CakePHP的模型关联 [英] Model associations for CakePHP

问题描述

我是CakePHP的新手，在弄清楚如何建立模型关联时遇到了一些麻烦。

I'm new to CakePHP and having a bit of trouble figuring out how to set up model associations.

说我有3个表：付款，预订和Reservation_details及其以下数据

Say I have 3 tables: payments, reservations and reservation_details with the following data

table reservations
id  |  confirmation_number   |   guest_id
1         123                        1

table reservation_details -a reservation can have multiple entries (multiple rooms)
id  |  reservation_id   |   date     |  time   | room_id    |   rate
 2           1            2014-18-04    13:00        1           9.99
 3           1            2014-18-04    13:00        2           4.99

table payments - many payments for one reservation can be made
id  |   reservation_id  |   payment_amount    |  payment_type   |   guest_id
 4            1                    14.98                Cash             1

这是我的模特协会

 //Reservation model
 public $hasMany = array('ReservationDetail', 'Payment');

 //ReservationDetail model
 public $belongsTo = array('Reservation');

 //Payment model
 public $belongsTo = array('Reservation');
 public $hasMany = array('ReservationDetail' => array('foreignKey' => 'reservation_id')); 

我要执行的操作是能够搜索付款，并会返回相应的该付款的Reservation和Reservation_details。因此，它将从reservation_details中获取共享相同reservation_id的所有记录。现在，预订已返回，但reserve_details返回为空

What I'm trying to do is be able to search for a payment and it would return the corresponding reservation and reservation_details for that payment. So it would grab any records from reservation_details that share the same reservation_id. Right now the reservation is returned, but the reservation_details is returned empty

以下搜索从付款和预订中返回信息，但从Reservation_details返回空数组。

The following search returns information from payments and reservations, but an empty array from reservation_details.

$payment = $this->Payment->find('all',array(
                                      'conditions' => array(
                                         'Payment.guest_id' => '1'
                                        )
                                ));

我几乎可以肯定，它正在加入payments.id = payments.reservation_id的reservation_details表而不是Payments.reservation_id = Reservation_details.reservation_id。当我手动将payments.id更改为1（reservation_id值）时，将返回reservation_details。

我相信我正在尝试的MySQL查询

I believe the MySQL query that I'm trying to achieve would be something like

SELECT reservations.*, reservation_details.*, payments.* from payments
INNER JOIN reservations on reservations.id = payments.reservation_id
INNER JOIN reservation_details on reservation_details.reservation_id = payments.reservation_ID
WHERE payments.guest_id = '1'

推荐答案

Payment.php

将可包含的行为添加为-

Payment.php

add containable behavior as-

public $actsAs = array('Containable');

PaymentsController.php

PaymentsController.php

$payments = $this->Payment->find('all', array(
                                            'conditions' => array(
                                                'Payment.guest_id' => '1'
                                            ),
                                            'contain' => array(
                                                'Reservation' => array(
                                                    'ReservationDetail'
                                                )
                                            )

));

debug($payments);

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