willingtolearn Posted May 25, 2012 Share Posted May 25, 2012 hello - i am working on a script that someone else has created and need to add the option for more images to be displayed - only if there is any.. the script is the same for all images 1-6, I have copied and pasted for the additional images but i only want them to display only if there is any added... how do i edit the below code to show this I am new to php, so please be kind - any help you can give would be greatly appreciated. ?> $query = "SELECT photo_name FROM photos WHERE PID=$id ORDER BY photo_name ASC LIMIT 5, 1"; $photos = mysql_query($query) or die ("Select Failed!"); $photo6 = mysql_fetch_array($photos); $photo6 = $photo6['photo_name']; ?> <a href="../smallphotos/<? echo $photo6; ?>.jpg" rel="lightbox[Villas]" > <img src="../smallphotos/<? echo $photo6; ?>.jpg" width="280" height="210" border="1"> </a> Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/ Share on other sites More sharing options...
QuickOldCar Posted May 25, 2012 Share Posted May 25, 2012 http://php.net/manual/en/function.file-exists.php Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/#findComment-1348532 Share on other sites More sharing options...
PravinS Posted May 25, 2012 Share Posted May 25, 2012 Add if condition before the image link tag, check the below code <?php if (file_exists("../smallphotos/".$photo6.".jpg")) { ?> <a href="../smallphotos/<? echo $photo6; ?>.jpg" rel="lightbox[Villas]" > <img src="../smallphotos/<? echo $photo6; ?>.jpg" width="280" height="210" border="1"> </a> <?php } ?> Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/#findComment-1348533 Share on other sites More sharing options...
willingtolearn Posted May 25, 2012 Author Share Posted May 25, 2012 thanks so much for replying - i really really appreciate it!! well, my head is exploding.. it doesn't work:-( i keep getting an error : Parse error: syntax error, unexpected '<' in /home/morairal/public_html/rentals/details100.php on line 219 which is: <?php if (file_exists("../smallphotos/".$photo7.".jpg")) { ?> the page link is http://www.morairaluxuryvillas.com/rentals/details100.php?id=45 <? $query = "SELECT photo_name FROM photos WHERE PID=$id ORDER BY photo_name ASC LIMIT 4, 1"; $photos = mysql_query($query) or die ("Select Failed!"); $photo7 = mysql_fetch_array($photos); $photo7 = $photo7['photo_name']; <?php if (file_exists("../smallphotos/".$photo7.".jpg")) { ?> <a href="../smallphotos/<? echo $photo7; ?>.jpg" rel="lightbox[Villas]" > <img src="../smallphotos/<? echo $photo7; ?>.jpg" width="280" height="210" border="1"> </a> <?php } ?> Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/#findComment-1348562 Share on other sites More sharing options...
cyberRobot Posted May 25, 2012 Share Posted May 25, 2012 Assuming the code appears as described above, there's an extra "<?php". $photo7 = $photo7['photo_name']; <?php if (file_exists("../smallphotos/".$photo7.".jpg")) { ?> Should be $photo7 = $photo7['photo_name']; if (file_exists("../smallphotos/".$photo7.".jpg")) { ?> Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/#findComment-1348568 Share on other sites More sharing options...
willingtolearn Posted May 25, 2012 Author Share Posted May 25, 2012 if i could kiss u i would thank u thank u thank u !! Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/#findComment-1348585 Share on other sites More sharing options...
cyberRobot Posted May 25, 2012 Share Posted May 25, 2012 if i could kiss u i would Hmm... Glad to help! Link to comment https://forums.phpfreaks.com/topic/263106-display-image-only-if-it-exists/#findComment-1348591 Share on other sites More sharing options...
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