simple upload form

Question

**when I worked on a simple upload form I found something wrong in the result

first my code was :**

$filename= $_FILES ['file']['name']; $filesize= $_FILES ['file']['size']; $tmpname= $_FILES ['file']['tmp_size']; $filetype= $_FILES ['file']['type']; $folder = "upload/"; if(isset($_POST['do']) and $_POST['do']== 'upload'){ if(empty($filename)){ echo "the file is not exist"; } else if ($filesize > 2048){ echo " the file is biger than 2 MB"; }else{ echo "the file is uploaded"; move_uploaded_file($tmpname, $folder); } } echo " <form action='upload.php' method='post' enctype='multipart/form-data'> file path : <input type='file' name='file'/> <input type='submit' name='do' value='upload'/> </form> "

when the file size is less than 2048 the result always be "the file is bigger than 2048" although I'm sure it's less than 2 MB when i makes it 100000 to see the result what would be the result was "the file is uploaded" but I couldn't find the file in the upload folder anyone can help me ? what is the wrong ?

file path : <input type='file' name='file'/> <input type='submit' name='do' value='upload'/> </form> " — Hanan
– Hanan, Commented Dec 22, 2010 at 19:30

John Parker · Accepted Answer · 2010-12-22 19:42:19Z

4

$_FILES['xxx']['size'] is in bytes, not megabytes. As such, unless the file is less than 2KB, it'll be deemed too large.

To check the file is less than 2MB, use:

$filesize > 2097152

Additionally, you're setting $tmpname incorrectly. It should be...

$tmpname = $_FILES['file']['tmp_name'];

edited Dec 22, 2010 at 19:42

answered Dec 22, 2010 at 19:36

John Parker

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2 Comments

Hanan Over a year ago

thanks alot :) the wrong was (tmp_name) I wrote it wrong . thanks for your help

John Parker Over a year ago

@beginer in php - No worries. Incidentally, if you mark your preferred answer as accepted (click on the grey tick to the left of the answer) both you and the person who wrote the answer we receive reputation points. (The effective "currency" of Stack Overflow.)

simshaun · Accepted Answer · 2010-12-22 19:36:53Z

1

The filesize reported by PHP is in bytes, not KB.

Try else if ($filesize > 2097152)

answered Dec 22, 2010 at 19:36

simshaun

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Comments

dbwebtek · Accepted Answer · 2010-12-22 19:37:47Z

One way to test early is to echo the values of all just before you process them like once you submit the form then do checking values of file size

Hamdi · Accepted Answer · 2010-12-22 19:38:16Z

You should include the filename in the destination string.

move_uploaded_file($tmpname, $folder . $filename);

Also, you are relaying in the browser calculated file size, you should use this instead:

$filesize= filesize($_FILES ['file']['tmp_size']) / 1024; //To be in kilobytes as you expect

Leave the rest as it is!

Second part is wrong. filesize() needs the filename as input, and should be ` * 1024 ` not ` / 1024 `
Oh really!? Try learn basic mathmatics before converting bytes to kilobytes and commenting others! You shall review this before posting random comments: php.net/manual/en/features.file-upload.post-method.php

JakeParis · Accepted Answer · 2010-12-22 19:43:11Z

Why you can't find the file: 2nd argument of move_uploaded_file() needs to be in format 'path/to/file.jpg' so it should be:

$folder = "upload/" . $filename; move_uploaded_file($tmpname, $folder );

sikas · Accepted Answer · 2010-12-22 19:45:36Z

0

you can find here a tutorial on how to create simple upload form

answered Dec 22, 2010 at 19:45

sikas

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