Question

我有一个名为POINTS的mysql表，如下所示：

userid1 | userId2 | points
---------------------------
    1   |    1    |   3
    1   |    1    |   2
    1   |    2    |   5
    2   |    1    |   4
    1   |    3    |   5

我正在尝试创建此表的视图，如下所示：

userId | gained | received | spent | current
--------------------------------------------
   1   |    5   |     4    |  10   |   -1
   2   |    0   |     5    |   4   |    1
   3   |    0   |     5    |   0   |    5

列的算法是：

userId，在userId1或userId2列中的POINTS表中找到的userIds的唯一值
获得，在POINTS表中userId1 = userId和userId2 = userId的点数总和
收到，在POINTS表中userId1！= userId和userId2 = userId的点数总和
花费，在POINTS表中userId1 = userId和userId2！= userId的点数总和
当前，获得+收到 - 花费

我是mysql的新手，我的问题是，如果我进入多个子查询，我不知道如何为不同的用户保存总和值。任何帮助将不胜感激

更新：示例中的数字是准确的

Answer 1

这是结构化查询语言的结构化部分的直接应用。

您需要两个关键组件子查询才能将它们组合在一起。一个是每个用户的点gained列表。这是这样的。（http://sqlfiddle.com/#!2/1b1884/1/0）

                      SELECT userid1 AS userid,
                             SUM(points) AS gained
                        FROM points
                       WHERE userid1 = userid2
                       GROUP BY userid1

同样，您可以生成支出和接收点数（http://sqlfiddle.com/#!2/1b1884/5/0）

                      SELECT userid1 AS userid,
                             SUM(points) AS spent
                        FROM points
                       WHERE userid1 != userid2
                       GROUP BY userid1

和（http://sqlfiddle.com/#!2/1b1884/7/0）

                      SELECT userid2 AS userid,
                             SUM(points) AS received
                        FROM points
                       WHERE userid1 != userid2
                       GROUP BY userid2

您还需要一个子查询，它会为您提供所有不同用户的列表。如果你有一个单独的user表，你在这里没有提到，那将比这更好。（http://sqlfiddle.com/#!2/1b1884/9/0）

                      SELECT DISTINCT userid1 userid FROM points
                       UNION
                      SELECT DISTINCT userid2 userid FROM points

现在，您需要将这四个子查询加在一起，结果就是这样。这是大纲。

SELECT a.userid, 
       IFNULL(g.gained,0) gained, 
       IFNULL(s.spent,0) spent,
       IFNULL(r.received,0) received,
       IFNULL(g.gained,0)-IFNULL(s.spent,0)+IFNULL(r.received,0) total
  FROM (   /* all users subquery */
       ) a
  LEFT JOIN (   /* gained subquery */
       ) g ON a.userid = g.userid
  LEFT JOIN (   /* received subquery */
       ) r ON a.userid = r.userid
  LEFT JOIN (   /* spent subquery */
       ) s ON a.userid = s.userid
 ORDER BY a.userid

最后，将它们全部放在一起（http://sqlfiddle.com/#!2/1b1884/13/0）

SELECT a.userid, 
       IFNULL(g.gained,0) gained, 
       IFNULL(s.spent,0) spent,
       IFNULL(r.received,0) received,
       IFNULL(g.gained,0)-IFNULL(s.spent,0)+IFNULL(r.received,0) total
  FROM (   /* all users subquery */
                      SELECT DISTINCT userid1 userid FROM points
                       UNION
                      SELECT DISTINCT userid2 userid FROM points
       ) a
  LEFT JOIN (   /* gained subquery */
                      SELECT userid1 AS userid,
                             SUM(points) AS gained
                        FROM points
                       WHERE userid1 = userid2
                       GROUP BY userid1
       ) g ON a.userid = g.userid
  LEFT JOIN (   /* received subquery */
                      SELECT userid2 AS userid,
                             SUM(points) AS received
                        FROM points
                       WHERE userid1 != userid2
                       GROUP BY userid2
       ) r ON a.userid = r.userid
  LEFT JOIN (   /* spent subquery */
                      SELECT userid1 AS userid,
                             SUM(points) AS spent
                        FROM points
                       WHERE userid1 != userid2
                       GROUP BY userid1
       ) s ON a.userid = s.userid
 ORDER BY a.userid

如果你认为它是一个四层俱乐部三明治，子查询作为奶酪片，JOIN条款作为面包片，它可能会有所帮助。我承认它是满口的！

Answer 2

您可以对条件表达式（或ifs）使用条件聚合来完成您想要的任务。

基本查询如下所示：

select 
    userid
    , sum(case when u.userid =  p.userid1 and u.userid =  p.userId2 then points else 0 end) gained
    , sum(case when u.userid <> p.userid1 and u.userid =  p.userId2 then points else 0 end) received
    , sum(case when u.userid =  p.userid1 and u.userid <> p.userId2 then points else 0 end) spent
    , sum(case when u.userid =  p.userid1 and u.userid =  p.userId2 then points else 0 end) 
    + sum(case when u.userid <> p.userid1 and u.userid =  p.userId2 then points else 0 end) 
    - sum(case when u.userid =  p.userid1 and u.userid <> p.userId2 then points else 0 end) _current
from 
  (select userid1 userid from points union select userid2 from points) u
  , points p
group by u.userid;

如果要基于它创建视图，from子句中的子查询也需要在视图中。在下面链接的小提琴中，代码已被重构为几个视图。

此示例SQL Fiddle显示了使用视图实现的视图，根据您的示例数据，输出：

| userid | gained | received | spent | current |
|--------|--------|----------|-------|---------|
|      1 |      5 |        4 |    10 |      -1 |
|      2 |      0 |        5 |     4 |       1 |
|      3 |      0 |        5 |     0 |       5 |

另一种选择，而不是条件聚合，将使用相关子查询：

select 
    userid
    , (select coalesce(sum(points),0) from points p where u.userid =  p.userid1 and u.userid =  p.userId2) gained
    , (select coalesce(sum(points),0) from points p where u.userid <> p.userid1 and u.userid =  p.userId2) received
    , (select coalesce(sum(points),0) from points p where u.userid =  p.userid1 and u.userid <> p.userId2) spent
from (select userid1 userid from points union select userid2 from points) u;

这会产生相同的结果（但我没有包含当前列，因为它只是重复以保持代码简短。）

多个条目的

2 个答案: