Warning: getimagesize

wilbur

hi,

I have an annoying problem with GetImageSize. I am simply trying to create thumbnail pictures of full size JPG's I have stored. The problem is that when I use the line:

$image_stats = getimagesize('/'.$picture);

I get the error message:

Warning: getimagesize: Unable to
open '/images/fullpictures/wedding/image2.jpg' for reading

Now a few points: I can access this file useing $picture elsewhere so I know that such a file exists and that there should not be a problem accessing it. and that I have GD 2 installed on the server. Can anyone suggest wha the heck I'm doing wrong??

Thanks in advance.

The_Chancer

Have you tried it without the first / ??

wilbur

yep, tried it with and without. Makes no difference. any other ideas?? I checked all the php boards for info on this problem and I could not find the solution.

bad76

Hi,
do you have some other handle to the file already open?
may be a file-share problem.

Also i look on the PHP manual tis two notice:

1) getimagesize do not need GD library

2) to READ a jpg you must enable it
--with-jpeg-dir=/percorso/per/jpeg-6b.

For more information

http://www.php.net/manual/en/function.getimagesize.php

and

http://www.php.net/manual/en/ref.image.php

good luck

wilbur

good point re: file already open, let me just check.

thx

superwormy

Try using the FULL PATH of the image, ie c:\path\to\file.jpg on Windows or /usr/local/www/path/to/file.jpg on *nix and see if that works...

wilbur

thanks for the suggestion, but no luck so far. I have tried inputting the full user directory which was 'http://www. etc, but with exactly the same result. any other ideas?

BigToach

could be a .bmp saved as a .jpg cause getimagesize() can't read the file when that happens.

superwormy

Can you post your code so we can see it?

wilbur

hi,

I have tried a number of different functions as well as getimagesize and seem to have the same problem. It just doesnt seem to be able to find the location of the file - but I can access the variable containing the address elsewhere in the code in order to display the picture without a problem. I am having exactly the same problem with the code below:

$src_img = imagecreatefromjpeg("/".$row_Recordset1['location']);
$origw=imagesx($src_img);
$origh=imagesy($src_img);
$new_w = $thumb_width;
$diff=$origw/$new_w;
$new_h=$new_w;
$dst_img = imagecreate($new_w,$new_h);
imagecopyresized($dst_img,$src_img,0,0,0,0,$new_w,$new_h,imagesx($src_img),imagesy($src_img));

each time, no matter what I seem to try I get the error message:

Warning: imagecreatefromjpeg: Unable to open '/images/fullpictures/wedding/image2.jpg' for reading

I'm not sure but might it make a difference, as was suggestted earlier by superwormy, to enter the direct path for the server. The problem with this is that I'm not sure if using the http://www.webpage.com/pathtoroute/images/fullpictures....etc is the same and will work or do I really need to find out the /usr/local/www/path/to/file.jpg directory - and just how do I find out this directory?!?!

In response to the request for the code - I've tried it out in several different code sets and am having the same problem, so I know its not the code. I've ruled out quite a few problems - and concluded it must be something to do with the location.

thanks,

superwormy

You may need to use /usr/local/www/path/to/file.jpg a path like that... but I'm nto sure. Worth a try though.

Telnet in and cd to directory where image is, and type 'pwd'