php - CakePHP3 在同一模型中将表链接在一起(内连接)

Question

我有一个名为“customers”的表，其架构如下(缩短):

id | customer_id | type | salutation | forename | surname | created | modified

在此表中，我想存储可能关联的人员:

id | customer_id | type    | salutation | forename | surname | created             | modified
 1 | NULL        | husband | Mr.        | John     | Doe     | 2016-01-05 10:00:00 | 2016-01-05 10:00:00
 2 |  1          | wife    | Mrs.       | Jane     | Doe     | 2016-01-05 10:01:00 | 2016-01-05 10:01:00
 3 |  1          | child   | Mr.        | Jim      | Doe     | 2016-01-05 10:02:00 | 2016-01-05 10:02:00

Customers with "customer_id" = NULL are the master customers, but #2 and #3 refers to #1.

I've created the table in phpmyadmin and do: "bin/cake bake all customers" without errors. Then I created the 'master customer'. When I create the second account, I expect that the select-field shows customer #1, but the dropdown-field is empty.

The model:

public function initialize(array $config)
{
    parent::initialize($config);

    $this->table('customers');
    $this->displayField('id');
    $this->primaryKey('id');

    $this->belongsTo('Customers', [
        'foreignKey' => 'customer_id',
        'joinType' => 'INNER'
    ]);
    $this->hasMany('Customers', [
        'foreignKey' => 'customer_id'
    ]);
}

如果您需要更多信息或更多代码，请告诉我。
提前谢谢了。

最诚挚的问候
马丁

<小时/>

重现 SQL

CREATE TABLE IF NOT EXISTS `customers` (
  `id` int(11) NOT NULL,
  `typ` varchar(10) NOT NULL,
  `customer_id` int(11) DEFAULT NULL,
  `salutation` varchar(4) NOT NULL,
  `prename` varchar(255) NOT NULL,
  `surname` varchar(255) NOT NULL,
  `created` datetime NOT NULL,
  `modified` datetime NOT NULL
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1;

INSERT INTO `customers` (`id`, `typ`, `customer_id`, `salutation`, `prename`, `surname`, `created`, `modified`) VALUES
(1, 'husband', NULL, 'Mr.', 'John', 'Doe', '2016-03-02 21:26:32', '2016-03-02 21:26:32'),
(2, 'wife', 1, 'Ms.', 'Jane', 'Doe', '2016-03-02 21:27:25', '2016-03-02 22:10:05'),
(3, 'child', 1, 'Mr.', 'Jim', 'Doe', '2016-03-02 21:27:41', '2016-03-02 22:10:15');

ALTER TABLE `customers` ADD PRIMARY KEY (`id`), ADD KEY `customers_fk0` (`customer_id`);
ALTER TABLE `customers` MODIFY `id` int(11) NOT NULL AUTO_INCREMENT,AUTO_INCREMENT=4;
ALTER TABLE `customers` ADD CONSTRAINT `customers_fk0` FOREIGN KEY (`customer_id`) REFERENCES `customers` (`id`);

然后，运行

bin/cake bake all customers

<小时/>

编辑 我查了一下，possible duplicate ，并改变了我的模型

$this->belongsTo('Customers', [
    'foreignKey' => 'customer_id',
    'joinType' => 'INNER'
]);
$this->hasMany('ChildCustomers', [ // <-- changed this line
    'className' => 'Customers',    // <-- added this line
    'foreignKey' => 'customer_id'
]);

现在，如果我尝试添加/编辑，则不会更改选择。但如果我查看现有客户，我会得到:

Error: SQLSTATE[42000]: Syntax error or access violation: 1066 Not unique table/alias: 'Customers'

这是显示选择的代码:

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

此代码是由“bin/cake eat allcustomers”生成的

<小时/>

编辑2

我会总结一下我目前的状态:

型号:

public function initialize(array $config)
{
    parent::initialize($config);

    $this->table('customers');
    $this->displayField('id');
    $this->primaryKey('id');

    $this->belongsTo('ParentCustomers', [
        'className' => 'Customers',
        'foreignKey' => 'customer_id',
        'joinType' => 'INNER'
    ]);
    $this->hasMany('ChildCustomers', [
        'className' => 'Customers',
        'foreignKey' => 'customer_id'
    ]);
}

public function buildRules(RulesChecker $rules)
{
    $rules->add($rules->existsIn(['customer_id'], 'ParentCustomers'));
    return $rules;
}

Controller :

public function view($id = null)
{
    $customer = $this->Customers->get($id, [
        'contain' => ['ParentCustomers']
    ]);

    $this->set('customer', $customer);
    $this->set('_serialize', ['customer']);
}

如果用户有关联(妻子、 child )，则 View 会在页面底部显示标题“相关客户”，但不会显示其他显示相关客户的表格。如果用户具有“customer_id = 0”，则 View 显示“在表“customers”中找不到记录”。

我还添加了

$this->set('customers', $this->Customers->find('list'));

到 add() 和 edit() 函数，但我发现没有办法允许空值。

澄清一下: 稍后，首页 (index()) 应该只列出 customer_id = 0 的“主客户”和一个小嵌套表，如果他有妻子和/或 child 的话。

我认为我走在正确的道路上......是吗？

再次感谢

Answer 1

您还应该使用 ParentCustomers 作为第一个关联。 Customers 重复自身。

$this->belongsTo('ParentCustomers', [ // <-- You need to change here
    'className' => 'Customers', // <-- You need to change here
    'foreignKey' => 'customer_id',
    'joinType' => 'INNER'
]);
$this->hasMany('ChildCustomers', [
    'className' => 'Customers',
    'foreignKey' => 'customer_id'
]);