Re: What in Hades?

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On Thu, Feb 26, 2009 at 9:14 AM, PJ <af.gourmet@xxxxxxxxxxxx> 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) {


// notice the $result, should be $result1

 echo("<P>Error performing query: " .
>       mysql_error() . "</P>");
>  exit();
> }
>

Hope that helps

Richard




>
> 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@xxxxxxxxxxxxx
>   http://www.ptahhotep.com
>   http://www.chiccantine.com
>
>
> --
> PHP General Mailing List (http://www.php.net/)
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>
>


-- 
Richard Whitney
phpMyDEV@xxxxxxxxx
http://phpmydev.com
602-288-5340
310-943-6498

"You come up with ideas, I come up with solutions."

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