Symfony2的原则选择IFNULL(symfony2 doctrine select IFNUL

2019-09-01 02:35发布

好吧,我有这样的代码:

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。 类似的东西?

        $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();

这个错误

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

Answer 1:

使用COALESCE代替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,如果为空和B不为空,则COALESCE(A,B)将返回B.



Answer 2:

有一种学说扩展,其中包括增加了这一点。

这是从IFNULL的DQL文件。 https://github.com/beberlei/DoctrineExtensions/blob/master/src/Query/Mysql/IfNull.php

本章介绍如何使用它们。 http://symfony.com/doc/2.0/cookbook/doctrine/custom_dql_functions.html



文章来源: symfony2 doctrine select IFNULL