Symfony Doctrine createQueryBuilder From() 别名不起作用

Question

当尝试执行这个查询时它会抱怨：

[2/2] QueryException: [Semantical Error] line 0, col 443 near 'User 用户左'：错误：未定义类“用户”。

我什至不知道为什么我需要在我的 from 中使用别名，但我在那里添加了它并且它不起作用。

实体类名是“用户”

    $query = $em->createQueryBuilder()
        ->from('User','user')
        ->select('user.email,
                person.givenName,
                person.familyName,
                contactPoint.contactInfo,
                organization.legalName,
                postalAddress.streetAddress1,
                postalAddress.streetAddress2,
                postalAddress.city,
                postalAddress.postalCode,
                country.name,
                region.name')

        ->leftJoin('person', 'WITH', "user.person_id=person.id")
        ->leftJoin('contactPoint', 'WITH', "person.contactPoint_id=contactPoint.id")
        ->leftJoin('organization', 'WITH', "person.organization_id=organization.id")
        ->leftJoin('postalAddress', 'WITH', "organization.postal_address_id=postalAddress.id")
        ->leftJoin('country', 'WITH', "postalAddress.country_id=country.id")
        ->leftJoin('region', 'WITH', "postalAddress.region_id=region.id")
        ->where("user.email = '$email' ")
        ->getQuery();
    dump($query->getArrayResult());die();

更新的工作查询 //////////////////

$user = $this->get('security.token_storage')->getToken()->getUser();

    $email = $user->getEmail();

    $query = $em->createQueryBuilder()
        ->from(User::class,'user')
        ->select('user.email,
                person.givenName,
                person.familyName,
                contactPoint.contactInfo,
                organization.legalName,
                postalAddress.streetAddress1,
                postalAddress.streetAddress2,
                postalAddress.city,
                postalAddress.postalCode,
                country.name,
                region.name')

        ->leftJoin(Person::class, 'person', 'WITH', "user.accountOwner=person.id")
        ->leftJoin(ContactPoint::class, 'contactPoint', 'WITH', "person.contactPoint=contactPoint.id")
        ->leftJoin(Organization::class, 'organization', 'WITH', "person.organization=organization.id")
        ->leftJoin(PostalAddress::class, 'postalAddress', 'WITH', "organization.postalAddress=postalAddress.id")
        ->leftJoin(Country::class, 'country', 'WITH', "postalAddress.country=country.id")
        ->leftJoin(Region::class, 'region', 'WITH', "postalAddress.region=region.id")
        ->where("user.email = '$email' ")
        ->getQuery();
    dump($query->getResult());die();

Answer 1

你应该使用完全限定的类名，试试：

    ->from(User::class,'user')

而不是

    ->from('User','user')

更笼统，参考 Symfony Doctrine 的documentation：

您需要考虑选择 PHP 对象，而不是行 在数据库中。因此，您从 AppBundle:Product 实体（可选的快捷方式） AppBundle\Entity\Product class)，然后将其别名为p。

所以你应该在DQL语句中使用类的字段而不是表的列名。

希望有帮助