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Re: [PHP-DB] Newbie alert: supplied argument is not a valid MySQL result resource
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| Niel Archer 2007-07-10, 3:58 am |
| Hi
On my way to bed so too tired too look closely, but at a guess I'd say
it's a logic error. I would guess this:
> $sid = mysql_fetch_array($query) || die ('Could not fetch from database:
> ' . mysql_error()); // <--- line 49
isn't evaluating in the order you expect.. You're using a logic
statement as an assignment, and the result of that get's assigned to your
variable not the resource you expect. Try it like this to see if it works:
$sid = mysql_fetch_array($query);
if ($sid === false) die ('Could not fetch from database: ' . mysql_error());
If that's not the problem, I haven't got a clue until after sleep and
coffee.
Niel
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| Hi
> On my way to bed so too tired too look closely, but at a guess I'd say
> it's a logic error. I would guess this:
>
>
> isn't evaluating in the order you expect.. You're using a logic
> statement as an assignment, and the result of that get's assigned to your
> variable not the resource you expect. Try it like this to see if it works:
>
> $sid = mysql_fetch_array($query);
> if ($sid === false) die ('Could not fetch from database: ' . mysql_error());
>
> If that's not the problem, I haven't got a clue until after sleep and
> coffee.
I've had some sleep, and coffee, and I see I missed something silly.
I should of suggested this
// get sid and write cookies
$query = mysql_query("SELECT MAX(SID) FROM Sessions");
if ($query === false) die ('Could not query database: ' . mysql_error());
$sid = mysql_fetch_array($query)
if ($sid === false) die ('Could not fetch from database: ' . mysql_error());
--
Niel Archer
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