Broken Image...display alternate image instead

[sgtpepper]

Hello,

I am grabbing images to display from another website like this

<img src="http://www.otherwebsite.com/<? echo $row['Id']; ?>.jpg" width="90" height="61">

It may be possible that the website will not have images for the specific Id.

Instead of a broken image displaying....how can I have it display a imageunavailable.jpg file hosted locally?

[kevinsequeira]

$file = "http://www.otherwebsite.com/".$row['Id'].".jpg";
if(fopen ($file, "r"))
//display image
else
//display default image

reg
kevin

[sgtpepper]

Hmmm, sorry for the beginner questions but now I have this:

<?
$file = "http://www.website.com/".$row['Id']."1.jpg";
if(fopen ($file, "r")) {
//display image
} else {
//display default image
<img src="http://www.defaultsite.com/logo.gif" border="0" width="142" height="65">
?>

I get a parse error on the default image line. I am sure, I havent placed that line correctly. Should I be using an echo?

[stevenmusumeche]

Yes...

echo "<img src=\"http://www.defaultsite.com/logo.gif\" border=\"0\" width=\"142\" height=\"65\">";

[sgtpepper]

The error now becomes

Warning: fopen("http://www.site/image.jpg", "r") - Success in /home/httpd/virtual/test.php on line 98

And it displays the default image....

Actual Code:

<?
$file = "http://www.XXX/".$row['Id']."1.jpg";
if(fopen ($file, "r")) {
//display image
} else {
//display default image

echo "<img src=\"http://XXX/logo.gif\" border=\"0\" width=\"142\" height=\"65\">";
}
?>

[sgtpepper]

What about if I want to find the image in two places either

<img src="http://www.xxxxxx.com/listing_img/<? echo $row['Id']; ?>1.jpg" width="100" height="65">
<img src="http://www.xxxxxx.com/propImgs/<? echo $row['Id']; ?>1.jpg" width="100" height="65">

How could I say something like.....

try 1st url.....if no image try second url