Question

来自$this->request-data的数据需要修改，因此我不能直接使用表单来填充所有表。在添加列＆＃34; state＆＃34;。

之前，所有代码都已被烘焙

order_id    int(11) PK
part_id int(11) PK
state   varchar(10)

我从$this->request-data获取ID并相应地修改它：

$this->request->data['parts']['_ids'][]=....

其余的按照手册进行：

$order = $this->Orders->newEntity();
$this->Orders->patchEntity($order, $this->request->data);
$this->Orders->save($order);

我无法修改传入数据以将其保存到状态列

编辑：

＆＃34;州＆＃34;字段仍为空值。

OrdersTable中

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

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

    $this->addBehavior('Timestamp');

    $this->belongsTo('Users', [
        'foreignKey' => 'user_id',
        'joinType' => 'INNER',
    ]);
    $this->belongsToMany('Parts', [
        'foreignKey' => 'order_id',
        'targetForeignKey' => 'part_id',
        'joinTable' => 'orders_parts',
        'through' => 'OrdersParts',

    ]);
    $this->belongsToMany('Sets', [
        'foreignKey' => 'order_id',
        'targetForeignKey' => 'set_id',
        'joinTable' => 'orders_sets',
    ]);
}

OrdersTable中

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

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

    $this->addBehavior('Timestamp');


    $this->belongsTo('Orders', [
        'foreignKey' => 'order_id',
        'joinType' => 'INNER',
    ]);

    $this->belongsToMany('Orders', [
        'foreignKey' => 'part_id',
        'targetForeignKey' => 'order_id',
        'joinTable' => 'orders_parts',

    ]);

    $this->belongsToMany('States', [
        'foreignKey' => 'part_id',
        'targetForeignKey' => 'state_id',
        'joinTable' => 'parts_states'
    ]);
}

插入

$order = $this->Orders->newEntity();
$order->status = 'new';
$order->user_id = $this->Auth->user('id');
$this->request->data['parts'][0]['_joinData']['state'] = "whatever";
$this->Orders->patchEntity(
$order,
$this->request->data,
['associated' => ['Sets','Parts._joinData']]
);

$this->Orders->save(
$order,
['associated' => ['Sets','Parts._joinData']]
);

我必须将aslo SetsModel包含在关联的中。

Answer 1

$ this-＆gt; request-＆gt; data ['parts'] ['_ ids']允许您连接表但不能以这种方式添加数据。为此，您的$ this-＆gt; request-＆gt; data ['parts']结构必须如下所示：

'parts' => [
    0 => [
        'id'=> 1,
        '_joinData' => [
            'state' => 'exampleSate'
        ]
    ]
]

确保您的'parts'数组中不再包含'_ids'数组，否则您将无法将数据保存到连接表中。
'id'字段对应于相关部分的id。设置它：

$this->request->data['parts'][0]['id'] = <idOfTheConcernedPart>;

要设置“州”字段，请执行以下操作：

$this->request->data['parts'][0]['_joinData']['state'] = "whatever";

最后你需要通过向patchEntity添加一些配置并保存方法来告诉cakephp你正在保存一些_joinData：

$this->Orders->patchEntity(
    $order, 
    $this->request->data, 
    ['associated' => ['Parts._joinData']]
);

$this->Orders->save(
    $order, 
    ['associated' => ['Parts._joinData']]
);

我希望它有所帮助，
欢呼声。

CakePHP 3在连接表中保存额外数据

1 个答案: