Question

大家好，我是stackoverflow新手。

如果我对提出这个问题有任何不妥之处，希望能够纠正我。   ：）

我想制作一个提问系统，我在MySQL中有两个表。

表＆＃39;问题＆＃39; ：它存储问题信息。
表＆＃39; question_communication＆＃39; ：它存储经理和用户之间的问题回复。

以下是详细表格。

question (Table)
    - question_id(INT)
    - uid(INT)
    - category(CHAR)
    - description(CHAR)
    - submit_time(DATETIME)

和

question_communication (Table)
    - question_reply_id(INT)
    - question_id(INT)
    - uid(ID)
    - reply(CHAR)
    - time(DATETIME)
    - seen(TINYINT)           --- Other side has seen the message or not.(0 is not seen, 1 is seen)

我希望查询结果包括：

question_id, uid, category, description, submit_time, seen(=1), seen(=0)

然后我尝试写下面的代码，

SELECT T1.question_id, T1.uid, T1.category, T1.description, DATE_FORMAT(T1.submit_time, '%Y/%m/%d %H:%i') submit_time, SUM(T2.seen = 1) seen, SUM(T2.seen = 0) notseen
  FROM question T1, question_communication T2
  WHERE T1.question_id = T2.question_id
    AND T2.uid != (Here is attribute.)
  ORDER BY submit_time DESC

但WHERE T1.question_id = T2.question_id此行可能在一种情况下不起作用。

当T1 (question)有问题时，

并且它在T2 (question_communication)中没有任何回复。

所以T1.question_id = T2.question_id会导致SQL JOIN超出我的预期。

我的问题摘要：

如何成功查询结果如：

question_id, uid, category, description, submit_time, seen(=1), seen(=0)

如果T1中有问题，但T2没有回复。 seen(=1)和seen(=0)必须为zero。

谢谢大家：）

来自@Gordon Linoff的回答，我添加了COALESCE（）：

SELECT T1.question_id, T1.uid, T1.category, T1.description,
       DATE_FORMAT(T1.submit_time, '%Y/%m/%d %H:%i') as submit_time,
       COALESCE(SUM(T2.seen = 1), 0) as seen, COALESCE(SUM(T2.seen = 0), 0) as notseen
FROM question T1 LEFT JOIN
     question_communication T2
     ON T1.question_id = T2.question_id 
GROUP BY T1.question_id, T1.uid, T1.category, T1.description,
         DATE_FORMAT(T1.submit_time, '%Y/%m/%d %H:%i') 
ORDER BY submit_time DESC;

Answer 1

我认为你只需要一个left join。实际上，您应该始终使用显式join语法。您的查询还需要group by。因此，查询看起来像：

SELECT T1.question_id, T1.uid, T1.category, T1.description,
       DATE_FORMAT(T1.submit_time, '%Y/%m/%d %H:%i') as submit_time,
       SUM(T2.seen = 1) as seen, SUM(T2.seen = 0) as notseen
FROM question T1 LEFT JOIN
     question_communication T2
     ON T1.question_id = T2.question_id 
GROUP BY T1.question_id, T1.uid, T1.category, T1.description,
         DATE_FORMAT(T1.submit_time, '%Y/%m/%d %H:%i') 
ORDER BY submit_time DESC;

我不知道T2.uid != (Here is attribute.)应该做什么。如果您在T2上有过滤条件，请将其放在ON子句中。

MySQL：使用SUM B加入（A，B），但在某些情况下B不存在行

1 个答案: