cakephp 3.x 保存多个实体 - newEntities

我在保存多条记录方面遇到了最困难的时期。我已经尝试了一百万次，但最终遇到了同样的问题：我的记录没有保存，而且我看不到任何错误。请记住，我是 cakephp 的新手，也是一名新手编码员。

我是否遗漏了一些明显且关键的东西？

Table:

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

    $this->belongsTo('Transactions', [
        'foreignKey' => 'transaction_id',
        'joinType' => 'INNER'
    ]);
    $this->belongsTo('Accounts', [
        'foreignKey' => 'account_credit_id',
        'joinType' => 'INNER'
    ]);

控制器：

    $splits = $this->Splits->newEntity();
    if ($this->request->is('post')) {
        $splits = $this->Splits->newEntities($this->request->data());

        debug($splits);

        foreach ($splits as $split){
            $this->Splits->save($split);
        }
   }

    $transactions = $this->Splits->Transactions->find('list', ['limit' => 200]);
    $accounts = $this->Splits->Accounts->find('list', ['limit' => 200]);
    $this->set(compact('split', 'transactions', 'accounts'));
    $this->set('_serialize', ['split']);

模板：

        echo $this->Form->input('Splits.1.transaction_id', ['options' => $transactions]);
        echo $this->Form->input('Splits.1.amount',  ['type' => 'float']);
        echo $this->Form->input('Splits.1.account_id', ['options' => $accounts]);
        echo $this->Form->input('Splits.2.transaction_id', ['options' => $transactions]);
        echo $this->Form->input('Splits.2.amount',  ['type' => 'float']);
        echo $this->Form->input('Splits.2.account_id', ['options' => $accounts]);
        echo $this->Form->input('Splits.3.transaction_id', ['options' => $transactions]);
        echo $this->Form->input('Splits.3.amount',  ['type' => 'float']);
        echo $this->Form->input('Splits.3.account_id', ['options' => $accounts]);

调试 $splits：

[
(int) 0 => object(App\Model\Entity\Split) {

    (int) 1 => [
        'transaction_id' => '108',
        'amount' => '100.33',
        'account_id' => '2'
    ],
    (int) 2 => [
        'transaction_id' => '108',
        'amount' => '50.22',
        'account_id' => '4'
    ],
    (int) 3 => [
        'transaction_id' => '108',
        'amount' => '65.22',
        'account_id' => '5'
    ],
    '[new]' => true,
    '[accessible]' => [
        '*' => true
    ],
    '[dirty]' => [
        (int) 1 => true,
        (int) 2 => true,
        (int) 3 => true
    ],
    '[original]' => [],
    '[virtual]' => [],
    '[errors]' => [],
    '[repository]' => 'Splits'

}
]

你在哪里看到过这个吗Table.index.field正在使用的风格，或者您只是尝试过一些东西并希望它能起作用？

当保存许多记录分别创建许多实体时，预期格式是一个数字索引数组，用于保存各个记录的数据，如文档中所示

Cookbook > 数据库访问和 ORM > 保存数据 > 转换多条记录 http://book.cakephp.org/3.0/en/orm/saving-data.html#converting-multiple-records

创建同时创建/更新多个记录的表单时，您 可以使用 newEntities():

[...]

在这种情况下，多篇文章的请求数据应该看起来 喜欢：

$data = [
    [
        'title' => 'First post',
        'published' => 1
    ],
    [
        'title' => 'Second post',
        'published' => 1
    ],
];

因此，您的输入不应使用表名称，而应仅使用索引和字段名称，例如

echo $this->Form->input('0.transaction_id', /* ... */);
echo $this->Form->input('0.amount', /* ... */);
echo $this->Form->input('0.account_id', /* ... */);
echo $this->Form->input('1.transaction_id', /* ... */);
echo $this->Form->input('1.amount', /* ... */);
echo $this->Form->input('1.account_id', /* ... */);
echo $this->Form->input('2.transaction_id', /* ... */);
echo $this->Form->input('2.amount', /* ... */);
echo $this->Form->input('3.account_id', /* ... */);