Re: Is this possible ?

One way: <? Function MyFunc() { if(isset($_POST[var])){ $sql = mysql_query("select * from table_name where field=\"$_POST[var]\""); $num = mysql_num_rows($sql); // I want to use this result outside this function. $returnsomething ="blah blah"; } return array($returnsomething, $num); } list($text, $numrows) = MyFunc(); ?> hth toby ----- Original Message ----- From: "Dave Carrera" <dave@ephgroup.com> To: <php-general@lists.php.net> Sent: Sunday, January 18, 2004 11:22 AM Subject: [PHP] Is this possible ? Hi List, I have a function that makes a call to mysql based on certain vars. ---example---- Function MyFunc(){ if(isset($_POST[var])){ $sql = mysql_query("select * from table_name where field=\"$_POST[var]\" "); $returnsomething ="blah blah"; } return $returnsomething; } And that all works fine no probs here but..... I want to use a result somewhere in my script that is not returned by return. Let me show you... ---example---- Function MyFunc(){ if(isset($_POST[var])){ $sql = mysql_query("select * from table_name where field=\"$_POST[var]\" "); $num = mysql_num_rows($sql); // I want to use this result outside this function. $returnsomething ="blah blah"; } return $returnsomething; } So $num contains a number that I want to use outside the function which is not covered by return. I know return stops a script and returns what I want it to return but how do I send out of the function the var I want. I have tried $GLOBAL[var]=$num; but that don’t work, but I thought I would'nt anyway just tried it and yes I know I have to declare it inside my new function using global $var; to use it. So I ask is this achiveable or how can I do this. Thank you in advance Dave C --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.560 / Virus Database: 352 - Release Date: 08/01/2004 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php