I'm using LAST_INSERT_ID() in php and mysql but i can't get a result. Why?

Question

$sql_comp5 ="INSERT INTO `tiquets` (`Id_Tiquet`) VALUES (NULL); SELECT LAST_INSERT_ID()";   
    $result8 = mysql_query($sql_comp5); 

     $flag_control=0;
    while ($row = mysql_fetch_assoc($result8, MYSQL_BOTH)) 
    {
        $flag_control=$flag_control+1;
             $id_t[$flag_control]=$row['LAST_INSERT_ID()'];              
    }


    for ($buc = 1; $buc <=$flag_control; $buc++)
    {
           $id_tiquet=$id_t[$buc];                
    }

I am doing the correct? Or i'm wrong?

Very thanks!!

[Please, don't use mysql_* functions in new code](http://bit.ly/phpmsql). They are no longer maintained [and are officially deprecated](http://j.mp/XqV7Lp). See the [red box](http://j.mp/Te9zIL)? Learn about [_prepared statements_](http://j.mp/T9hLWi) instead, and use [PDO](http://php.net/pdo) or [MySQLi](http://php.net/mysqli) - [this article](http://j.mp/QEx8IB) will help you decide which. If you choose PDO, [here is a good tutorial](http://j.mp/PoWehJ). — Luigi Siri, May 25 '13 at 15:39

score 0 · Answer 1 · answered May 25 '13 at 15:39

You can't do two queries use mysql_* functions. You need mysqli::multi_query() for that. To get the last insert ID using mysql_* use mysql_insert_id():

$sql_comp5 ="INSERT INTO `tiquets` (`Id_Tiquet`) VALUES (NULL);";
$result8 = mysql_query($sql_comp5); 
$id      = mysql_insert_id($result8);

score 0 · Answer 2 · answered May 25 '13 at 19:02

0

Remove the second query - and use php's mysql_insert_id()

$sql_comp5 ="INSERT INTO tiquets (Id_Tiquet) VALUES (NULL)";
$result8 = mysql_query($sql_comp5); $insertedId = mysql_insert_id();

answered May 25 '13 at 19:02

YomY

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I'm using LAST_INSERT_ID() in php and mysql but i can't get a result. Why?

2 Answers2