查询生成器/ DQL无法使用INNER JOIN - 语法问题

Ela*_*hts 5 doctrine inner-join dql query-builder symfony

我知道我有一个语法,但是我无法弄明白.我正在尝试对5个表进行SELECT和INNER JOIN,但Symfony抱怨JOIN中的实体在定义之前使用.

实际错误如下: [Semantical Error] line 0, col 121 near 'I ON C.id = ': Error: Identification Variable MySiteBundle:Items used in join path expression but was not defined before.

这是PHP代码.

注意:我已将此查询缩短为两列,两个表和一个联接,以使问题简单并显示我的观点.实际查询更长,并产生相同的错误.

$em = $this->getDoctrine()->getEntityManager();
$query = $em->createQuery(
    'select C.name as CName, I.id as IId
    FROM MySiteBundle:Categories C
    INNER JOIN MySiteBundle:Items I ON C.id = I.category_id');
$result = $query->getResult();
Run Code Online (Sandbox Code Playgroud)

更新

正如所建议的那样,我已经废除了DQL代码并使用了Query Builder代码.我得到一个非常类似的错误'Categories c': Error: Class 'Categories' is not defined.我的QB代码如下.

$em = $this->getDoctrine()->getEntityManager();
$qb = $em->createQueryBuilder()
        ->select('c.name, i.id, i.image, i.name, i.description, m.id, m.quantity, m.value, m.qty_received, m.custom_image, m.custom_name, m.custom_description, u.user1fname, u.user1lname, u.user2fname, u.user2lname')
        ->from('Categories', 'c')
        ->innerJoin('Items', 'i', 'ON', 'c.id = i.category_id')
        ->innerJoin('MemberItems', 'm', 'ON', 'i.id = m.item_id')
        ->innerJoin('User', 'u', 'ON', 'm.memberinfo_id = u.id')
        ->where('u.id = ?', $slug)
        ->orderBy('c.id', 'ASC')
        ->getQuery();

$memberItems = $qb->getResult();
Run Code Online (Sandbox Code Playgroud)

有什么建议?

Cer*_*rad 9

路易斯在我打字时发布了.那好吧.

DQL根据您的关联为您处理联接详细信息.通常,您只需要拼出FROM类名称.就像是:

'select C.name as CName, I.id as IId
FROM MySiteBundle:Categories C
INNER JOIN C.items');
Run Code Online (Sandbox Code Playgroud)

并最终使用查询生成器.

================================================== ===========================

以下是在Symfony 2中使用查询构建器的示例.

public function getAccounts($params = array())
{
    // Build query
    $em = $this->getEntityManager();
    $qb = $em->createQueryBuilder();

    $qb->addSelect('account');
    $qb->addSelect('accountPerson');
    $qb->addSelect('person');
    $qb->addSelect('registeredPerson');
    $qb->addSelect('projectPerson');

    $qb->from('ZaysoCoreBundle:Account','account');

    $qb->leftJoin('account.accountPersons',  'accountPerson');
    $qb->leftJoin('accountPerson.person',    'person');
    $qb->leftJoin('person.registeredPersons','registeredPerson');
    $qb->leftJoin('person.projects',         'projectPerson');
    $qb->leftJoin('projectPerson.project',   'project');

    if (isset($params['accountId']))
    {
        $qb->andWhere($qb->expr()->in('account.id',$params['accountId']));
    }
    if (isset($params['projectId']))
    {
        $qb->andWhere($qb->expr()->in('project.id',$params['projectId']));
    }
    if (isset($params['aysoid']))
    {
        $qb->andWhere($qb->expr()->eq('registeredPerson.regKey',$qb->expr()->literal($params['aysoid'])));
    }
    $query = $qb->getQuery();

  //die('DQL ' . $query->getSQL());
    return $query->getResult();
}
Run Code Online (Sandbox Code Playgroud)


Lou*_*eau 3

DQL 不使用这样的联接。它们有点简化。然而我也发现它们的记录不足。

    $em = $this->getDoctrine()->getEntityManager();
    $query = $em->createQuery(
        'select C.name as CName, I.id as IId
        FROM MySiteBundle:Categories C
        INNER JOIN C.items I');
    $result = $query->getResult();
Run Code Online (Sandbox Code Playgroud)

实际使用的关系取决于您的模型。

我通常使用查询生成器。

    $em = $this->getEntityManager();
    $request = $em->getRepository('MySiteBundle:Categories');

    $qb = $request->createQueryBuilder('C');
    $query = $qb 
        ->select('C.name, I.id')
        ->innerJoin('C.items', 'I')
        ->getQuery();
Run Code Online (Sandbox Code Playgroud)