Hmm, interesting.. i dunno why it doesn't work for me, here is a bit of the
code, $var isn't getting the output of $id, any ideas?
$var = &$id;
if( ! $forg = resizer_main("image","image_$var", blah, blah........");
$org = getimagesize( "$root/$forg" );
$result = @mysql_query("INSERT INTO images blah blah.....");
$id = mysql_insert_id();
cheers,
Sebastian
----- Original Message -----
From: "Marcus Rasmussen" <[EMAIL PROTECTED]>
Putting an & sign in front of the $id in the first line should do the trick:
$variable = &$id;
A short example:
$bar = 0;
$foo = &$bar;
$bar = 2;
print $foo; //prints 2
______________________
Marcus Rasmussen
[EMAIL PROTECTED]
www.marcusr.dk
-------------------------------------------------------------
On 31-03-2003 at 19:56 Sebastian wrote:
-------------------------------------------------------------
hello all,
$variable = $id;
// some other stuff....
@mysql_query here....
$id = mysql_insert_id();
to the question: How do I get $id from insert_id() to pass to $variable
above? Hard to explain the situation i am in, but the query has to be below
$variable, is it possible to 'globalize' $id so it passes to the top
$variable?
first time i ever ran into this problem.
Thanks in advanced.
cheers,
- Sebastian
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