How to make users able to upload images in a form? - php

Question

I have a classifieds website where users must fill in a form in order to put a ad. The form consists of name, password, category, specifications etc etc.
Now, I need to add a image upload function into this form, which must have the following:
1- Upload up to 5 images.
2- A 'remove image link' beneath each image if the user wants another image instead.
How would you do this?
Thanks
Best would be if there was a plugin or something to Jquery which is easy to modify...

Answer 1

jQuery Multiple File Upload
You can limit the number of uploads using the max option or passing a number as the only parameter. More info on the Examples tab.

Answer 2

If you want to upload it as a file to an SQL database, have something like this in the form (in php echo format):
echo "<input type=\"hidden\" name=\"MAX_FILE_SIZE\" value=\"2000000\">
<input name=\"userfile\" type=\"file\" id=\"userfile\">";
Then in your form's receiving php page put something like this:
if ($_FILES['userfile']['size'] > 0) {
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$fp = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}
$query = "INSERT INTO files (name, size, type, content ) VALUES ('$fileName', '$fileSize', '$fileType', '$content')";
mysql_query($query) or die('Error, query failed');
$thisq = mysql_query("SELECT * FROM `files` WHERE 1 ORDER BY `id` DESC LIMIT 1");
$fileidnumber= mysql_fetch_array($thisq);
}
That will store the file to a database and then return the key for you to save or use however you'd like. You can then create a page to download the files like this:
<?php
import_request_variables(gp);
if(isset($_GET['id'])) {
// if id is set then get the file with the id from database
$id = $_GET['id'];
$query = "SELECT name, type, size, content " .
"FROM `files` WHERE id = '$id'";
$result = mysql_query($query) or die('Error, query failed');
list($name, $type, $size, $content) = mysql_fetch_array($result);
header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $content;
exit;
}
?>