我在CakePHP 3.0中进行了一些有点复杂的嵌套连接：
$query = $this->find()
->contain([
'A',
'A.B',
'A.B.C',
'A.B.C.D',
'A.B.C.D.Z'
]);
A，B，C，D和Z之间的关系如下：
B hasMany A B hasMany C C hasMany B C hasMany D D hasMany C Z hasMany D 在contain()数组中，我可以A，A.B，A.B.C和A.B.C.D，但不 { {1}}。给出的错误消息是：
A.B.C.D.Z我没有找到D is not associated with Z (InvalidArgumentException)
Could this be caused by using AutoTables?
Please try correcting the issue for the following table aliases:
• D
的任何拼写错误的引用。
如何在此加入声明中包含D？
我的Z型号：
C我的<?php
namespace App\Model\Table;
use Cake\ORM\Table;
class CTable extends Table
{
public function initialize(array $config)
{
$this->table('c');
$this->primaryKey('id');
$this->hasMany('D', [
'className' => 'D',
'foreignKey' => false,
'conditions' => ['D.id' => 'C.d_id']
]);
$this->hasMany('B', [
'className' => 'B',
'foreignKey' => false,
'conditions' => ['B.id' => 'C.b_id']
]);
}
}
型号：
D我的<?php
namespace App\Model\Table
use Cake\ORM\Table;
class DTable extends Table
{
public function initialize(array $config)
{
$this->table('z');
$this->primaryKey('id');
$this->hasMany('C', [
'className' => 'C',
'foreignKey' => 'd_id'
]);
$this->belongsTo('Z', [
'className' => 'Z',
'foreignKey' => 'z_id'
]);
}
}
型号：
Z我的<?php
namespace App\Model\Table;
use Cake\ORM\Table;
class ZTable extends Table
{
public function initialize(array $config)
{
$this->table('z');
$this->primaryKey('id');
$this->hasMany('D', [
'className' => 'D',
'foreignKey' => 'z_id'
]);
// Unrelated to the join statement, but in here nevertheless
$this->belongsTo('X', [
'className' => 'X',
'foreignKey' => 'x_id',
]);
}
}
表格列：
C我的id | b_id | d_id
表格列：
D我的id | z_id | data
表格列：
Z答案 0 :(得分：0)
您的文件需要分别命名为DTable.php CTable.php和类class DTable extends Table。
答案 1 :(得分：0)
据我所知，问题是通过“hasMany”关系在 C 和 D 之间进行递归。 也许将这种关系在 C 中重命名如下：
$this->hasMany('aliasForD', [
'className' => 'D',
'foreignKey' => false,
'conditions' => ['aliasForD.id' => 'C.d_id']
]);