cakephp3两个外键同型号

Question

Tabel coding

id |  level1  |  level2  |  title
_____________________________________
1  |  100     |  1       |  numberOne
2  |  100     |  2       |  NumberTwo
3  |  101     |  1       |  Compay
4  |  101     |  2       |  association

表Organizations

id  | name  |  type  | level
____________________________
1   | intel |  3     |  1 
2   | IBM   |  4     |  2

模型Organizations

class organizationsTable extends table
{    
    public function initialize(array $config)
    {       
        $this->table('organizations');
        $this->primaryKey('id');

        $this->belongsTo('coding', [
            'foreignKey' => 'type',
            'joinType' => 'INNER',      
            ],
            'coding', [
            'foreignKey' => 'level',
            'joinType' => 'INNER']);

    }
}

此代码$organization->find('all')->contain(['coding'])仅将level设置为foreignKey，但我希望将level和type设置为foreignKey for table organizations

像我这样的SQL查询

select * from organizations inner join coding on organizations.type=coding.id inner join coding ccc on organizations.level=ccc.id

如何创建这样的连接？

修改

我写这个，但说错误：organizations is not associated with coding

$this->belongsTo('yyyy', [
    'foreignKey' => 'type',
    'className' => 'coding' 
    ],
    'xxxx', [
    'foreignKey' => 'level',
    'className' => 'coding'
]);

Answer 1

在cakephp2 Multiple relations to the same model

中

但是在cakephp3中尝试这样

$this->belongsTo('aliasName1', [
    'foreignKey' => 'type',
    'className' => 'coding'
    ]);
$this->belongsTo('aliasName2', [
    'foreignKey' => 'level',
    'className' => 'coding' 
    ]);

控制器

$organization->find('all')->contain(['aliasName1'])->contain(['aliasName2']);