Question

我有这两张桌子：

table1
uid   points
5     13
7     9
12    5
17    3
1     1
2     2
3     1 

table2
uid   points
9     21 
13    17
15    11
17    7
12    6
2     2
1     3
22    1

我需要一个查询来返回前5位用户有分数

目标结果：

uid   points
9     21
13    17
5     13
12    11
15    11

我尝试了什么：

select uid, count(points) c from table1 order by c limit 5 
union all
select uid, count(points) c from table2 order by c limit 5

但我没有得到我想要的东西。

Answer 1

SELECT al.uid as UID , SUM(al.points) AS total_points FROM (SELECT points, uid FROM table1
                             UNION ALL
          SELECT points,uid FROM table2) al group by al.uid

Answer 2

试试这个

select uid, (table1.points + table2.points) as c from table1
Left join table2 on table1.uid = table2.uid
order by (table1.points + table2.points) desc limit 5

Answer 3

您可以尝试此查询

  Select 
      t3.uid , SUM(t3.points) As points  
  From 
      (SELECT * from Table1 UNION ALL SELECT * from Table2) t3 
  GROUP by 
       t3.uid 
  Order by 
       SUM(t3.points) DESC 
  LIMIT 5

演示http://sqlfiddle.com/#!2/5605d/23

Answer 4

试试这个：

SELECT Uid,SUM(Points)
FROM
(
   SELECT Uid,Points FROM Table1
   UNION ALL
   SELECT Uid,Points FROM Table2
) as T
GROUP BY Uid
ORDER BY SUM(Points) DESC
LIMIT 5

SQLFiddle demo

Answer 5

如果您想要排名前5位，则需要使用订单并限制子查询：

select uid, sum(points) points from (
  select uid, points from table1 
  union all
  select uid, points from table2
) x
group by uid
order by sum(points) desc
limit 5

来自不同mysql表中不同字段的总和

5 个答案: