caldridge1 Posted March 4, 2006 Share Posted March 4, 2006 Hi all, This is probably very simple to do, but I haven't got a clue how to do it.I have a MySQL database that stores information for car classified ads. One of the fields in the database is image, which stores the path of the image for the ad - something simple like "/classpics/1.jpg".Now, not all of the ads have images, but because of my somewhat amateur coding, the path to where the image should be is always inserted into the database, even if no image has been uploaded.So, what I need is a simple piece of code that will see if the image in the database actually exists, and if it doesn't, it displays another image that says "no image to display". I am using HTML script to display the image.ie: if !<img src=classpics/1.jpg> <img src=classpics/nopic.jpg>Basically, if the link to the original image is broken, display the nopic.jpg image instead.Any ideas?Thanks Quote Link to comment https://forums.phpfreaks.com/topic/4089-image-displayno-image-easy-if-you-know-how/ Share on other sites More sharing options...
AndyB Posted March 4, 2006 Share Posted March 4, 2006 More or less, this should point the way ..[code]if (file_exists($img_url)) { echo "<img src='". $img_url. "' border='0' alt=''>";} else { echo "<img src='classic/nopic.jpg'>";}[/code] Quote Link to comment https://forums.phpfreaks.com/topic/4089-image-displayno-image-easy-if-you-know-how/#findComment-14196 Share on other sites More sharing options...
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