Try doing print_r($row); in your while loop and see what exactly is in
that array.
Ken
On Tue, Apr 29, 2008 at 1:46 PM, Jon L. <jonllmsed@xxxxxxxxx> wrote:
> I could be wrong, but I don't think table aliases continue to exist in PHP;
> only the column names.
>
> $g_name = $row["gigName"];
> $vname = $row["venueName"];
> $genre = $row["name"];
>
>
> You may also consider revising the query to only grab the columns you need,
> using an alias for at least genre.name.
>
> $query = 'SELECT gig.gigName AS gig, venue.venueName AS venue, genre.name AS
> genre FROM `gig` LEFT JOIN genre ON gig.genreId=genre.genreId LEFT JOIN
>
> venue ON gig.venueID = venue.vid where gig.gigid = '.$gigDetail.' ORDER BY
> gig.gigid';
>
> $result = mysql_query($query) or die(mysql_error());
> while ($row = mysql_fetch_array($result))
> {
> echo "Program is in While loop";
> $g_name = $row["gig"];
> $vname = $row["venue"];
> $genre = $row["genre"];
>
> echo("Gig Name: ".$g_name);
> }
>
>
> - Jon L.
>
>
> On Tue, Apr 29, 2008 at 1:27 PM, Nasreen Laghari <nasreen_laghari@xxxxxxxxx>
> wrote:
>
>
>
> > Hi,
> > Why my program is not going in while loop? When I run the same query in
> > SQL cmd, it brings result but here when I print $result gives Resouce ID
> > number means $result has data then why $row is empty.
> > $query = 'SELECT * FROM `gig` LEFT JOIN genre ON gig.genreId=genre.genreId
> > LEFT JOIN venue ON gig.venueID = venue.vid where gig.gigid = '.$gigDetail.'
> > ORDER BY gig.gigid';
> >
> > $result = mysql_query($query) or die(mysql_error());
> > while ($row = mysql_fetch_array($result))
> > {
> > echo "Program is in While loop";
> > $g_name = $row["gig.gigName"];
> > $vname = $row["venue.venueName"];
> > $genre = $row["genre.name"];
> > echo("Gig Name: ".$g_name);
> > }
> > Regards
> > Nasreen
> >
> >
> >
> > ____________________________________________________________________________________
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> >
>
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