function problem - Dev Shed

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November 3rd, 2003, 02:52 PM

wannabe

=) wannabe?

Join Date: Jul 2002

Location: florida

Posts: 2,153

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function problem

this just prints Array ( ). i dont know why.

PHP Code:


		
			
function build_pages($table,$array)



{







    $$array = Array();







    $getcats = mysql_query("SELECT ID,name FROM $table");







    $mi = 0;



    while ($cat = mysql_fetch_array($getcats))



    {



        $$array["$mi"][0] = $cat['ID'];



        $$array["$mi"][1] = $cat['name'];



    



        $mi++;



    }



    return $$array;







}











$ar = build_pages('cats','cats');



print_r($ar);

i'm not really looking for an alternative. just wondering what it's not working.

thank you

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November 3rd, 2003, 03:02 PM

cranium

Contributing User

Join Date: Jun 2002

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Posts: 1,029

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Hi,

You are using a variable variable ($$array), therefore when you return $$array, you are only returning the reference of the stuff that it is pointing to, which an array.

You would need to make it $array and do some array manipulation to get the correct output.

I hope that helps,
cranium

November 3rd, 2003, 03:08 PM

Kaleina

The Linux-Girl

Join Date: Dec 2001

Posts: 396

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Re: function problem

Quote:

Originally posted by wannabe
this just prints Array ( ). i dont know why.

i'm not really looking for an alternative. just wondering what it's not working.

thank you

I'm not sure what you need the $array for in your function.. Remove that and the assignment when calling the function Also remove the $$ for array... what's that for? $array=array() will work..

I ran the script on a similar table on my database

print_r($ar) returns the following...

Array ( [0] => Array ( [0] => 1 [1] => Superior Room ) [1] => Array ( [0] => 2 [1] => Deluxe Room ) [2] => Array ( [0] => 3 [1] => Superior City View Studio w/Kitchenette ) [3] => Array ( [0] => 4 [1] => Superior Partial Ocean View Studio w/Kitchenette )

etc...