Re: What in Hades?

On Thu, 2009-02-26 at 11:14 -0500, PJ wrote: > Something is rotten here. Twist it any way you like and it just does not > work. > I am trying to figure out the working of SELECT LAST_INSERT_ID() and > cannot get beyond the first $sql - But it works fine from command-line! > > <? > include ("lib/db1.php"); // Connect to database > > $sql1 = "INSERT INTO test (name, age) VALUES ('Joe Blow', '69')"; > $result1 = mysql_query($sql1,$db); > if ( !$result) { > echo("<P>Error performing query: " . > mysql_error() . "</P>"); > exit(); > } > > THIS IS WHERE IT STOPS WITH ERROR.$sql1 just refuses to work. db > connection is fine and the INSERT works from xterm on remote puter. > Can't figure this out. > ============ > > > $sql = "SELECT LAST_INSERT_ID() FROM test"; > $result = mysql_query($sql,$db); > if (!$result) { > echo("<P>Error performing query: " . > mysql_error() . "</P>"); > exit(); > } > while ( $row = mysql_fetch_array($result) ) { > echo("<P>" . $row["id"] . "</P>"); > } > ?> > > What am I missing here? Or how can I trace the error? > > -- > > Phil Jourdan --- pj@ptahhotep.com > http://www.ptahhotep.com > http://www.chiccantine.com > > PHP has a built-in function for retrieving the last inserted auto id, called mysql_insert_id (http://uk2.php.net/mysql_insert_id) which is a lot easier than using another query to do the same. So in your above line, you would replace $sql = "SELECT LAST_INSERT_ID() FROM test"; with: $sql = mysql_insert_id($result1); and then remove the $result line, as you don't need to execute a query for it now. Ash www.ashleysheridan.co.uk