Question

我在Symfony2中遇到一些问题，我有3个问题（它们有点简化）：

class Group
{
    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    protected $id;

    /**
     * @var string
     *
     * @ORM\Column(name="name", type="string", length=255)
     */
    protected $name;

    /** @ORM\OneToMany(targetEntity="GroupMember", mappedBy="family") */
    protected $groupMembers;


class GroupMember
{
    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    protected $id;

    /**
     * @ORM\Column(name="group_member_type", type="integer")
     */
    protected $groupMemberType;

    /**
     * @ORM\ManyToOne(targetEntity="Group", inversedBy="groupMembers")
     * @ORM\JoinColumn(name="group_id", referencedColumnName="id", nullable=false)
     */
    protected $group;

/**
 * @ORM\ManyToOne(targetEntity="Test\UserBundle\Entity\User", inversedBy="memberships")
 * @ORM\JoinColumn(name="user_id", referencedColumnName="id", nullable=false)
 */
protected $user;

USER - ＆gt;标准FOSuserbundle用户

我想进行查询，返回当前用户为组成员的所有组。有人可以告诉我该怎么做吗？

这是我的尝试：

SELECT g FROM (group g INNER JOIN groupmember gm ON g.id=gm.group_id) INNER JOIN fos_user u ON gm.user_id = u.id WHERE u.id = 6

但它返回

Unknown column 'g' in 'field list'

Answer 1

使用查询构建器，假设您已提取用户并将其存储在$userObject中，而您的包名称为BundleName：

$groups = $this->getDoctrine()->getManager()->createQueryBuilder()
    ->select('g')
    ->from('BundleName:Group', 'g')
    ->innerJoin('g.groupMembers', 'gm')
    ->innerJoin('gm.user', 'u')
    ->where('u = :user')
    ->setParameter('user', $userObject)
    ->getQuery()
    ->getResult();

您还需要确保关系映射是正确的。

/** @ORM\OneToMany(targetEntity="GroupMember", mappedBy="family") */
protected $groupMembers;

应该是

/** @ORM\OneToMany(targetEntity="GroupMember", mappedBy="group") */
protected $groupMembers;

Answer 2

这是正确的查询：

SELECT g.* FROM (group g INNER JOIN groupmember gm ON g.id=gm.group_id) INNER JOIN fos_user u ON gm.user_id = u.id WHERE u.id = 6

在Symfony2中加入3个表

2 个答案: