Hi people!
I'm having a problem with a file upload script that I'm writing. It
works with Firefox but not Internet explorer.
When the form is submitted(to it's self) it check for a file upload
using this:
$room_type = $_POST['room_type'];
$sleeps = $_POST['sleeps'];
$description = $_POST['description'];
$count = $_POST['count'];
$cost_per_night = $_POST['cost_per_night'];
// ==============
// Configuration
// ==============
$uploaddir = "../account_images"; // Where the files
to upload
//$thumbs_dir ="account_images/thumbs"; // Where the
thumbnail uploads to
// ==============
// Upload Part
// ==============
if(is_uploaded_file($_FILES['file']['tmp_name']))
{
if(($_FILES['file']["type"] == "image/jpeg") ||
($_FILES['file']["type"] == "image/gif") ||
($_FILES['file']["type"] == "image/png"))
{ //200kb
move_uploaded_file($_FILES['file']
['tmp_name'],$uploaddir.'/'.$_FILES['file']['name']);
$current_file_uploaded = $_FILES['file']
['name'];
$picture_name = $current_file_uploaded;
}
}
Then it runs a function which inputs the info from the form into a db
table like so:
add_room_type_to_db($room_type,$sleeps,$descriptio n,$count,
$cost_per_night,$picture_name,$connection);
It works fine when i test it in FF, the image is uploaded and the
image name is inserted into the database, but when i use IE, i get an
E_NOTICE :
"Notice: Undefined variable: picture_name in C:\webserver\webroot
\booking_system\staff\controls.php on line 208"
Line 208 being the line that calls the function "add_room_type_to_db".
Any ideas why this may be happening?
Please help!
Paul 3 2264
On Tue, 14 Aug 2007 01:05:47 +0200, macca <pt*******@googlemail.comwrote:
Hi people!
I'm having a problem with a file upload script that I'm writing. It
works with Firefox but not Internet explorer.
When the form is submitted(to it's self) it check for a file upload
using this:
$room_type = $_POST['room_type'];
$sleeps = $_POST['sleeps'];
$description = $_POST['description'];
$count = $_POST['count'];
$cost_per_night = $_POST['cost_per_night'];
// ==============
// Configuration
// ==============
$uploaddir = "../account_images"; // Where the files
to upload
//$thumbs_dir ="account_images/thumbs"; // Where the
thumbnail uploads to
// ==============
// Upload Part
// ==============
if(is_uploaded_file($_FILES['file']['tmp_name']))
{
if(($_FILES['file']["type"] == "image/jpeg")||
($_FILES['file']["type"] == "image/gif") ||
($_FILES['file']["type"] == "image/png"))
{ //200kb
move_uploaded_file($_FILES['file']
['tmp_name'],$uploaddir.'/'.$_FILES['file']['name']);
$current_file_uploaded = $_FILES['file']
['name'];
$picture_name = $current_file_uploaded;
}
}
add_room_type_to_db($room_type,$sleeps,$descriptio n,$count,
$cost_per_night,$picture_name,$connection);
It works fine when i test it in FF, the image is uploaded and the
image name is inserted into the database, but when i use IE, i get an
E_NOTICE :
"Notice: Undefined variable: picture_name in C:\webserver\webroot
\booking_system\staff\controls.php on line 208"
Line 208 being the line that calls the function "add_room_type_to_db".
Any ideas why this may be happening?
The conditional is not run, either because there's no transfer, or the
mimetype is something you don't expect. print_r($_FILES) to find out.
Also, a 'type' in an upload is something provided by the user and highly
unreliable. For images, I usually use getimagesize() to check wether it's
recognized as one.
--
Rik Wasmus
Hi thanks for the quick reply.
You are right. it was the "type" causing it. I used print_r and it
gave me this:
Array ( [file] =Array ( [name] =DSCF0495.JPG [type] =image/pjpeg
[tmp_name] =C:\WINDOWS\TEMP\php11.tmp [error] =0 [size] =>
115829 ) )
The type was coming out as "image/pjpeg" for some reason.
I changed it to this and it now works fine. Thanks.
if(is_uploaded_file($_FILES['file']['tmp_name']))
{
print_r($_FILES);
$img_mime_type = getimagesize($_FILES['file']
['tmp_name']);
if(
($img_mime_type[2] == IMAGETYPE_GIF) ||
($img_mime_type[2] == IMAGETYPE_JPEG)||
($img_mime_type[2] == IMAGETYPE_PNG)
)
{ //200kb
move_uploaded_file($_FILES['file']
['tmp_name'],$uploaddir.'/'.$_FILES['file']['name']);
$current_file_uploaded = $_FILES['file']
['name'];
$picture_name = $current_file_uploaded;
//CREATE THUMBNAIL OF UPLOADED IMAGE
========================= JPG PNG
//createthumb($uploaddir.'/'.
$current_file_uploaded,$uploaddir.'/thumbs/thumb_'.
$current_file_uploaded,150,150);
} else {
die('You may only upload file types
JPEG/JPG, PNG or GIF.');
}
}
Regards,
Paul
On Tue, 14 Aug 2007 01:51:41 +0200, macca <pt*******@googlemail.comwrote:
Hi thanks for the quick reply.
You are right. it was the "type" causing it. I used print_r and it
gave me this:
Array ( [file] =Array ( [name] =DSCF0495.JPG [type] =image/pjpeg
[tmp_name] =C:\WINDOWS\TEMP\php11.tmp [error] =0 [size] =>
115829 ) )
The type was coming out as "image/pjpeg" for some reason.
I changed it to this and it now works fine. Thanks.
if(is_uploaded_file($_FILES['file']['tmp_name']))
{
print_r($_FILES);
$img_mime_type = getimagesize($_FILES['file']
['tmp_name']);
if(
($img_mime_type[2] == IMAGETYPE_GIF) ||
($img_mime_type[2] == IMAGETYPE_JPEG)||
($img_mime_type[2] == IMAGETYPE_PNG)
)
I'd use:
if($img_mime_type && ($img_mime_type[2] == IMAGETYPE_GIF ||
$img_mime_type[2] == IMAGETYPE_JPEG || $img_mime_type[2] == IMAGETYPE_PNG))
As getimagesize() will return false for something not recognized as an
image.
--
Rik Wasmus This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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