Doctrine QueryBuilder需要为leftJoin更改ON条件

问题描述 投票:0回答:1

当有on或condition时,我有左连接的本机sql查询,如何在查询生成器中表示它?

 $query = "  SELECT  te.id
            FROM    task_executions AS te
            INNER JOIN tasks AS t ON t.id = te.task_id
            LEFT JOIN cost_objects AS co ON co.id = t.cost_object_id
            LEFT JOIN cost_object_managers AS com ON com.cost_object_id = co.id OR com.cost_object_id = co.parent_id

我需要在查询生成器中表示它。但是在User实体中我有ManyToMany关系,没有单独的表,当我尝试左连接WITH条件时,这与我需要的不一样。我需要改变ON的关系

LEFT JOIN cost_object_managers AS com ON com.cost_object_id = co.id OR com.cost_object_id = co.parent_id

用户实体

class User
{
...
/**
 * @ORM\ManyToMany(targetEntity="CostObject", mappedBy="users")
 */
private $costObjects;
}

CostObject实体

class CostObject
{
    /**
 * @var CostObject
 *
 * @ORM\ManyToOne(targetEntity="CostObject", inversedBy="children")
 * @ORM\JoinColumns({
 *   @ORM\JoinColumn(name="parent_id", referencedColumnName="id", onDelete="CASCADE")
 * })
 */
private $parent;

    /**
 * @var ArrayCollection
 *
 * @ORM\ManyToMany(targetEntity="User", inversedBy="costObjects")
 * @ORM\JoinTable(name="cost_object_managers",
 *      joinColumns={@ORM\JoinColumn(name="cost_object_id", referencedColumnName="id", onDelete="CASCADE")},
 *      inverseJoinColumns={@ORM\JoinColumn(name="user_id", referencedColumnName="id", onDelete="CASCADE")}
 * )
 */
private $users;

和我的查询生成器没有条件

        $qb->select('te')
        ->from('AppBundle:TaskExecution', 'te')
        ->innerJoin('te.task', 't')

        ->leftJoin('t.costObject', 'co')
        ->leftJoin('co.users', 'com')

这是$query->getSQL()

SELECT some_name FROM task_executions t0_ INNER JOIN tasks t1_ ON t0_.task_id = t1_.id LEFT JOIN cost_objects c2_ ON t1_.cost_object_id = c2_.id LEFT JOIN cost_object_managers c4_ ON c2_.id = c4_.cost_object_id LEFT JOIN users u3_ ON u3_.id = c4_.user_id ORDER BY t0_.execution_start DESC

在这个例子中,我看到ON关系条件LEFT JOIN users u3_ ON u3_.id = c4_.user_id。并且需要像在本机sql中一样进行更改

我现在有

        $qb->select('te')
        ->from('AppBundle:TaskExecution', 'te')
        ->innerJoin('te.task', 't')
        ->leftJoin('t.costObject', 'co')
        ->leftJoin(
            'co.users',
            'com',
            Join::ON,
            $qb->expr()->orX(
                'co = com.costObjects',
                'co.parent = com.costObjects'
            )
        )

但是得到了错误

[Syntax Error] line 0, col 112: Error: Expected end of string, got 'ON'

如果我使用WITH条件,在我的sql表示我仍然有id的关系,我不需要那个

            ->leftJoin(
            'co.users',
            'com',
            Join::WITH,
            $qb->expr()->orX(
                'co MEMBER OF com.costObjects',
                'co.parent MEMBER OF com.costObjects'
            )
        )

LEFT JOIN users u3_ ON u3_.id = c4_.user_id AND (EXISTS (SELECT 1 FROM cost_object_managers c5_ INNER JOIN cost_objects c6_ ON c5_.cost_object_id = c6_.id WHERE c5_.user_id = u3_.id AND c6_.id IN (c2_.id)) OR EXISTS (SELECT 1 FROM cost_object_managers c5_ INNER JOIN cost_objects c6_ ON c5_.cost_object_id = c6_.id WHERE c5_.user_id = u3_.id AND c6_.id IN (c2_.parent_id)))

我的意思是users u3_ ON u3_.id = c4_.user_id AND但在原生查询中我们只有LEFT JOIN cost_object_managers AS com ON com.cost_object_id = co.id OR com.cost_object_id = co.parent_id

如何使用ON条件类型在Query Builder中重现?

php mysql sql-server symfony doctrine-orm
1个回答
0
投票

如果您有查询并且它适用于您,则无需完成将其转换为DQL或QueryBuilder编程语法的所有工作。您可以使用Doctrine的Native Query,然后 - 如果需要 - 将结果映射到您的对象。只需创建一个自定义存储库,并在其中创建一个大致如下所示的新方法:

public function findTaskExecutionBy...()
{
    $query = $this->entityManager->createNativeQuery('SELECT te.id FROM ...');

    return $query->getSingleScalarResult(); // If it's just one id you expect
}

如果您希望返回多个id,也可以使用$query->getResult()。或者,如果您想要整个Task-object,请使用ResultSetMapping:

$rsm = new ResultSetMappingBuilder($this->entityManager);
$rsm->addRootEntityFromClassMetadata('App\Entity\Task', 'te');

$query = $this->entityManager->createNativeQuery(
    'SELECT te.* FROM ...',

您还可以查看Doctrine文档以获取更详细的说明和更多示例:http://docs.doctrine-project.org/projects/doctrine-orm/en/latest/reference/native-sql.html $ rsm);

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