Ash, Martin,
Seems you are both wandering around the obvious problem...
I suspect that $tipo (in the next line) is *supposed* to be $type - sounds like
a partial Italian translation to me...
So given that $type="imagecreatefrompng" (for example, if the mime check returns
'png' - not very reliable, I suspect),
then
$immagine = $type($this->updir.$id.'.png')
should create a GD resource from the file, but the image appears to be empty.
My take on this is:
OP says he gets the same result from
$immagine = imagecreatefromjpeg(this->updir.$id.'.png')
- well I might expect to get an error message if I loaded a PNG expecting it to
be a JPEG, but I certainly wouldn't expect an image.
On some basic tests, I find that mime_content_type() is not defined on my
system, so ignoring that and trying to load a PNG file using imagecreatefromjpeg
results in pretty much the same result as the OP...
Conclusions:
First: if you use Italian for your variable names, don't change half of their
instances to English...
Second: Make sure you actually know the mime type of a file before you try to
load it into GD...
My version of this would test against known image types to try the GD function:
foreach (Array('png','jpeg','gif') as $typeName)
{
$type = 'imagecreatefrom'.$typeName;
$immagine = $type(this->updir.$id.'.png'le);
if (is_resource($immagine))
{
header('Content-type: image/jpeg');
imagejpeg($immagine,null,100);
imagedestroy($immagine);
break;
}
}
header('HTTP/1.0 500 File is not an allowed image type');
--
Peter Ford phone: 01580 893333
Developer fax: 01580 893399
Justcroft International Ltd., Staplehurst, Kent
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