当图像不可用时显示另一个图像

Question

i have a code which gets images from a database. I need to tell it to to display another image (which says no image available) if an image is not found. How would I do this????

Your suggestions would be very much appreciated

<a class="thumbimage" href="<?PHP mrd("$MyProductTitle", "$row[LID]", "$_GET[category]", "$rowxxx[MR]", "index.php?page=detail"); ?>"><img src="images/thumb/<?php echo "$row[IMAGENAME]"; ?>.jpg" border="1" /></a>

Answer 1

Try this:

<?php
    $currentImage = "images/thumb/".$row[IMAGENAME].".jpg";
    if(!file_exists($currentImage))
    {
        $currentImage = "PATH_TO_IMAGE_UNAVAILABLE";
    }
?>

<a class="thumbimage" href="<?PHP mrd("$MyProductTitle", "$row[LID]", "$_GET[category]", "$rowxxx[MR]", "index.php?page=detail"); ?>"><img src="<?=$currentImage?>" border="1" /></a>