Question

这是我的table

http://www.sqlfiddle.com/#!2/47e2d/8

╔════╦══════════╦════════╦═══════╦════════╗
║ id ║ category ║ userid ║ users ║ points ║
╠════╬══════════╬════════╬═══════╬════════╣
║ 1  ║    2     ║  91    ║ user1 ║  100   ║
║ 2  ║    3     ║  92    ║ user2 ║   80   ║
║ 3  ║    4     ║  93    ║ user3 ║   10   ║
║ 4  ║    4     ║  91    ║ user1 ║   12   ║
╚════╩══════════╩════════╩═══════╩════════╝

如何获得每个类别中得分最高的用户的计数结果？

在上面的例子中，我想得到

user1: 2
user2: 1

在stackoverflow中搜索类似的问题一段时间之后，我想出了

SELECT users, COUNT(*) 
FROM `table` 
WHERE points=(SELECT MAX(points) FROM `table`) 
GROUP BY points;

但它不会产生我想要的东西。

Answer 1

试试这个：

  SELECT 
  users,
  COUNT(*) 
FROM
  (SELECT 
    users,
    COUNT(*) COUNT
  FROM
    `table` 
  WHERE points IN 
    (SELECT 
      MAX(points) 
    FROM
      `table` 
    GROUP BY category) 
  GROUP BY points) temp 
GROUP BY users ;

Answer 2

试试这个：

SELECT
  t1.users,
  COUNT(t2.category)
FROM `table` AS t1
INNER JOIN
(
  SELECT category, MAX(points) MAxpoints
  FROM `table`
  GROUP BY category
) AS t2 ON t1.category = t2.category AND t1.points = t2.MAxpoints
GROUP BY t1.users;

加入的子查询：

  SELECT category, MAX(points) MAxpoints
  FROM `table`
  GROUP BY category

会为每个类别提供最大points，然后我们通过将该子查询与原始表一起加入那些最大点的用户，然后我们使用COUNT来计算这些类别的每个用户的最高分数。

Updated SQL Fiddle Demo

这会给你：

╔════════╦═══════╗
║ USERS  ║ COUNT ║
╠════════╬═══════╣
║ user1  ║     2 ║
║ user2  ║     1 ║
╚════════╩═══════╝

Answer 3

试试这个：

SELECT users, count(*) 
FROM `TABLE`, 
  (SELECT category, max(points) as max_points FROM `TABLE` GROUP BY category) as maxes
WHERE maxes.category = `TABLE`.category AND maxes.max_points = `TABLE`.points
GROUP BY users

这里的关键是创建一个临时表，其中包含每个类别的最大点

SELECT category, max(points) as max_points FROM `TABLE` GROUP BY category

然后我使用它来加入现有的TABLE以找到具有最大分数的正确用户。

按类别计算MySQL中具有最高值的结果

3 个答案: