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
- Re: [PHP-DB] pass by reference Jeremy Shovan
- Re: [PHP-DB] pass by reference Chris Boget
- Re: [PHP-DB] pass by reference Jennifer Goodie
- Re: [PHP-DB] pass by reference Ignatius Reilly