【发布时间】:2017-12-29 11:01:10
【问题描述】:
当有条件或条件时,我有带有左连接的本机 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,这与我需要的不同。我需要更改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;
}
成本对象实体
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;
和我的无条件查询生成器
$qb->select('te')
->from('AppBundle:TaskExecution', 'te')
->innerJoin('te.task', 't')
->leftJoin('t.costObject', 'co')
->leftJoin('co.users', 'com')
这是$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 中一样更改它
现在我有
$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'
)
)
但出现错误
[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 中表示我仍然有id 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 但在本机查询中我们只有 LEFT JOIN cost_object_managers AS com ON com.cost_object_id = co.id OR com.cost_object_id = co.parent_id
它如何在具有ON 条件类型的查询生成器中重现?
【问题讨论】:
-
我用过 Symfony。我需要查询生成器代表
标签: php mysql sql-server symfony doctrine-orm