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 --- [email protected]
> 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
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