使用三个表的Mysql连接查询

我用了三张桌子。第一个表中记录了客户问题

  customer_id   |   question   | question_id
----------------------------------------------
       58       |   question 4 |      4
       41       |   question 5 |      5
       37       |   question 3 |      3 

SELECT question, count(*) as total_comments, T1.user 
FROM T1, T2, T3
WHERE T1.customer_id = T3.user 
  AND T3.status = 1,
  AND T1.question_id = T2.question_id
GROUP BY T2.user 
ORDER BY question_id DESC
LIMIT 5

第二个表中记录了与第一个teble中的问题相关的评论

    comment   | question_id   |  user
---------------------------------------
    comment 1 |      4        |    41  
    comment 2 |      5        |    58         
    comment 3 |      4        |    41  
    comment 4 |      5        |    58  
    comment 5 |      3        |    23  

第三个表中列出了有关该网站用户的数据

    user      | status
--------------------------------
     58       |      1
     41       |      1
     37       |      0 
     23       |      0 

我如何进行一个查询，结果是最后五个问题的列表，按问题id和每个问题相关的评论总数排序。这些问题和意见只能由第三个表中状态为“1”的用户提出

本例中的结果应如下所示：

  question   | total comments  |   user
-----------------------------------------------
 question 5  |      2          |    41
 question 4  |      2          |    58    

---

快看一下，我会这样做：

SELECT q.question_id, count(c.comment) from questions as q left join 
comments as c on q.question_id=c.question_id left join user as u on 
c.user=u.user where u.status=1 group by q.question_id

---

我有理由肯定这是正确的，尽管我这里没有数据来验证

SELECT questions.question, 
       count(comments.question_id) as 'total comments', 
       user.user
FROM user 
  JOIN questions on user.user = questions.customer_id
  LEFT JOIN comments on questions.question_id = comments.question_id
WHERE user.status = 1
GROUP BY comments.question_id,  questions.question, users.user
ORDER BY questions.question
LIMIT 5

问题的左连接将允许问题被表示，即使没有对其留下任何评论

如果你有任何问题，请告诉我

  customer_id   |   question   | question_id
----------------------------------------------
       58       |   question 4 |      4
       41       |   question 5 |      5
       37       |   question 3 |      3 

SELECT question, count(*) as total_comments, T1.user 
FROM T1, T2, T3
WHERE T1.customer_id = T3.user 
  AND T3.status = 1,
  AND T1.question_id = T2.question_id
GROUP BY T2.user 
ORDER BY question_id DESC
LIMIT 5

---

T1、T2、T3是有问题的表格，它们的显示顺序是正确的。

因此，我在问题列中得到空值。使用上述问题中的确切表格？对不起，凯文。在测试您的查询时，我没有输入列的专有名称。这次我正确地测试了它，它在注释表中不计算空值。此查询仅统计有注释的问题的注释。那些没有评论的问题不算在内。啊-谢谢评论。我不知道你需要这些。你知道我怎么做吗？我需要在这个查询中更改什么？@cb0-它不计算注释表中的空值

SELECT      question_id as question, count(c.comment) as 'total comments', q.customer_id as user
FROM        (SELECT * 
            FROM question
            ORDER BY question_id DESC
            LIMIT 5) as q LEFT JOIN 
            comment as c using(question_id) LEFT JOIN
            user as u using(user) 
WHERE       u.status=1 
GROUP BY    q.question_id
ORDER BY    question_id DESC