Alternate image not working inPHP

Question

I'm trying to show an image if the intended image cannot be found. The images are taken from an RSS feed. The original code is last.

Ive tried a few things. But doesnt seem to be working. Its only showing the backup images: blank_newprod.png.

New code:

foreach($channel->item as $item){
            //if($i==6){break;}
            if($item->prodimg=="~"){break;}
            if($item->prodpage=="~"){break;}
            $i += 1;
            echo '<span ';
            switch ($i) {
                case 1: echo 'style="margin-right:18px;">'; break;
                case 2: echo 'style="margin-right:18px;">'; break;
                case 3: echo '>'; break;
                case 4: echo 'style="margin-right:18px;">'; break;
                case 5: echo 'style="margin-right:18px;">'; break;
                case 6: echo '>'; $i = 0; break;
            }
            $filename = '.$item->prodimg.';

            if (file_exists($filename)) {
            echo '<a href="'.$item->prodpage.'" title="'.$item->$title.'"><img src="'.$item->prodimg.'" width="181" height="100" alt="'.$item->$title.'" /></a></span>';
            } else {
            echo '<img src="../includes/php/blank_newprod.png"/>';
            }
        }

Original code:

foreach($channel->item as $item){
            //if($i==6){break;}
            if($item->prodimg=="~"){break;}
            if($item->prodpage=="~"){break;}
            $i += 1;
            echo '<span ';
            switch ($i) {
                case 1: echo 'style="margin-right:18px;">'; break;
                case 2: echo 'style="margin-right:18px;">'; break;
                case 3: echo '>'; break;
                case 4: echo 'style="margin-right:18px;">'; break;
                case 5: echo 'style="margin-right:18px;">'; break;
                case 6: echo '>'; $i = 0; break;
            }
            echo '<a href="'.$item->prodpage.'" title="'.$item->$title.'"><img src="'.$item->prodimg.'" width="181" height="100" alt="'.$item->$title.'" /></a></span>';
        }

score 1 · Accepted Answer · edited May 23 '17 at 11:53

1

This doesn't look like a valid file name:

$filename = '.$item->prodimg.';

Surely you meant to just use the value itself?:

$filename = $item->prodimg;

Of course, if that value is a URL (as implied by its later use) then to check for the corresponding file you'd need to convert it to a file system path. Fortunately, that's been asked already.

edited May 23 '17 at 11:53

Community

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answered Mar 07 '17 at 15:26

David

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Alternate image not working inPHP

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