doctrine ORM count arraycollection in where condition

时间:2015-12-09 09:02:36

标签: symfony orm doctrine-orm arraycollection

Symfony2.8Doctrine 2.5一起使用,我希望在Doctrine ORM查询中过滤所有数组,其中arraycollection恰好包含3个元素。

$em = $this->getDoctrine()->getManager();
$query = $em->getRepository("AppBundle:EduStructItem")
->createQueryBuilder('e')
 ->addSelect('COUNT(e.preconditions) AS HIDDEN numberpre')
 ->having('numberpre = 3')
->getQuery();
$res = $query->getResult();
dump($res);
foreach ($res as $entity){
    print "title:".$entity->getTitle()."<br>";
    dump($entity->getPreconditions()->toArray());
}

preconditions是一个包含前置条件集合的arraycollection。

最后,我想让所有结果都有3个前提条件。 另外,通过arraycollection中的值的数量(类似order by Count(e.preconditions))来排序也很棒。

由于使用了另一个捆绑包,我将原则从2.5.2降级到2.5.0。我不认为这是我的问题的原因,但为了完整起见,这是我的作曲家秀的教义部分:

data-dog/pager-bundle                v0.2.4             Paginator bundle for symfony2 and doctrine orm, allows customization with filters and sorters
  doctrine/annotations                 v1.2.7             Docblock Annotations Parser
  doctrine/cache                       v1.5.2             Caching library offering an object-oriented API for many cache backends
  doctrine/collections                 v1.3.0             Collections Abstraction library
  doctrine/common                      v2.5.2             Common Library for Doctrine projects
  doctrine/data-fixtures               v1.1.1             Data Fixtures for all Doctrine Object Managers
  doctrine/dbal                        v2.5.2             Database Abstraction Layer
  doctrine/doctrine-bundle             1.6.1              Symfony DoctrineBundle
  doctrine/doctrine-cache-bundle       1.2.2              Symfony Bundle for Doctrine Cache
  doctrine/doctrine-fixtures-bundle    2.3.0              Symfony DoctrineFixturesBundle
  doctrine/doctrine-migrations-bundle  1.1.1              Symfony DoctrineMigrationsBundle
  doctrine/inflector                   v1.1.0             Common String Manipulations with regard to casing and singular/plural rules.
  doctrine/instantiator                1.0.5              A small, lightweight utility to instantiate objects in PHP without invoking their constructors
  doctrine/lexer                       v1.0.1             Base library for a lexer that can be used in Top-Down, Recursive Descent Parsers.
  doctrine/migrations                  v1.1.0             Database Schema migrations using Doctrine DBAL
  doctrine/orm                         v2.5.0             Object-Relational-Mapper for PHP             

这是一个测试实体:

<?php
// src/AppBundle/Entity/EduStructItem.php
namespace AppBundle\Entity;


use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;

/**
 * @ORM\Entity
 * @ORM\Table(name="test_edustructitemcollection")
 */
class EduStructItem
{
    /**
     * @ORM\Column(type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    protected $id;


    /**
     * @var string
     * @Assert\NotBlank()
     * @ORM\Column(type="string", length=255, nullable=false)
     */
    private $title;


    /**
     * Preconditions are EduStructItems referencing to an EduStructItem.
     * For a single EduStructItem its empty (which have no subelements).
     * A join table holds the references of a main EduStructItem to its sub-EduStructItems (preconditions)
     *
     * @ORM\ManyToMany(targetEntity="EduStructItem",indexBy="id", cascade={"persist"})
     * @ORM\JoinTable(name="test_edustructitem_preconditioncollection",
     *     joinColumns={@ORM\JoinColumn(name="edustructitem_id", referencedColumnName="id")},
     *     inverseJoinColumns={@ORM\JoinColumn(name="edustructitem_precondition_id", referencedColumnName="id")}
     * )
     */
    public $preconditions;

    public function __construct()
    {
        $this->preconditions = new ArrayCollection();
    }

    public function getTitle()
    {
        return $this->title;
    }

    public function setTitle($title)
    {
        $this->title = $title;
    }


    public function getPreconditions()
    {
        return $this->preconditions;
    }

    public function addPrecondition(\AppBundle\Entity\EduStructItem $precondition)
    {
        $this->preconditions->add($precondition);
    }

    public function removePrecondition(\AppBundle\Entity\EduStructItem $precondition)
    {
        $this->preconditions->removeElement($precondition);
    }

}
?>

最后我总是得到错误: [语义错误]第0行,第18行附近'preconditions)':错误:无效的PathExpression。期望StateFieldPathExpression或SingleValuedAssociationField。

现在我尝试了你的新解决方案:

$em = $this->getDoctrine()->getManager();
$query = $em->getRepository("AppBundle:EduStructItem")
    ->createQueryBuilder('e')
    ->addSelect('COUNT(e.preconditions) AS HIDDEN countpre')
    ->join('e.preconditions', 'precondition', Join::WITH)
    ->having('countpre = 1')
    ->getQuery();

