Specify it only at function definition:
myFunction( &$var1, &$var2 ) {
// ...
}
$var = myFunction( $var1, $var2 ) ;
Ignatius
_
- Original Message -
From: "Jeremy Shovan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, October 09, 2003 9:54 PM
Subjec
> How do you pass by reference? I tried
> $var = myFunction(&$var1, &$var2);
> but it gave me a warning saying that pass by reference is depreciated
>
Call time pass by refrence is depreciated you should read the manual to see
how it should be done.
http://www.php.net/manual/en/language.references
> How do you pass by reference? I tried
> $var = myFunction(&$var1, &$var2);
> but it gave me a warning saying that pass by reference is depreciated
function myFunction( &$var1, &$var2 ) {
}
You now have to do it in the function definition, not the function call.
Chris
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