Foreign key constraint when using drop down in CakePHP - php

Question

I am using a drop down box with a foreign key relationship. I have got the drop down filling in the correct values but the only problem is when I add a user there is a foreign key constraint. But I can make users if I just use the normal input box and type an id that exists in the other table.
For example when I enter the id with this in my add.ctp, it works:
echo $this->Form->input('location');
but when I use this it doesn't
echo $this->Form->input('location_id', array('type' => 'select', 'options' => $CompanyLocations));
This is my add function in my UsersController
public function add()
{
$user = $this->Users->newEntity();
if ($this->request->is('post')) {
$user = $this->Users->patchEntity($user, $this->request->data);
if ($this->Users->save($user)) {
$this->Flash->success(__('The user has been saved.'));
return $this->redirect(['action' => 'index']);
}
$this->Flash->error(__('The user could not be saved. Please, try again.'));
}
$CompanyLocations= $this->Users->CompanyLocations->find('list');
$this->set(compact('CompanyLocations'));
$this->set(compact('user'));
$this->set('_serialize', ['user']);
This is in my UsersTable
$this->belongsTo('CompanyLocations');
and my CompanyLocationsTable
public function initialize(array $config)
{
parent::initialize($config);
$this->table('company_locations');
$this->displayField('location_name');
$this->primaryKey('location_id');
$this->belongsTo('Locations', [
'foreignKey' => 'location_id',
'joinType' => 'INNER'
]);
}
and my MySQL code
CREATE TABLE IF NOT EXISTS southpac_team.company_locations (
location_id INT NOT NULL AUTO_INCREMENT,
location_name VARCHAR(45) NULL,
PRIMARY KEY (location_id))
ENGINE = InnoDB;
DROP TABLE IF EXISTS southpac_team.users ;
CREATE TABLE IF NOT EXISTS southpac_team.users (
id INT NOT NULL AUTO_INCREMENT,
username VARCHAR(20) NOT NULL,
password VARCHAR(255) NOT NULL,
name VARCHAR(255) NOT NULL,
department INT NULL,
mobile VARCHAR(255) NULL,
email VARCHAR(255) NULL,
extension INT NULL,
lame_number INT NULL,
spa_auth_number VARCHAR(15) NULL,
creation_date DATE NULL,
picture VARCHAR(255) NULL,
employed TINYINT(1) NOT NULL,
location INT NOT NULL,
PRIMARY KEY (id),
INDEX get location_idx (location ASC),
CONSTRAINT get location
FOREIGN KEY (location)
REFERENCES southpac_team.company_locations(location_id)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;

Answer 1

Naming conventions
You are not following the naming conventions, by default the foreign key name for a belongsTo association is the singular underscored variant of the association alias, postfixed with _id so in the case of CompanyLocations that would be company_location_id, not just location.
echo $this->Form->input('company_location_id');
Also the variable holding the list should use camel casing, then you don't even need to specify it via the options argument:
$companyLocations= $this->Users->CompanyLocations->find('list');
$this->set(compact('companyLocations'));
Change the association defaults
If you are working with a legacy database that you cannot modify, then you need to configure CakePHP accordingly, ie specify the custom foreign key via the options argument of Table::belongsTo().
$this->belongsTo('CompanyLocations', [
'foreignKey' => 'location'
]);
Bake gets confused
The belongsTo association in CompanyLocationsTable looks fishy too, unless you really have a LocationsTable that should be associated with CompanyLocationsTable via:
company_locations.location_id > locations.primary_key
I guess you've created the model via bake, which treated location_id as a foreign key since it matches the default foreign key naming scheme for a belongsTo association.
See also
Cookbook > CakePHP at a Glance > Conventions > Model and Database Conventions
Cookbook > Database Access & ORM > Associations - Linking Tables Together > BelongsTo Associations
Cookbook > Views > Helpers > Form > Creating Form Controls
Cookbook > Views > Helpers > Form > Creating Inputs for Associated Data