is in the set defined by the array.

That will never work

If you are tring to do that, try this piece of code instead:

if (in_array(substr($_SERVER["SCRIPT_NAME"],1),$array))){
It's in the array
else
not in the array

Anikin
Hugo Alexandre Dias
Web-Programmer
anikin@anikin-skywalker.com