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Need some help with code in displayimage.php (FIXED)

Started by adriano, February 18, 2004, 03:58:18 PM

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adriano

I would add in the displayarea under the image a link to a external place for a hires picture.

$info[$lang_picinfo['File Size']] = '<a href="http://myexternalurl/$number$/' . substr($CURRENT_PIC_DATA['filename'], 0, 4) . '/' . $CURRENT_PIC_DATA['filename'] . '>Download</a>';

In frontend I see no generated output!

I used the nonused variable "File Size" in displayimage.php and would receive from the script a URL like this:

http://myexternalurl/$number$/folder/filename.jpg

The name of my pictures represent in the first four chars the foldername (0010.0999.jpg -> URL should be "http://myexternalurl/$number$/0010/0010.0999.jpg) where the pictures will be found. This function for separating the foldername from the filename works. It seems, that anycase the URL I placed in front does not work or the not defined variable $numbers$ which I need, brings a wrong result.

Has anyone here a tipp for me how write the right code?

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I found the wrong part - Thanks!
[size=10]Fotos von Norddeutschland und der Nordseeküste
Photos of Northern Germany and the North Sea Coast:
http://www.wackernah.net[/size]