cakephp 3.x保存多个实体 - newEntities

Question

我在保存多个记录方面遇到了最困难的时间.我已经尝试了一百万件事,但我最终遇到了同样的问题:我的记录没有保存,我也看不到任何错误.请记住,我是cakephp和新手编码器的新手.

我错过了一些明显和关键的东西吗？

表:

    $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]);

在$ splits上调试:

[
(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'

}
]

Answer 1

你有没有看到这种Table.index.field风格被使用过,或者你刚试过一些东西并希望它会起作用？

保存分别创建许多实体的许多记录时,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>转换多个记录

创建一次创建/更新多个记录的表单时,可以使用newEntities():

[...]

在这种情况下,多篇文章的请求数据应如下所示:

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

所以你的输入不应该使用表名,而只是索引和字段名,比如

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', /* ... */);