Problem was a fat-finger - meant 'bid' not 'bids'... Doh!!!
Sparky - Thanks again.
----- Original Message -----
From: "Shena Delian O'Brien" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, July 01, 2003 16:01
Subject: [PHP] Re: Nested mysql_query()'s
> Sparky Kopetzky wrote:
> > Hi!
> >
> > I've got two nested queries where one looks up data based on values
returned from the first.
> >
> > $lot_query = "SELECT * FROM LOT WHERE lot_category_id=" .
$lot_category .
> > " AND lot_close_time>" . time();
> > $lot_result = mysql_query($lot_query, $CONNECT_ID);
> >
> > if ($lot_result)
> > {
> > while ($row = mysql_fetch_array($lot_result))
> > {
> > blah-blah-blah...
> >
> > $bid_query = "select * from bids where bid_lot_id=" .
$lot_id .
> > " order by bid_amount";
> > $bid_result = mysql_query($bid_query, $CONNECT_ID);
> > $bid_count = mysql_num_rows($bid_result);
> >
> > blah-blah-blah...
> >
> > I get this error message: "Warning: mysql_num_rows(): supplied
> > argument is not a valid MySQL result resource " from the second
> > query. Do I need a second $CONNECT_ID for a second link to
> mysql_query #2??
>
> No... you don't even need to use a connect ID. It should work fine if
> you just use mysql_query($bid_query)
>
> The problem is likely that the query is failing for some reason and thus
> not generating a proper result. Do an echo mysql_error($CONNECT_ID) to
> figure out if there's an error in your syntax, etc.
>
>
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