echo "<img src=\"".$row["img"]."\">"; But you must change one of the row variable, there are two while one inside the other using the same var name. ""Ross"" <ross@xxxxxxxxxxxxx> escreveu na mensagem news:0F.DB.48682.57313F44@xxxxxxxxxxxxxxx >I have retireved the unique gallery and all the data from the row. I now >need to output the data ($row['bin_data']) as a jpg. > > <? > include("includes/config.php"); > > $link = mysql_connect($host, $user, $password) or die ('somethng went > wrong:' .mysql_error() ); > mysql_select_db($dbname, $link) or die ('somethng went wrong, DB error:' > .mysql_error() ); > > $query = "SELECT DISTINCT gallery FROM thumbnails"; > $result = @mysql_query( $query,$link ); > > while ($row = @mysql_fetch_assoc($result) ) { > > $gallery_id=$row['gallery']; > > $query2 = "SELECT * FROM thumbnails WHERE gallery ='$gallery_id' LIMIT 1"; > $result2 = @mysql_query($query2); > > while ($row = @mysql_fetch_array($result2, MYSQL_ASSOC)){ > echo $id=$row['id']; > > //i want to output the jpeg here > > > } > > } -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
