Uploaded files appear as Array in database and userid isn't saving - php

Question

Wondering if you can help?
I am currently working on a script that allows an admin to upload a file to the server. When the file uploads it takes information from the form and stores it in the database, to later allow users who have permissions to download it.
I have managed to make the form:
Select File To Upload
<div style="width:100%; margin:5px;">
<form action="uploadclientfile.php" method="post" enctype="multipart/form-data" name="upload" style="margin:5px;">
<label> File</label><br /><br />
<input name="uploaded_file" type="file" class="input-xlarge" required/>
<input type="hidden" name="MAX_FILE_SIZE" value="1000000" /><br /><br />
<label>Select Users Who Can Download The File</label><br /><br />
<select name="users[]" multiple="multiple" style="width:300px">
<?php
$useruploadids = mysql_query("SELECT member_id, firstname, lastname FROM members");
while ($row = mysql_fetch_assoc($useruploadids)) {
$user_id = $row['userid'];
$firstname = $row['firstname'];
$lastname = $row['lastname'];
?>
<option value=""<?php echo $user_id ?>"><?php echo $firstname ?><?php echo $lastname ?></option>
<?php } ?>
<input name="Upload" type="submit" value="Upload" class="btn" />
</form>
So this will create a multiselect box with all users from the database to select more than one if required. It calls their userid first name and last name. But it doesn't display the user id. Which is fine.
The next file is the uploading file, the issue I am encountering is in the database. It isn't recording the userids selected and the filename is always coming up as array. Although it does save the file location correctly.
Any idea what is wrong?
<?php
$target = "clientfiles/";
$target = $target . basename( $_FILES['uploaded_file']['name']);
$userid=$_POST['users[]'];
$file=($_FILES['uploaded_file']);
mysql_connect("localhost", "username", "password") or die(mysql_error()) ;
mysql_select_db("database") or die(mysql_error()) ;
mysql_query("INSERT INTO `cfiles` VALUES ('$userid', '$file', '$target')") ;
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploadedfile']). " has been uploaded, and your information has been added to the directory";
}
else {
echo "Sorry, there was a problem uploading your file.";
}
?>
Any clues?
Thank you for reading

Answer 1

Given that your allowing the user to select multiple userids with the form notation [] the following $userid=$_POST['users[]']; will be an array. Given $userid is an array when you storing it as a string the value Array() will be stored instead of the actual user_id value. If you database table field user_id is of type INT then string value of Array() can't even be stored it will result in 0.
If you goal is to store multiple values, converting into multiple records you should iterate through the $user_id array and store them as follows
foreach ($userid as $id) {
mysql_query("INSERT INTO `cfiles` VALUES ('$id', '$file', '$target')")
}