Question

我想弄清楚用户最常回答的语言，并按user_id返回，language_id他们最多回答，有多少次回答。

我从SELECT开始返回这些结果的表/子表：

Table: `sub-selected`
`user_id`    `language_id`    `answers`
  1               1               1
  2               1               1
  1               2               5
  2               2               2
  1               4               3
  1               5               1

此表返回user_id，language_id以及用户已回答language_id的次数。我用这个查询得到它：

SELECT t1.user_id, t2.to_language_id, COUNT(t2.to_language_id) as answers
FROM translation_results as t1
LEFT JOIN translations as t2
ON t2.translation_id = t1.translation_id
GROUP BY t2.to_language_id, t1.user_id

表结构是：

Table: `translations`
`translation_id`    `from_phrase_id`    `to_language_id`

Table: `translation_results`
`translation_id`    `result_id` PRI-AI    `user_id`

translations表存储了所有请求的翻译，translation_results表存储了这些翻译的答案以及相应的user_id。

因此，为了总结表并获取user_id，他们回答最多language_id，以及他们在language_id中回答了多少次，我使用了：

SELECT t1.user_id, t1.to_language_id, MAX(t1.answers)
FROM (
    //The sub-table
    SELECT t1.user_id, t2.to_language_id, COUNT(t2.to_language_id) as answers
    FROM translation_results as t1
    LEFT JOIN translations as t2
    ON t2.translation_id = t1.translation_id
    GROUP BY t2.to_language_id, t1.user_id
) as t1
GROUP BY t1.user_id, t1.to_language_id

但是这不会将表格折叠成所需的结构而是返回：

Table: `sub-selected`
`user_id`    `language_id`    `answers`
  1               1               1
  1               2               5
  1               4               3
  1               5               1
  2               1               1
  2               2               2

我知道它受two clauses组的影响，但如果我只按user_id分组并且在我选择的列中没有包含to_language_id，我就无法知道哪个相应的language_id最多回答。我也尝试过子查询和一些连接，但我发现无论选择哪一列，我总是需要使用MAX(t1.answers)，因此破坏了我正确地整理group by的希望。如何正确折叠查询，而不是让group by找到MAX()和user_id的所有唯一to_language_id组合？

Answer 1

获得：

user_id，他们回答最多language_id，以及他们多少次用那个language_id回答

你可以使用变量：

SELECT user_id, language_id, answers
FROM (
  SELECT user_id, language_id, answers,
         @rn:= IF(@uid = user_id,
                  IF(@uid:=user_id, @rn:=@rn+1, @rn:=@rn+1),
                  IF(@uid:=user_id, @rn:=1, @rn:=1)) AS rn
  FROM (SELECT t1.user_id, t2.to_language_id AS language_id, 
               COUNT(t2.to_language_id) as answers     
        FROM translation_results as t1 
        LEFT JOIN translations as t2 
           ON t2.translation_id = t1.translation_id
        GROUP BY t2.to_language_id, t1.user_id 
       ) t
  CROSS JOIN (SELECT @rn:=0, @uid:=0) AS vars
  ORDER BY user_id, answers DESC
) s
WHERE s.rn = 1

上述查询中存在一个限制：如果有多个language_id共享user_id的最大答案数，则只会返回一个。

Demo here

另一种方法是使用两次查询作为派生表：

SELECT t1.user_id, language_id, t1.answers
FROM (SELECT t1.user_id, t2.to_language_id AS language_id, 
             COUNT(t2.to_language_id) as answers
      FROM translation_results as t1
      LEFT JOIN translations as t2
         ON t2.translation_id = t1.translation_id
      GROUP BY t2.to_language_id, t1.user_id ) t1
INNER JOIN (      
   SELECT user_id, MAX(answers) AS answers
   FROM (SELECT t1.user_id, t2.to_language_id, 
                COUNT(t2.to_language_id) as answers
         FROM translation_results as t1
         LEFT JOIN translations as t2
            ON t2.translation_id = t1.translation_id
         GROUP BY t2.to_language_id, t1.user_id 
        ) t
   GROUP BY user_id ) t2
ON t1.user_id = t2.user_id AND t1.answers = t2.answers

此查询没有上一个查询的限制，但与前一个查询相比可能效率较低。

Demo here

Answer 2

如果我不明白你的问题，你应该用子查询的结果定义一个临时表或派生表，让我们调用sub_selected，然后你应该这样做：

SELECT t1.user_id, t1.to_language_id, answers
FROM sub_selected as t1
WHERE t1.answers = 
   (SELECT MAX(answers)
     FROM sub_selected t2
     WHERE t1.user_id = t2.user_id and t1.to_language_id = t2.language_id)

在同一查询中使用MAX（）和COUNT（）

2 个答案: