Question

通过使用COALESCE，我可以像这样创建一个名为comment_votes的临时变量：

SELECT comments.*, COALESCE(rs_reputations.value, 0) AS comment_votes FROM `comments` 
LEFT JOIN rs_reputations ON comments.id = rs_reputations.target_id AND 
rs_reputations.target_type = 'Comment' AND rs_reputations.reputation_name = 
'comment_votes' AND rs_reputations.active = 1 WHERE (impression_id = 1)

我想在来源查询中创建一个名为impression_votes的第二个变量。我试图这样做：

SELECT comments.*, COALESCE(rs_reputations.value, 0) AS comment_votes 
FROM 'comments' 
    LEFT JOIN rs_reputations ON 
        comments.id = rs_reputations.target_id AND 
        rs_reputations.target_type = 'Comment' AND 
        rs_reputations.reputation_name = 'comment_votes' AND 
        rs_reputations.active = 1 
SELECT comments.*, COALESCE(rs_reputations.value, 0) AS impression_votes 
FROM 'comments' 
    LEFT JOIN rs_reputations ON 
    comments.id = rs_reputations.target_id AND
    rs_reputations.target_type = 'Comment' AND
    rs_reputations.reputation_name = 'impression_votes' AND 
    rs_reputations.active = 1 
WHERE

这会导致错误：

You have an error in your SQL syntax

我正在尝试甚至可能吗？如果是这样，我似乎错误地桥接了两个SELECT / COALESCE语句。我该怎么写呢？

Answer 1

MySQL COALESCE函数实际上是一个内置函数，它返回第一个非空值 - 它不是一个变量，它是一个在各种数据库系统中实际支持的函数。

例如，使用下表：

| Id    | Name      | Counter    |
| 1     | lolcat    | NULL       |
| 2     | codez     | 1          |

sql语句：

SELECT Id, Name, COALESCE(counter, 0) AS NonNullCounter FROM table

将返回结果：

| Id    | Name      | NonNullCounter |
| 1     | lolcat    | 0              |
| 2     | codez     | 1              |

在这种情况下，NULL值已被0替换。

这对您很有用，因为如果您在评论中的行的rs_reputations中还没有匹配的行，LEFT JOIN将为NULL列返回rs_repuations.value，然后由0替换为COALESCE。

如果您不熟悉JOIN，那么有一个很棒的visual guide by Jeff Atwood。

您的第一个查询实际上可以是：

SELECT     comments.*, 
           COALESCE(rs_reputations.value, 0) AS comment_votes 
FROM       comments 
LEFT JOIN   rs_reputations ON comments.id = rs_reputations.target_id 
                           AND rs_reputations.reputation_name = 'comment_votes' 
WHERE       impression_id = 1;

选择1 - 联盟

您有几个选择 - 您可以像这样UNION结果：

SELECT     comments.*, 
           COALESCE(rs_reputations.value, 0) AS votes,
           'comment_votes' AS vote_type 
FROM       comments 
LEFT JOIN   rs_reputations ON comments.id = rs_reputations.target_id 
                           AND rs_reputations.reputation_name = 'comment_votes' 
WHERE       impression_id = 1

UNION

SELECT     comments.*, 
           COALESCE(rs_reputations.value, 0) AS votes,
           'impression_votes' as vote_type 
FROM       comments 
LEFT JOIN   rs_reputations ON comments.id = rs_reputations.target_id 
                           AND rs_reputations.reputation_name = 'impression_votes' 
WHERE       impression_id = 1;

在这种情况下，您的结果将如下所示：

|comments_columns|votes|vote_type       |
| *              |12   |comment_vote    |
| *              |2    |impression_vote |

选择2 - 加入相同的表格两次

或者您可以通过using the same table name but a different alias两次自行加入同一张桌子：

SELECT     comments.*, 
           COALESCE(CommentRep.value, 0) AS comment_votes,
           COALESCE(ImpressionRep.value, 0) AS impression_votes,
FROM       comments 
LEFT JOIN   rs_reputations AS CommentRep ON comments.id = CommentRep.target_id 
                           AND CommentRep.reputation_name = 'comment_votes' 
LEFT JOIN   rs_reputations AS ImpressionRep ON comments.id = ImpressionRep.target_id 
                           AND ImpressionRep.reputation_name = 'impression_votes'
WHERE       CommentRep.impression_id = 1
AND         ImpressionRep.impression_id = 1

在这种情况下，您的结果将如下所示：

|comments_columns|comment_votes|impression_votes|
| *              |12           |0               |
| *              |2            |6               |

最后（phew）原始SQL中出现错误的原因是您将两个SELECT语句链接在一起而没有实际关联它们 - 在这个实例中，SQL实际上没有意义，因为您需要在逻辑上将它们关联起来（或者通过UNION或者按照上面的重复连接。

MySQL：使用COALESCE创建多个变量

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