Getting names to show in dropdown list with cakephp

Question

I want to show the names of all our project leaders in a dropdown list.

The project leaders are only some of the employees that work in the company.

Here are my tables:

project_leaders
,----,----------------,
| id | hr_employee_id |
|----|----------------|
| 1  |  18            |
'----'----------------'

projects
,----,------,-------------------,
| id | name | project_leader_id |
|----|------|-------------------|
| 1  | cake |         1         |
'----'------'-------------------'

hr_employees
,----,------,---------,
| id | name | surname |
|----|------|---------|
| 18 | joe  | dirt    |
'----'------'---------'

My ProjectsController looks like this:

public function add() {
    if ($this->request->is('post')) {
        $this->Project->create();
        if ($this->Project->save($this->request->data)) {
            $this->_sendProjectRequestEmail($this->Project->getLastInsertID() );
            $this->Session->setFlash(__('The project has been saved'));
            $this->redirect(array('action' => 'index'));
        } else {
            $this->Session->setFlash(__('The project could not be saved. Please, try again.'));
        }
    }
    $this->set('projectLeaders',$this->Project->ProjectLeader->find('list');
}

This only returns the id of the Project Leaders, not the name and surname. So instead of Joe Dirt, it returns 1.

I've tried doing $this->Project->ProjectLeader->HrEmployee->find('list') but that lists all the employees.

I've also tried specifying the fields, but it returns a unknown field error.

What am I doing wrong?

Answer 1

$this->set('projectLeaders',$this->Project->ProjectLeader->find('list'));

This will just list the records from the project_leaders table and most likely, as the table doesn't itself contain a name/title field (which cake would pick up automatically as the displayField) be like so:

array(
    1 => 1
)

Use an appropriate find

To get a meaningful list of project leaders, you need to ensure you get a join between project_leaders and hr_employees one way to do that is using the containable behavior and simply specifying which fields to use in the list:

$projectLeaders = $this->Project->ProjectLeader->find('list', array(
    'contain' => array(
        'HrEmployee'
    ),
    'fields' => array(
        'ProjectLeader.id',
        'HrEmployee.name',
    )
));
$this->set(compact('projectLeaders'));

Is your db structure appropriate for your needs?

Having a table equivalent to saying "this user is admin" may not be the best idea - it would be easier to avoid data problems, and give you simpler queries, if the table project_leaders was (only) a boolean field on your hr_employees table and project_leader_id pointed at the hr_employee table, not some other data-abstraction.

Of course I don't know your whole schema, there very well may be a good reason for having a seperate project_leaders table.

Alternatively - denormalize

If you add a name field to project_leaders - you don't need a join to know the name of a project leader or any funkiness:

alter table project_leaders add name varchar(255) after id;
update project_leaders set name = (select name from hr_employees where hr_employees.id = hr_employee_id);

In this way you can know who is the relevant project lead with one query/join instead of needing to do two joins to get an employee's name.

Answer 2

You forgot to define the displayField:

The default case is

public $displayField = 'name';

And only for an existing "name"/"title" field in this table cake will automatically know what your drop-downs should get as label text. To use the fields of other tables, make sure you pass fields in your find() options:

'fields' => array('Project.id', 'Project.name');

etc. Cake will then know: the first one is the key, and the second one the display value.

If you need combined/custom fields/texts in your dropdowns use virtualFields

public $virtualFields = array(
    'full_name' => 'CONCAT(...)'
);

and set the displayField to this full_name virtual field then or use fields as explained above.

Answer 3

You could add a virtual field to your ProjectLeader Model:

public $virtualFields = array (
'name' => 'SELECT name FROM hr_employees AS Employee WHERE Employee.id = ProjectLeader.employee_id'
);

Also, add this virtual field as your displayField:

public $displayField = 'name';