CODEWITHSUNDEEP
phpmysqlcakephp
harryg
harryg

Reputation: 24077

CakePHP output a select input for picking a value from a joined table

I have managed to join 2 of my tables clients and risk_codes to each other where clients.risk_code_id is a foreign key for risk_codes.id.

In my edit view for Clients I ouput a form using the HTML Helper. E.g. to add an input to edit clients.name I would use echo $this->Form->input('name');

Given that risk_codes is a separate table/model how would I output a select dropdown with the options being risk_codes.name and the values being risk_codes.id?

The tables are linked like so:

Client belongsTo RiskCode
RiskCode hasMany Client

Upvotes: 0

Views: 975

Answers (1)

ndm
ndm

Reputation: 60463

In your RiskCode model make sure that displayField is either set to null or name (the latter is one of the defaults):

public $displayField = 'name'; // or null;

In the controller set a list of risk codes for the view:

$this->set('riskCodes', $this->Client->RiskCode->find('list'));

And in the view simply reference the appropriate foreign key field name:

echo $this->Form->input('risk_code_id');

CakePHP will automatically create an appropriate select list, using the models id and displayField field values from the list set as riskCodes.

ps. Many questions like this are answered in the Cookbook, and can also be figured out by using CakePHP to bake the controllers and views.

Upvotes: 1

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