替代图像在PHP中不起作用

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>';
        }

Answer 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.