Why I have empty record in MySQL? - php

Question

In our system they asked us to add research interest but when we add new user, we should assign an interest to him/her. It's adding successfully to the database table, we have problem which is we get zero record in the interest_id database table, why?
Here all research interest added successfully to database:
As you see in picture below when admin add a new user he chooses multiple interests and assign them to a user:
Here is screenshot of user profile as you can see interests that admin assigned to new user:
The problem is that it don't write interest id to database?
Below you can see the codes and SQL table:
<p>
<label>Research Areas</label>
<select name="interest_id[]" class="small-input" multiple>
<?php
$query = "SELECT * FROM research_interest ORDER BY name ASC";
$result = mysqli_query($link, $query);
while($row = mysqli_fetch_object($result))
{
?>
<option value="<?php echo $row->id; ?>"><?php echo $row->name; ?></option>
<?php
}
?>
</select>
</p>
Here is also codes for MySQL:
if(isset($_POST['action'])) {
switch($_POST['action']) {
case 'user_add':
$user_id = mysqli_real_escape_string($link,$_POST['user_id']);
$email = mysqli_real_escape_string($link,$_POST['email']);
$password = rand_pass(8);
$title_id = intval($_POST['title_id']);
$user_type_id = intval($_POST['usertype_id']);
$department_id = intval($_POST['department_id']);
$view_publication = mysqli_real_escape_string($link,$_POST['publication']);
$referee = mysqli_real_escape_string($link,$_POST['referee']);
$gender = mysqli_real_escape_string($link,$_POST['gender']);
$creation_date = date('Y.m.d');
//Interest
if(!empty($_POST['interest_id']) && is_array($_POST['interest_id'])) {
foreach ($_POST['interest_id'] as $interest)
{
$query = "INSERT INTO user_interest (user_id, interest_id)
VALUES ('$user_id', '$interest')";
$result = mysqli_query($link,$query);
}
}
//SQL Query
$query = "INSERT INTO user (user_id, email, title, user_type_id, department_id, referee, gender, password, view_publication, creation_date)
VALUES ('$user_id', '$email', '$title_id', '$user_type_id', '$department_id', '$referee', '$gender', '$password', '$view_publication', '$creation_date')";
$result = mysqli_query($link,$query);
echo '<script type="text/javascript">';
echo 'window.location = "mail.php?action=user_mail&user_id='.$user_id.'"';
echo '</script>';
break;
}
}
And this is a user_interest table shows that these three research IDs assign to user 1
but here it show that interest_id is zero , it don't shows ids of research interest that added to user 1

Answer 1

You are not including interest_id in your Insert query to the users table. So the table is assigning a default value of 0.

Answer 2

You're not inserting anything into the "interest_id" column in the second query so it's defaulting to zero. It looks like you're trying to set up a many to many relationship (one user can have many interests and one interest can belong to many users). In that case you'll need to set up a junction table (http://en.wikipedia.org/wiki/Junction_table) to handle the relationship.