问题描述
我正在尝试为我的表对象创建一个抽象对象.
I'm trying to create an abstract object for my Table Objects.
今天我有很多对象,例如:CategoriaTable
、FornecedoresTable
等,它们实现了 $this->tableGateway->insert()
、$this->tableGateway->update()
等
Today I have lots of object like: CategoriaTable
, FornecedoresTable
, etc that implement $this->tableGateway->insert()
, $this->tableGateway->update()
, etc
我创建了一个包含大部分功能的 TableAbstract
,但我遇到了一个问题:
I created an TableAbstract
that contains most of those functionallities, but I stuck on one problem:
// In CategoriaTable my table id is named cat_id
$this->tableGateway->update($object->getArrayCopy(),array('cat_id' => $object->getId()))
// But in FornecedoresTable my table id is named for_id
$this->tableGateway->update($object->getArrayCopy(),array('for_id' => $object->getId()))
如何从 tableGateway 获取表的 ID?有更好的方法来做我想做的事吗?
How can I get from tableGateway the id of an table? There is an better way to do what I want?
我想我可以在我的对象中注入 id 名称,但我不认为这是一个很好的方法......
I guess I could inject the id name in my object but I don't thing this is a good way to do that...
推荐答案
您可以创建新的 TableGateway 类参数.(在我的例子中,我创建了 $this->primary;)
You can create new TableGateway class parameter.(In my case I created $this->primary;)
如果未设置,请使用 Zend\Db\Metadata\Metadata 直接从数据库结构中查找.
And if it is not set use Zend\Db\Metadata\Metadata to find it straight from db structure.
<?php
//...
use Zend\Db\TableGateway\AbstractTableGateway;
use Zend\Db\Metadata\Metadata;
class AbstractTable extends AbstractTableGateway
{
protected $primary;
public function getPrimary()
{
if (null === $this->primary) {
$metadata = new Metadata($this->adapter);
$constraints = $metadata->getTable($this->getTable()->getTable())
->getConstraints();
foreach ($constraints AS $constraint) {
if ($constraint->isPrimaryKey()) {
$primaryColumns = $constraint->getColumns();
$this->primary = $primaryColumns;
}
}
}
return $this->primary;
}
}
?>
这篇关于使用 tableGateway 返回表主键名称的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!