RE: [PHP] get image size function
| From: | Jack Dempsey | Date: | Wed, 09 Apr 2003 20:21:23 +0000 |
| Subject: | RE: [PHP] get image size function | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-143189@lists.php.net to get a copy of this message | ||
hey tyler,
> When I echo/print the variables everything goes well until I get to
> $photos = file($data_file);
>
> That's when I get a value of array.
> Anybody got any idea why this is doing this?
well, that's different than the line you listed before.
$photos is obviously going to be an array because file() returns one:
array file ( string filename [, int use_include_path])
as far as your errors before, if you would give the output of what i
suggested,
maybe we could figure things out...the url to the site to see if the actual
images are where
php's looking could help too...
jack
>
>
>
>
>
> >From: John Dempsey <dempsejn@gusun.georgetown.edu>
> >To: Tyler Durdin <linuxgeek23@hotmail.com>
> >CC: php-general@lists.php.net
> >Subject: Re: [PHP] get image size function
> >Date: Wed, 9 Apr 2003 15:38:47 -0400 (EDT)
> >
> >ok, so that looks like its not finding the file given in your photo
> >array...not to doubt you, but are you sure you're getting the full
> >filename, extension included, in the array, and that the file's actually
> >in the same directory? try a test with file_exists to make sure
> its there,
> >and print out what the trim($photos[$i]) value is to double check what's
> >happening...
> >
> >jack
> >
> >On Wed, 9 Apr 2003, Tyler Durdin wrote:
> >
> > >
> > >
> > >
> > >
> > > I ran the print_r and sure enough all of the image values
> were there. So
> >now
> > > I am stuck again. This is the error message I am getting:
> > >
> > > Warning: getimagesize(Mvc-038f.jpg) [function.getimagesize]: failed to
> > > create stream: No such file or directory in
> > > /www/servers/comgo_test/gallery.php on line 47
> > >
> > > And this is line 47
> > >
> > > $image_size = getimagesize(trim($photos[$i]));
> > >
> > > These are the variables the line is referring to
> > > $photos = file($data_file);
> > > ** the datafile just has the names of 3 jpg images **
> > >
> > >
> > > >From: dempsejn@georgetown.edu
> > > >To: "Tyler Durdin" <linuxgeek23@hotmail.com>
> > > >CC: php-general@lists.php.net
> > > >Subject: Re: [PHP] get image size function
> > > >Date: Wed, 09 Apr 2003 14:17:06 -0400
> > > >
> > > >hey tyler,
> > > >
> > > >just looking at that quickly, i'd try testing it with print_r to see
> > > >the structure; if you're getting the string "Array" then its
> obviously
> > > >an array that you have to then select something from. print_r($var)
> > > >should make it clear.
> > > >
> > > >jack
> > > >
> > > >----- Original Message -----
> > > >From: "Tyler Durdin" <linuxgeek23@hotmail.com>
> > > >Date: Wednesday, April 9, 2003 2:13 pm
> > > >Subject: [PHP] get image size function
> > > >
> > > > > When I run this script:
> > > > >
> > > > > <?php
> > > > > $size = getimagesize ("pictures/pictures/bobby.jpg");
> > > > > echo "<img src=\"pictures/pictures/bobby.jpg\"
> > > > > {$size[3]}>";
> > > > > ?>
> > > > >
> > > > > I get Array (the word Array) and a picture of the image. Why am I
> > > > > not
> > > > > getting the image size? Did I not configure php to use the image
> > > > > functions?
> > > > > If that's the case how can I correct this problem? If it's not
> > > > > the
> > > > > case what
> > > > > is going on here? Please help!!!
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
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