Question

我不是MySQL的专家，今天已证明这一点很明显，但我需要帮助按销售顺序订购用户。

我有两张桌子。一个叫用户

+------------------------+
+ id | fname | lname     +
+------------------------+
+  1 | bob   | french    +
+  2 | fred  | smith     +
+  3 | ted   | nugent    +
+  4 | kyle  | frank     +
+------------------------+

和另一个销售

+------------------------------------------------------------+
+ id | date        | commission  | lister      | seller      +
+------------------------------------------------------------+
+  1 | 2017-11-01  | 2200        | 2           | 2           +
+  2 | 2018-01-15  | 1800        | 1           | 1           +
+  3 | 2017-11-07  | 3600        | 2           | 1           +
+  4 | 2017-11-30  | 1252        | 4           | 1           +
+------------------------------------------------------------+

佣金由卖家和卖家分成50/50。列表和卖家列对应于用户ID

我需要找到两件事。

A）一个人（例如Bob French）他的佣金总额，这个月的销售额

因此鲍勃应该拥有销售额＃3的50％和销售＃4的50％。

销售＃2不在本月

＃p>所以销售＃3（1800）的一半和销售＃3（626）的一半是2426

这个月的排名应该

Fred @ 4000总佣金
Bob @ 2426总佣金
Kyle @ 626总佣金
Ted @ 0总佣金

本月我需要返回RANK（在这种情况下为2）和TOTAL COMMISSION（在这种情况下为2426）（本例中为bob）（本例中为11月）

B）我需要在不同的声明中显示整个上表（对于主管来说，看到每个人都排名。用户只能匿名查看他们的排名。

对于两者，它可能是相同的SQL查询，然后我只是从结果集中通过id从用户中提取用户，除非有更有效的方法。

我尝试了什么

SELECT x.id, x.fname, x.lname, y.commission,
FIND_IN_SET( commission , ( 
    SELECT GROUP_CONCAT( commission ORDER BY commission DESC )
    FROM sales
) ) AS rank

FROM users x
JOIN sales y ON x.id IN (lister, seller)
ORDER BY rank ASC

这是有效的，但不限于月份。我试过添加

WHERE (date between DATE_FORMAT(NOW() ,'%Y-%m-01') AND NOW() )

但似乎没有效果，因为我尝试将日期限制在2010年，结果无论如何都会回来。

此外，上述查询是由获得单一最高佣金的人返回RANK，而不是他们的佣金总和。

请帮忙。

Answer 1

您可以使用UNION ALL：

select x.id, x.name, sum(commision) as commision from 
(select a.date as date, b.id as id, b.fname as name, a.commision/2 as commision
from sales a join users b
on a.lister = b.id
union all
select a.date as date, b.id as id, b.fname as name, a.commision/2 as commision
from sales a join users b
on a.seller = b.id) as x
where x.date between '2017-11-01' and '2017-11-30'
group by x.id, x.name

这里的小提琴：http://sqlfiddle.com/#!9/8ae69a/7

对于您的过滤器WHERE和订购，您可以添加：

select x.id, x.name, sum(commision) as commision from 
(select a.date as date, b.id as id, b.fname as name, a.commision/2 as commision
from sales a join users b
on a.lister = b.id
union all
select a.date as date, b.id as id, b.fname as name, a.commision/2 as commision
from sales a join users b
on a.seller = b.id) as x
where (x.date between DATE_FORMAT(NOW() ,'%Y-%m-01') AND NOW() )
group by x.id, x.name
order by commision

- 编辑 -

添加排名:(小提琴：http://sqlfiddle.com/#!9/8ae69a/30）

SET @rank = 0;

SELECT @rank := @rank + 1 AS rank,
       y.id,
       y.name,
       y.commision
FROM   (SELECT x.id,
               x.name,
               Sum(commision) AS commision
        FROM   (SELECT a.date          AS date,
                       b.id            AS id,
                       b.fname         AS name,
                       a.commision / 2 AS commision
                FROM   sales a
                       JOIN users b
                         ON a.lister = b.id
                UNION ALL
                SELECT a.date          AS date,
                       b.id            AS id,
                       b.fname         AS name,
                       a.commision / 2 AS commision
                FROM   sales a
                       JOIN users b
                         ON a.seller = b.id) AS x
        WHERE  ( x.date BETWEEN Date_format(Now(), '%Y-%m-01') AND Now() )
        GROUP  BY x.id,
                  x.name
        ORDER  BY x.commision) AS y

Answer 2

选择@a：= @ a + 1作为排名，＆＃39; @＆＃39;，u.fname，COALESCE（总和（佣金），0）佣金，＆＃39;总佣金＆＃39; from（SELECT sub.dates，（commission * 2）commission，sub.lister as id FROM（SELECT s.dates，（s.commission / 4）as commission，s.lister，（select @a：= 0）rank FROM sales s where s.dates 在2017-01-01＆＃39;之间和＆＃39; 2017-12-31＆＃39;）sub 联合所有 SELECT sub.dates，（佣金* 2）卖方，sub.seller FROM （SELECT s.dates，（s.commission / 4）as commission，s.sellerFROM sales s where s.dates 在2017-01-01＆＃39;之间和＆＃39; 2017-12-31＆＃39;）sub）main 在u.id = main.id上加入用户u 按sum（佣金）desc

分组

MYSQL按性能

2 个答案: