Re: Why $row is EMPTY
| From: | Jon L. | Date: | Tue, 29 Apr 2008 18:46:56 +0000 |
| Subject: | Re: Why $row is EMPTY | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-44922@lists.php.net to get a copy of this message | ||
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@yahoo.com>
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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