我的应用程序管理电子产品车间的维修订单。 当我尝试保存几个相同的项目（具有相同的ID）时，cake仅保留最后一个条目，而忽略先前的条目。 （在我的示例中，最后一个产品）
我试图在模型中配置“ saveStrategy”，并尝试使用“ append”，但是没有任何结果。当我输入不同的产品（具有不同的ID）时，它可以毫无问题地保存它。
门票
| id | client_id | reference |
产品
| id | code | description |
产品门票
| id | ticket_id | product_id | qty | serial_number | notes |
debug($this->request->getData());
[
'client_id' => '1',
'reference' => 'Client reference',
'products' => [
(int) 0 => [
'id' => '1',
'_joinData' => [
'qty' => '1',
'serial_number' => '123',
'notes' => 'Not working'
]
],
(int) 1 => [
'id' => '1',
'_joinData' => [
'qty' => '1',
'serial_number' => '345',
'notes' => 'Not Working too'
]
],
(int) 2 => [
'id' => '1',
'_joinData' => [
'qty' => '1',
'serial_number' => '567',
'notes' => 'Note number 3'
]
],
(int) 3 => [
'id' => '1',
'_joinData' => [
'qty' => '1',
'serial_number' => '978',
'notes' => 'Another note'
]
]
]
]
它应该保存的是所有数据，而不仅仅是最后一个值。
我该如何解决这个问题？
答案 0 :(得分：1)
如果我正确理解您的业务案例，则说明您正在为客户创建票证，并且在同一笔交易中向票证添加了4种产品。因此，您要添加带有4个关联的product_tickets（票据上的项目）的票据。
在模型ProductsTable.php
中 public function initialize(array $config)
{
parent::initialize($config);
$this->setTable('products');
$this->setDisplayField('id');
$this->setPrimaryKey('id');
$this->addBehavior('Timestamp');
$this->hasMany('ProductTickets', [
'foreignKey' => 'product_id'
]);
}
在TicketsTable.php模型中
public function initialize(array $config)
{
parent::initialize($config);
$this->setTable('tickets');
$this->setDisplayField('id');
$this->setPrimaryKey('id');
$this->addBehavior('Timestamp');
$this->hasMany('ProductTickets', [
'foreignKey' => 'ticket_id'
]);
}
在模型ProductTicketsTable.php
中 public function initialize(array $config)
{
parent::initialize($config);
$this->setTable('product_tickets');
$this->setDisplayField('id');
$this->setPrimaryKey('id');
$this->addBehavior('Timestamp');
$this->belongsTo('Tickets', [
'foreignKey' => 'ticket_id',
'joinType' => 'INNER'
]);
$this->belongsTo('Products', [
'foreignKey' => 'product_id',
'joinType' => 'INNER'
]);
}
然后，您只能使用Tickets控制器，并与关联的ProductTickets（工单行项目）一起保存。
因此，您可以在src\Template\Tickets\add.php中设置如下形式：
<?= $this->Form->create($ticket) ?>
<fieldset>
<legend><?= __('Add Ticket') ?></legend>
<?php
echo $this->Form->control('client');
echo $this->Form->control('reference');
?>
</fieldset>
<fieldset>
<legend><?= __('Add Product_Tickets (Ticket items)') ?></legend>
<table>
<thead>
<tr>
<th><?= __('Product') ?></th>
<th><?= __('Qty') ?></th>
<th><?= __('Serial Number') ?></th>
<th><?= __('Notes') ?></th>
</tr>
</thead>
<tbody>
<tr>
<td>
<?php echo $this->Form->control("product_tickets.0.product_id", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.0.qty", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.0.serial_number", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.0.notes", []) ?>
</td>
</tr>
<tr>
<td>
<?php echo $this->Form->control("product_tickets.1.product_id", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.1.qty", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.1.serial_number", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.1.notes", []) ?>
</td>
</tr>
<tr>
<td>
<?php echo $this->Form->control("product_tickets.2.product_id", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.2.qty", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.2.serial_number", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.2.notes", []) ?>
</td>
</tr>
<tr>
<td>
<?php echo $this->Form->control("product_tickets.3.product_id", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.3.qty", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.3.serial_number", []) ?>
</td>
<td>
<?php echo $this->Form->control("product_tickets.3.notes", []) ?>
</td>
</tr>
</tbody>
<tfoot>
</tfoot>
</table>
</fieldset>
<?= $this->Form->button(__('Submit')) ?>
<?= $this->Form->end() ?>
*（注意，还有其他/更好的方法来创建具有多个关联记录的表单。）
在Tickets控制器中，在add方法中：
public function add()
{
$ticket = $this->Tickets->newEntity();
if ($this->request->is('post')) {
$data = $this->request->getData();
$ticket = $this->Tickets->newEntity($data, [
'associated' => ['ProductTickets']
]);
$ticket = $this->Tickets->patchEntity($ticket, $this->request->getData(),['associated' => 'ProductTickets'
]);
if ($this->Tickets->save($ticket)) {
$this->Flash->success(__('The ticket has been saved.'));
return $this->redirect(['action' => 'index']);
}
$this->Flash->error(__('The ticket could not be saved. Please, try again.'));
}
$products = $this->Tickets->ProductTickets->Products->find('list');
$this->set(compact('ticket', 'products'));
}
这对我有用，应该保存与ticket_id相关的product_id的任何组合。例如，此请求数据将按预期保存：
debug($this->request->getData());
[
'client' => '1',
'reference' => 'reference 1',
'product_tickets' => [
(int) 0 => [
'product_id' => '1',
'qty' => '12',
'serial_number' => 'abc1234',
'notes' => 'note1'
],
(int) 1 => [
'product_id' => '2',
'qty' => '5',
'serial_number' => '4321cba',
'notes' => 'note2'
],
(int) 2 => [
'product_id' => '1',
'qty' => '6',
'serial_number' => 'a4b3c21',
'notes' => 'note3'
],
(int) 3 => [
'product_id' => '3',
'qty' => '21',
'serial_number' => '4c3b2a1',
'notes' => 'note4'
]
]
]
根据手册，Saving Has Many Associations
重要说明，如果您计划保存在故障单上的4种产品中的任何一种为NULL，则需要从请求数据中删除该产品。您可以使用ProductTicketsTable模型中的beforeMarshal方法来完成此操作。
希望这是您的初衷，对您有所帮助。
干杯
D。