再次出现错误: [语义错误]第0行,第18行附近'preconditions)':错误:无效的PathExpression。期望StateFieldPathExpression或SingleValuedAssociationField。我在HIDDEN之前写了别名我也得到了:[语义错误]第0行,第53页'FROM AppBundle \ Entity \ EduStructItem'附近:错误:未定义类'FROM'。考虑它是一个自我反思的关系,只有一个实体,但有两个表。正如您在我的实体的注释中所看到的,自我关系被保存在test_edustructitem_preconditioncollection中 - 由于注释而由学说生成的表。

我尝试了你最新的解决方案:

$qb = $em->getRepository("AppBundle:EduStructItem")
    ->createQueryBuilder('item');
$qb->addSelect('COUNT(precondition.id) AS countpre HIDDEN ')
    ->join('item.preconditions', 'precondition', Join::WITH)
    ->having('countpre = 1');

当我在HIDDEN之前有countpre时,我总是得到这个错误: [语义错误]第0行,第56页'FROM AppBundle \ Entity \ EduStructItem'附近:错误:未定义类'FROM'。

但是当我在HIDDEN之后放了countpre时:

$qb = $em->getRepository("AppBundle:EduStructItem")
            ->createQueryBuilder('item');
        $qb->addSelect('COUNT(precondition.id) AS HIDDEN countpre')
            ->join('item.preconditions', 'precondition', Join::WITH)
            ->having('countpre = 1');

我收到错误: 执行'SELECT t0_.id AS id_0时发生异常,t0_.title AS title_1,COUNT(t1_.id)AS sclr_2 FROM test_edustructitemcollection t0_ INNER JOIN test_edustructitem_preconditioncollection t2_ ON t0_.id = t2_.edustructitem_id INNER JOIN test_edustructitemcollection t1_ ON t1_.id = t2_.edustructitem_precondition_id拥有sclr_2 = 1':

SQLSTATE [42S22,207]:[Microsoft] [SQL Server的ODBC驱动程序11] [SQL Server]UngültigerSpaltenname'sclr_2'。

500内部服务器错误 - DBALException

1个链接的异常:SQLSrvException»

请考虑只有一个具有自引用的实体,并且有这两个表:

USE [easylearndev4_rsc]
GO
/****** Object:  Table [dbo].[test_edustructitemcollection]    Script Date: 14.12.2015 09:31:55 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[test_edustructitemcollection](
    [id] [int] IDENTITY(1,1) NOT NULL,
    [title] [nvarchar](255) NOT NULL,
PRIMARY KEY CLUSTERED 
(
    [id] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO

USE [easylearndev4_rsc]
GO
/****** Object:  Table [dbo].[test_edustructitem_preconditioncollection]    Script Date: 14.12.2015 09:32:21 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
REATE TABLE [dbo].[test_edustructitem_preconditioncollection](
    [edustructitem_id] [int] NOT NULL,
    [edustructitem_precondition_id] [int] NOT NULL,
PRIMARY KEY CLUSTERED 
(
    [edustructitem_id] ASC,
    [edustructitem_precondition_id] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
ALTER TABLE [dbo].[test_edustructitem_preconditioncollection]  WITH CHECK ADD  CONSTRAINT [FK_34E716A81B7A6CEB] FOREIGN KEY([edustructitem_precondition_id])
REFERENCES [dbo].[test_edustructitemcollection] ([id])
GO
ALTER TABLE [dbo].[test_edustructitem_preconditioncollection] CHECK CONSTRAINT [FK_34E716A81B7A6CEB]
GO
ALTER TABLE [dbo].[test_edustructitem_preconditioncollection]  WITH CHECK ADD  CONSTRAINT [FK_34E716A85D864668] FOREIGN KEY([edustructitem_id])
REFERENCES [dbo].[test_edustructitemcollection] ([id])
GO
ALTER TABLE [dbo].[test_edustructitem_preconditioncollection] CHECK CONSTRAINT [FK_34E716A85D864668]
GO

最后,我自己找到了解决方法:

$em = $this->getDoctrine()->getManager();
$qb = $em->getRepository("AppBundle:EduStructItem")
    ->createQueryBuilder('e');
$qb->join('e.preconditions', 'p', Join::WITH)
    ->groupBy('e.id, e.title')
    ->having('count(p.id) = 1');

但是我对此并不满意,因为数组收集已经是聚合数据,我为什么要加入,再次计数和分组! 这不是主义的想法! 有谁知道更好的解决方案?

2 个答案:

答案 0 :(得分:2)

试试这个:

$qb = $this->getDoctrine()->getManager()->getRepository("MyBundle:Groups")
    ->createQueryBuilder('g')
    ->having('count(g.members) = 3')
    ->orderBy('g.members', 'DESC')
; 

答案 1 :(得分:1)

$qb = $this->getDoctrine()->getManager()->getRepository"MyBundle:Groups")
    ->createQueryBuilder('g')
    ->addSelect('COUNT(g.members) AS count HIDDEN')
    ->having('count = 3')
    ->orderBy('count', 'DESC')
    ;

修改

在您更新问题后,应该清楚上述解决方案不起作用,因为您需要计算关系对象而不是单个字段。

$qb = $em->getRepository("AppBundle:EduStructItem") //Selfreferencing ManyToMany
         ->createQueryBuilder('item');
$qb->addSelect("COUNT(precondition.id) AS count HIDDEN")
   ->join('item.preconditions', 'precondition', Join::WITH)
   ->having('count = 3')
   ->orderBy('count');