I get a warning message:
Warning: Unable to open <!-- m --><a class="postlink" href="http://www.xyz.com/dirctory2/member/approvedpic/abc.jpg">http://www.xyz.com/dirctory2/member/approvedpic/abc.jpg</a><!-- m --> in ../imagesizer.php3 on line 29
(This is line 29):
//$fd = fopen($filename, "r");
$imageInfo = getimagesize($filename);
//fclose ($fd);
Here is what is happening:
I have an image sizer which resizes all pictures, and it works fine under the original directory (directory1).
But the image sizer is not working when I duplicated (directory1) and named it (directory2).
The image sizer will work if I copied all the images to the approvedpic in (directory2). However this is something I do not want to do which will take too much space on the server.
Would anyone know what I will have to write in the code above so the image sizer can identify the image from a different directory?
Regards,
Christine
Warning: Unable to open <!-- m --><a class="postlink" href="http://www.xyz.com/dirctory2/member/approvedpic/abc.jpg">http://www.xyz.com/dirctory2/member/approvedpic/abc.jpg</a><!-- m --> in ../imagesizer.php3 on line 29
(This is line 29):
//$fd = fopen($filename, "r");
$imageInfo = getimagesize($filename);
//fclose ($fd);
Here is what is happening:
I have an image sizer which resizes all pictures, and it works fine under the original directory (directory1).
But the image sizer is not working when I duplicated (directory1) and named it (directory2).
The image sizer will work if I copied all the images to the approvedpic in (directory2). However this is something I do not want to do which will take too much space on the server.
Would anyone know what I will have to write in the code above so the image sizer can identify the image from a different directory?
Regards,
Christine