Question

我遵循带注释的过滤器的实现，以仅加载一个用户关注的源。非常有用，可以正常工作。

https://api-platform.com/docs/core/filters/#using-doctrine-filters

但是：我想在我的实体中使用 ManyToMany 关系执行此操作，而不是像这样的ManyToOne：

<?php

namespace App\Entity;

use ApiPlatform\Core\Annotation\ApiResource;
use ApiPlatform\Core\Annotation\ApiSubresource;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
use Doctrine\Common\Collections\ArrayCollection;
use App\Annotation\UserAware;

/**
 *
 * @ApiResource(routePrefix="/api")
 * @ORM\Entity
 * @ORM\Table(name="albums")
 * @UserAware(userFieldName="user_owner")
 * @UserAware(userFieldName="album_access_user")
 */
class Album
{
    //...

    /**
     * @ORM\ManyToOne(targetEntity="User")
     * @ORM\JoinColumn(nullable=false, name="user_owner", referencedColumnName="id")
    */
    public $userOwner;

    /**
     * @ORM\ManyToMany(targetEntity="User", inversedBy="albums")
     * @ORM\JoinTable(name="album_access")
     * @ORM\JoinTable(
     *  name="album_access",
     *  joinColumns={
     *      @ORM\JoinColumn(name="album_access_album", referencedColumnName="id")
     *  },
     *  inverseJoinColumns={
     *      @ORM\JoinColumn(name="album_access_user", referencedColumnName="id")
     *  }
     * )
    */
    private $accesses;

    //...
}

由于现在有一个JoinTable，而不仅仅是JoinColum，我必须调整UserFilterConfigurator.php事件，但是我不知道该怎么做才能给他表而不是列...

这是我得到的逻辑错误，因为我现在请求实体表而不是联接表：

An exception occurred while executing 'SELECT DISTINCT o0_.id AS id_0, o0_.id AS id_1 FROM albums o0_ WHERE (o0_.album_access_user = '4') ORDER BY o0_.id ASC LIMIT 30':\n\nSQLSTATE[42703]: Undefined column: 7 ERROR:  column o0_.album_access_user does not exist\nLINE 1: ...id_0, o0_.id AS id_1 FROM albums o0_ WHERE (o0_...

如何为ManyToMany做到这一点<=>如何传递良好的（联接）表+（联接）列??

多对多的Api平台过滤器注释

0 个答案: