Posted by Dan McCullough on 10/11/05 20:45

$result = mysql_query($query) or die(mysql_error());

On 10/11/05, Silvio Porcellana <sporc@tin.it> wrote:
> aaronjw@martekbiz.com wrote:
> > <?php
> >
> > include ("../utils.inc");
> >
> > $date = date("U");
> > $dateExpire = $date + 90 * 86400;
> >
> > $code = "jack";
> >
> > $query = "INSERT INTO CouponTable VALUES
> > ('','$date','0','$code','preset','$dateExpire','3.75')";
> >
> > $result = mysql_query($query);
> >
> > echo mysql_error($result);
> >
> > ?>
> >
> > This SHOULD error out but I'm getting the error instead.
> >
> > It's just a test page to test my logic...
> >
>
> If you read carefully this page:
>
> http://php.net/mysql_error
>
> you will notice that it says that you need to pass to mysql_error *the
> MySQL connection*. Not the resource returned by mysql_query.
>
> HTH, cheers
> Silvio
>
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