Uploading a file --> FTP

[eXcellion]

Hi

 

I'm supposed to make a site (webshop) for school and it has to be done by tomorrow. There aren't any problems APART FROM this one thing..

 

So there's a page that will display the things people can 'buy', but I don't just want to add them in the html etc.

So I made an adminCP where I should be able to add items.. (the image of the article is uploaded to /images/bestellingen - or atleast it SHOULD be) and the url is put in the database with the rest of the information..

Naam - prijs - beschrijving - afbeelding

(equals name - price - description - image)

 

It all works apart from the image uploading --> ftp.. (the url IS put into the database correctly..)

I always get the reply 'fout' as a result of the if... else...

 

And yes the CHMOD of the /images/bestellingen/ folder is put to 777

 

Here you have the code.

 

HTML:

<table>
<form enctype="multipart/form-data" action="insert.php" method="post" >
<tr><td>Afbeelding:</td><td><input name="image" type="file"></td></tr>
<tr><td>Naam:</td><td><input name="naam" type="text" /></td></tr>
<tr><td>Prijs:</td><td><input name="prijs" type="text" /></td></tr>
<tr><td>Omschrijving:</td><td><textarea name="omschrijving" ></textarea></td></tr>
<tr><td></td><td><input value="Submit" type="submit"></td></tr>
</form>
</table>

 

Insert.php

<?php
$host="localhost";
$name = "correct";
$pass = "correct";
$dbname = "correct";

$dbi = mysql_connect($host, $name,$pass) or
die("Kan niet verbinden met de database. Error :" . mysql_error());
mysql_select_db($dbname,$dbi);

	$folder = "images/bestellingen/"; // folder where the images will be saved
	$file_name = base64_encode(rand().rand().rand()); // generate random name

	$path_info = pathinfo($_FILES['image']['name']); // Find extension
	$file_extension = $path_info['extension']; // put extension in var

	$target = $folder . $file_name .".". $file_extension; // where and with which filename will it be saved

	$source = $_FILES['image']['tmp_name']; // uploaded file (I think that the failure is located here..) 

	if(move_uploaded_file($source, $target)){  
		echo "afbeelding is geupload!";
	} else {
		echo "fout";
	} // Save file with new name on new location



      $naam = $_POST['naam'];
  $prijs = $_POST['prijs'];
  $omschrijving = $_POST['omschrijving'];
  
	$query = "INSERT INTO bestellingen (naam,prijs,beschrijving,image) VALUES
('$naam','$prijs','$omschrijving','$target')"; 
      $results = mysql_query($query, $dbi);
      

mysql_close($dbi);
?>

[grissom]

Try putting this line of code in your HTML

 

<INPUT TYPE = "HIDDEN" NAME = "MAX_FILE_SIZE" VALUE = "500000">

 

ie maybe like this

<tr><td><INPUT TYPE = "HIDDEN" NAME = "MAX_FILE_SIZE" VALUE = "500000">Afbeelding:</td><td><input name="image" type="file"></td></tr>

 

Next thing !! : Check the value of the variable $target.  Use an echo statement to print $target on the screen, it might not be what you want it to be.