當在cakephp3中不起作用

Question

Organizatinos與organType具有多對一關系

Tabel組織

id   |  name  | organType_id

1    |  a     | 1

2    |  b     | 1

3    |  c     | 2

4    |  d     | 3

表OrganType

id   |  name  

1    |  aa     

2    |  bb    

3    |  cc 

我在belongsTo=>'organType'模型中使用belongsTo=>'organType'

示范組織

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

        $this->belongsTo('organType', [
           'foreignKey' => 'organType_id',            
        ]);

    }
}

模型風琴類型

class organTypeTable extends table
{
    public function initialize(array $config)
    {    
        $this->table('organType');
        $this->primaryKey('id');
    }
}

調節器

$organization=TableRegistry::get('organizations');
$organization=$organization->find('all')->offset(0)->limit(30) ;  
debug($organization);

但結果是僅返回organizations表，我希望organization_type內部連接organs

不好的結果：

'sql' => 'SELECT organizations.id AS `organizations__id`, organizations.name AS `organizations__name`, organizations.organType_id AS `organizations__organType_id` FROM organizations organizations LIMIT 30 OFFSET 0',

我想要這個結果：

 'sql' => 'SELECT organizations.id , organizations.name , organizations.organType_id FROM organizations inner join organType on organizations.organType_id=organType.id

Answer 1

book.cakephp.org

用於從cakephp中的外部表中進行選擇3個繆斯使用了contain(['organType'])

$organization=$organization->find('all')->contain(['organType'])->offset(0)->limit(30) ;