Image dimensions

[Adam W]

I am trying to get the dimensions of an image by using GetImageSize()

 

I am using this code

$query = "SELECT data,filetype FROM uploads where id=$id";
$result = MYSQL_QUERY($query);
$data = MYSQL_RESULT($result,0,"data");
$type = MYSQL_RESULT($result,0,"filetype"); 
$size = GetImageSize($data);

 

I dont understand what is wrong but I get this error messsage

 

Warning: getimagesize(): Unable to access ÿØÿá8EExif in /usr/local/psa/home/vhosts/......../Members/upload.php on line 16

 

Any ideas to help?

 

Thanks

[aim25]

Well, first off php is case sensitive, so i don't think MYSQL_QUERY is the same as mysql_query.

[mkoga]

make sure that $data contains a correct path to the image. try printing it out to the screen just to check.

[Adam W]

Thanks

 

Changed to all lower case - no difference

 

If I print $data I get goobbledegook unless I use

Header( "Content-type: image/jpeg");

 

so I geuss  $data is not holding the file in a way GetImageSize can interpret it but how do I extract the image file from the MySql database?

 

Thanks

[chocopi]

can you post your whole code please

[Adam W]

Sure

 

<?php
require '../config.php';
require '../OpenDb.php';
$query = "SELECT data,filetype FROM uploads where id=$id";
$result = mysql_query($query);
$data = mysql_result($result,0,"data");
$type = mysql_result($result,0,"filetype");
//Header( "Content-type: $type");
//echo $data;
$size = GetImageSize($data);
if($size[0] > $size[1])
{
$thumbnail_width = 100;
$thumbnail_height = (int)(100 * $size[1] / $size[0]);
}
else
{
$thumbnail_width = (int)(100 * $size[0] / $size[1]);
$thumbnail_height = 100;
}
echo "thumbnail_width = $thumbnail_width thumbnail_height = $thumbnail_height";
print "To upload another file <a href=http://www.mikesierra.co.uk/Members/formloader.html> Click Here</a>";
?>

[chocopi]

try putting the image stuff along with the header inside another file and then include it

[redarrow]

[code]
giv that a go please.

<?php

Header( "Content-type: $type");

 

require '../config.php';

require '../OpenDb.php';

$query = "SELECT data,filetype FROM uploads where id='$id'";

$result = mysql_query($query);

$data = mysql_result($result,0,"data");

$type = mysql_result($result,0,"filetype");

 

 

$size = GetImageSize($data);

if($size[0] > $size[1])

{

$thumbnail_width = 100;

$thumbnail_height = (int)(100 * $size[1] / $size[0]);

}

else

{

$thumbnail_width = (int)(100 * $size[0] / $size[1]);

$thumbnail_height = 100;

}

echo "thumbnail_width = $thumbnail_width thumbnail_height = $thumbnail_height";

print "To upload another file <a href=http://www.mikesierra.co.uk/Members/formloader.html> Click Here</a>";

?>

[/code]

[Adam W]

Still get a similar error message

 

getimagesize(): Unable to access ÿØÿá8EExif in <b>/usr/local/psa/home/vhosts/etc

 

I am sucessfully retrieving a file which is an image file but it seems to be in the wrong format for getimagesize() to read

[corbin]

It appears to me that the actual image data is being stored in the database.

 

array getimagesize ( string $filename [, array &$imageinfo] )

 

Hmmm....

 

So you either need to find a function that can take the image contents (what you're pulling from your DB) and return the dimensions, or you're going to have to write temp files everytime you want to see the dimensions.

[Adam W]

I think writing temp files is about the only way I'm going to be able to do this - bit long winded but it should work!