SkyRanger Posted January 19, 2013 Share Posted January 19, 2013 $mysqli = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $phtocat = $_GET['cid']; $query = "SELECT * FROM `gallery_photos` WHERE photo_category='$phtocat' and `photo_owner`='$client'"; $result = $mysqli->query($query) or die($mysqli->error.__LINE__); //Define number of columns to use $maxColumns = 5; $count = 0; //Var to count records $categoryList = ''; //Var to hold HTML output while($row = $result->fetch_assoc()) { $count++; //Increment count //Open new row if first record in row if($count % $maxColumns == 1) { $categoryList .= "<tr>\n"; } $image_dir = "members/$client/photos"; $pid = $row['photo_id']; //Output current record $categoryList .= "<td align=center width=20%> <a href={$image_dir}/{$row[photo_filename]} rel=lightbox title='{$row[photo_caption]}'><img src={$image_dir}/tb_{$row[photo_filename]} border='0' alt='{$row[photo_caption]}' /></a> <br> <A HREF=\"javascript:popUpPic('printimage.php?pid={$pid}&client={$client}')\"><img src=images/print.jpg></a> <a href=\"downloadimage.php?file={$image_dir}/{$row[photo_filename]}\" onclick=\"this.blur();\"><img src=images/download.jpg></a> <a href=removephoto.php?pid=$pid&photo_owner=$client><img src=images/delete2.jpg></a> </td>\n"; //Close row if last record in row if($count % $maxColumns == 0) { $categoryList .= "</tr>\n"; } } //Close last row if needed if($maxColumns % $count != 0) { $categoryList .= "</tr>\n"; } ?> <table align="center" width="95%" border="0"> <?php echo $categoryList; ?> </table> Error I am getting: Warning: Division by zero in showphoto.php on line 259 I am not sure how to make it say "There are no pictures in this category" Could somebody please give me a hand. I have tried an else { echo "There are no......."; but with no luck I still get the error. Quote Link to comment Share on other sites More sharing options...
jazzman1 Posted January 19, 2013 Share Posted January 19, 2013 Where is the showphoto.php file and the line #259 ? Quote Link to comment Share on other sites More sharing options...
SkyRanger Posted January 19, 2013 Author Share Posted January 19, 2013 Sorry knew I forgot something if($maxColumns % $count != 0) { $categoryList .= "</tr>\n"; } It works no problem when there is something in the database for this category but when it is empty I get the error. Quote Link to comment Share on other sites More sharing options...
Jessica Posted January 19, 2013 Share Posted January 19, 2013 Check if $count is not 0 first. Quote Link to comment Share on other sites More sharing options...
SkyRanger Posted January 19, 2013 Author Share Posted January 19, 2013 Thanks Jessica, got it to work, I just added this if ($result->num_rows == 0) { $categoryList .= "<tr><td>No Images</td></tr>"; } else { original while code } Quote Link to comment Share on other sites More sharing options...
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