Areric wrote:
Ok all. I have a series of images stored in a db. Im trying to work on
a script that will let me scale them based on user input. Ive hit a bit
of a roadblock on this line.
$orig =
imagecreatefromjpeg("DisplayImage.php?imageId=".$t his->mImage->GetImageId());
Specifically it cant sopen that file.
DisplayImage.php is pretty simple, it just displays an image from the
db without writing to a file.
Im thinking the problem is in how im pathing to DisplayImage.php.
This file sits one level lower than most things so it should be
../displayImage, but i tried that and no luck.
Any ideas whats wrong with that line?
There is nothing wrong with that line except that it is NOT the path to your
image.
You are confusing:
- the path to your image (nonexistent)
- with a php-script delivering the image.
Here is the difference: The is no such thing as the path to your image. Only
a URI (http) that points to your image.
The difference is that in the URI situation you invoke a script that
produces the image, while imagecreatefromjpeg WANTS A PATH, as can be read
in the documentation at
www.php.net.
NOT a URI, unless:
from
http://nl2.php.net/manual/en/functio...tefromjpeg.php
--------------
Tip: You can use a URL as a filename with this function if the fopen
wrappers have been enabled. See fopen() for more details on how to specify
the filename and Appendix M for a list of supported URL protocols.
-----------------
So check that. :-)
And also remember that ../bla.php is NOT a good URI, it is just a part.
If you use URI, make sure they are full/well formed, like:
http://bla.bla.com/myImagescript.php?id=324
If you cannot make it work with fopenwrappers, just safe them from your
script in a directory, named after their ID's or something like that.
Regards,
Erwin Moler
