Why this error comes as per i upload multiple images? - image

Question

When i upload multiple images from my code, everything is uploaded to mysql database, but displays error with saying success :
I just don't want this error.[Error is :Notice: Undefined variable: user_id in C:\wamp\www\swaminarayant\multiple_file_up.php on line 13]
Can anyone help me out of this?
My code is:
<?php include 'connect.php';?>
<?php
if(isset($_FILES['files'])){
$errors= array();
foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
$file_name = $key.$_FILES['files']['name'][$key];
$file_size =$_FILES['files']['size'][$key];
$file_tmp =$_FILES['files']['tmp_name'][$key];
$file_type=$_FILES['files']['type'][$key];
if($file_size > 2097152){
$errors[]='File size must be less than 2 MB';
}
$query="INSERT into upload_data (`USER_ID`,`FILE_NAME`,`FILE_SIZE`,`FILE_TYPE`) VALUES('$user_id','$file_name','$file_size','$file_type'); ";
$desired_dir="user_data";
if(empty($errors)==true){
if(is_dir($desired_dir)==false){
mkdir("$desired_dir", 0700); // Create directory if it does not exist
}
if(is_dir("$desired_dir/".$file_name)==false){
move_uploaded_file($file_tmp,"$desired_dir/".$file_name);
}else{ // rename the file if another one exist
$new_dir="$desired_dir/".$file_name.time();
rename($file_tmp,$new_dir) ;
}
mysql_query($query);
}else{
print_r($errors);
}
}
if(empty($error)){
echo "Success";
}
}?
<form action="" method="POST" enctype="multipart/form-data">
<input type="file" name="files[]" multiple/>
<input type="submit"/>
</form>

Answer 1

Look at your Code around Line 13:
$query="INSERT into upload_data (`USER_ID`,`FILE_NAME`,`FILE_SIZE`,`FILE_TYPE`) VALUES
('$user_id','$file_name','$file_size','$file_type'); ";
Within the VALUES part you try to include a variable $user_id which was never defined above.
Youre setting the other values properly:
$file_name = $key.$_FILES['files']['name'][$key];
$file_size =$_FILES['files']['size'][$key];
$file_type=$_FILES['files']['type'][$key];
But you are missing a definition for $user_id. If you insert for example:
$user_id = "foo";
Or
$user_id = 1;
In case it must be a number you should be all good.