Symfony2的原则选择IFNULL

好吧，我有这样的代码：Symfony2的原则选择IFNULL

SELECT 
IFNULL(s2.id,s1.id) AS effectiveID, 
IFNULL(s2.status, s1.status) AS effectiveStatus, 
IFNULL(s2.user_id, s1.user_id) as effectiveUser, 
IFNULL(s2.likes_count, s1.likes_count) as effectiveLikesCount 

FROM statuses AS s1 
LEFT JOIN statuses AS s2 ON s2.id = s1.shared_from_id 
WHERE s1.user_id = 4310 
ORDER BY effectiveID DESC 
LIMIT 15

，我需要重写它的QueryBuilder。像那样的东西？

 $fields = array('IFNULL(s2.id,s1.id) AS effectiveID','IFNULL(s2.status, s1.status) AS effectiveStatus', 'IFNULL(s2.user_id, s1.user_id) as effectiveUser','IFNULL(s2.likes_count, s1.likes_count) as effectiveLikesCount'); 

    $qb=$this->_em->createQueryBuilder() 
      ->select($fields) 
      ->from('WallBundle:Status','s1') 
      ->addSelect('u') 
      ->where('s1.user = :user') 
      ->andWhere('s1.admin_status = false') 
      ->andWhere('s1.typ_statusu != :group') 
      ->setParameter('user', $user) 
      ->setParameter('group', 'group') 
      ->leftJoin('WallBundle:Status','s2', 'WITH', 's2.id=s1.shared_from_id') 
      ->innerJoin('s1.user', 'u')    
      ->orderBy('s1.time', 'DESC') 
      ->setMaxResults(15); 

    var_dump($query=$qb->getQuery()->getResult());die();

此错误是

[Syntax Error] line 0, col 7: Error: Expected known function, got 'IFNULL'

来源

2013-04-24 EnchanterIO

是的，就像错误信息说的 - 原则查询语言不知道这样的功能。找到类似问题的答案http://stackoverflow.com/a/9110213/1727046 – gatisl 2013-04-24 21:35:15

@gatisl是的先生，我看到它，但我不知道...我怎么能“安装”它？ – EnchanterIO 2013-04-24 22:56:30

使用COALESCE，而不是IFNULL这样

$fields = array('COALESCE(s2.id,s1.id) AS effectiveID','COALESCE(s2.status, s1.status) AS effectiveStatus', 'COALESCE(s2.user_id, s1.user_id) as effectiveUser','COALESCE(s2.likes_count, s1.likes_count) as effectiveLikesCount');

COALESCE返回列表中的非空的值，因此，如果是空， B不为空，则COALESCE(A,B)将返回B.

来源

2013-08-28 09:43:06 Thierry