help creating thumbnails

php_beginner_83

Hi All

I'm trying to create thumbnails on the fly using the gd library. I'm looping through images stored in a database and create thumbnails from these images. However, I can't get this to work. When I explicity type in the name of the image it works.

This is my code....

$result = mysql_query("SELECT * FROM pictures
INNER JOIN pics_in_albums
ON pictures.ID = pics_in_albums.PicID
INNER JOIN albums
ON pics_in_albums.AlbumID = albums.ID
WHERE albums.ID = 1") or die(mysql_error());

$paths = array();

$counter = 0;
//fetch tha data from the database
while ($row = mysql_fetch_assoc($result))
{

   $currentImage = "images//" . substr(strrchr($row['Path'],92),1);

   $counter++;

   if ($counter % 3 == 0)
   {
        echo '<a href=#><img src="thumb.php?image=$currentImage alt="" /></a><br/>';
   }
   else
   {
        echo '<a href=#><img src="thumb.php?image=$currentImage" alt="" /></a>';

   }

}
?>

So when I use this, it works ..

<img src="thumb.php?image=beach.jpg alt="" />

Thanks 😕

johanafm

echo '<a href=#><img src="thumb.php?image=$currentImage';

output:

<a href=#><img src="thumb.php?image=$currentImage

echo '<a href=#><img src="thumb.php?image=' . $currentImage;

toxic_brain

check the resulting html to see what is getting generated in src

paulnaj

According to this line ...
```
$currentImage = "images//" . substr(strrchr($row['Path'],92),1); 
```
... you expect the images to be in the 'images' folder.

When you wrote the file in explicitly, you left out any reference to the 'images' folder.

Why have you used two forward slashes in the above?
Do yourself a massive favour and don't embed variables into strings. For example,
```
echo '<a href=#><img src="thumb.php?image=$currentImage alt="" /></a><br/>';
```
is better written as:
```
echo '<a href="#"><img src="thumb.php?image='.$currentImage.'" alt="" /></a><br/>';
```
You might then have noticed that you left off a double-quote (I've added it back in).

Paul.