On Wed, 28 Jul 2004 20:47:37 -0700, Daevid Vincent <[EMAIL PROTECTED]> wrote: > Linux. PHP5. > > Why does this fail when using an array element, but using a variable will > work? Why should PHP care what the variable is I'm trying to store into? > > list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth); > > But this works: > > list($foo, $CompanyDB) = SQL_ROW($sth); > > And of course I'd have the extra... > $foo = $bar['CompanyCode']; >
Because list() is a language construct, not a function. It assumes what you give it is a normal variable, it doesn't understand arrays. Better IMHO to use $row = mysql_fetch_assoc() and access the array it returns directly. -- DB_DataObject_FormBuilder - The database at your fingertips http://pear.php.net/package/DB_DataObject_FormBuilder paperCrane --Justin Patrin-- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
