$ _POST PHP UPDATE错误而不是发送选项

I am trying $_POST an UPDATE to the following mysql table:
Note: there are 3 dropdowns: status, category and access

app_generalData
---------------
app_id
table
status_id
category_id
tags
access_id

Resulting tests with print_r($_POST) and echo(s) 's ensure the form $_POST is working:

Array ( 
        [MAX_FILE_SIZE] => 100000
        [app_id] => 1 
        [title] => Nothing Special 
        [status] => 
        [category] => 
        [tags] => new tag 
        [access] => 
        [update] => Update ) 

Data from the form.

• ID: 1
• Title: Nothing Special
• Status:
• Category:
• Tags: new tag
• Access:

Resulting Error message:

Error querying database for General Data.

Desired result(s):

1. $_POST items updating in the db

Snippet for UPDATE:

// Post the UPDATE to app_generalData
if (isset($_POST['update'])) {
  // print_r($_POST);
  // echo '<br />';

// Connect to the database
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

// Grab the data from the POST

  // General Data
  $app_id = mysqli_real_escape_string($dbc, trim($_POST['app_id']));
  $title = mysqli_real_escape_string($dbc, trim($_POST['title']));
  $status = mysqli_real_escape_string($dbc, trim($_POST['status']));
  $category = mysqli_real_escape_string($dbc, trim($_POST['category']));
  $tags = mysqli_real_escape_string($dbc, trim($_POST['tags']));
  $access = mysqli_real_escape_string($dbc, trim($_POST['access']));

// Confirm success with the user

    echo '<h1>Data from the form.</h1><br />';
    echo 'ID: ' . $app_id .'<br />';
    echo 'Title: ' . $title .'<br />';
    echo 'Status: ' . $status .'<br />';
    echo 'Category: ' . $category .'<br />';
    echo 'Tags: ' . $tags .'<br />';
    echo 'Access: ' . $access .'<br />';

    echo '<br />';

// Write the data to the database
  $query = "UPDATE      app_generalData
            SET         title = $title,
                        status_id = $status,
                        category_id = $category,
                        tags = $tags,
                        access_id = $access
            WHERE       app_id = $app_id
            ";

  mysqli_query($dbc,$query)
        or die('Error querying database for General Data.');

// close MySQL
  mysqli_close($dbc);

  exit();
}
else {
  echo 'Please enter all details below.';
}