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 > > > > ____________________________________________________________________________________ > Be a better friend, newshound, and > know-it-all with Yahoo! Mobile. Try it now. > http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ >