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我有一个表，其中 order_buyer_id 作为交易的 id，createdby 作为买方的 id，createdAt 作为交易发生的日期，数量作为每笔交易的重量。

在我的桌子上，我将买家分为 3 种类型：

- new buyer
- unique buyer
- existing buyer

这是找出新买家的语法，我称之为 A（新买家）：

select 
       count(distinct om.createdby) as count_buyer
from (select count(xx.count_) as count_
          from (select count(createdby) as count_ from order_match
                where order_status_Id in (4, 5, 6, 8)
                 group by createdby
                 having count(createdby) = 1) xx
        ) x1,
        (select createdby
           from order_match
          group by createdby
          having count(createdby) = 1) yy,
        order_match om
 where yy.createdby = om.createdby and
 order_status_id in (4, 5, 6, 8)
 and om.createdAt >= paramdatefrom
   and om.createdAt <= paramdateto
   and NOT EXISTS (select 1 from order_match om2
                where om.createdby = om2.createdby
               and order_status_id in (4, 5, 6, 8)
                  and om2.createdAt < paramdatefrom);

这是找出重复买家的语法，称为 B（唯一买家）：

    select
           count(distinct om.createdby) as count
   from (select count(xx.count_) as count_
          from (select count(createdby) as count_ from order_match
                where order_status_Id in (4, 5, 6, 8)
                 group by createdby
                 ) xx
        ) x1,
        (select createdby
           from order_match
          group by createdby
          ) yy,
        order_match om
 where yy.createdby = om.createdby and
 order_status_id in (4, 5, 6, 8)
 and om.createdAt >= paramdatefrom
   and om.createdAt <= paramdateto;

;

这是找出现有买家的语法，称为 C（现有买家）：

select
  count(distinct om.createdby) as count
from
  order_match om
  where om.order_status_id in (4,5,6,8)
  and om.createdAt <= paramdateto
  and om.createdAt >= paramdatefrom
  and EXISTS (select 1 from order_match om2
  where om.createdby = om2.createdby
  and om2.createdAt < paramdatefrom and
  om2.order_status_id in (4, 5, 6, 8)) 
  ;

基本上我希望所有这些语法都变成变量 A、B、C 所以我可以根据我的解释计算我需要的百分比，预期的结果就像这样

select (A (the result of syntax new Buyer) : B (the result of syntax unique buyer)) * 100 as percentage_1

和select (100 - percentage_1) as percentage_2

关键是如何使语法的每个结果都变成可变的，这样我就可以像预期的结果一样计算 percent_1 和 percent_2 。

4

1 回答 1

1

要测试更大的查询，您必须提供一些数据，以正确测试查询，请参阅

我在您的描述中找不到为什么需要 result_c，但您现在可以使用它。

顺便说一下，这是算法或查询，而不是语法。.

SELECT 
    result_a / result_b * 100 AS percentage_1,
    100 - (result_a / result_b * 100) AS percentage_2
FROM
    (SELECT 
        (SELECT 
                    COUNT(DISTINCT om.createdby) AS count_buyer
                FROM
                    (SELECT 
                    COUNT(xx.count_) AS count_
                FROM
                    (SELECT 
                    COUNT(createdby) AS count_
                FROM
                    order_match
                WHERE
                    order_status_Id IN (4 , 5, 6, 8)
                GROUP BY createdby
                HAVING COUNT(createdby) = 1) xx) x1, (SELECT 
                    createdby
                FROM
                    order_match
                GROUP BY createdby
                HAVING COUNT(createdby) = 1) yy, order_match om
                WHERE
                    yy.createdby = om.createdby
                        AND order_status_id IN (4 , 5, 6, 8)
                        AND om.createdAt >= paramdatefrom
                        AND om.createdAt <= paramdateto
                        AND NOT EXISTS( SELECT 
                            1
                        FROM
                            order_match om2
                        WHERE
                            om.createdby = om2.createdby
                                AND order_status_id IN (4 , 5, 6, 8)
                                AND om2.createdAt < paramdatefrom)) result_a,
            (SELECT 
                    COUNT(DISTINCT om.createdby) AS count
                FROM
                    (SELECT 
                    COUNT(xx.count_) AS count_
                FROM
                    (SELECT 
                    COUNT(createdby) AS count_
                FROM
                    order_match
                WHERE
                    order_status_Id IN (4 , 5, 6, 8)
                GROUP BY createdby) xx) x1, (SELECT 
                    createdby
                FROM
                    order_match
                GROUP BY createdby) yy, order_match om
                WHERE
                    yy.createdby = om.createdby
                        AND order_status_id IN (4 , 5, 6, 8)
                        AND om.createdAt >= paramdatefrom
                        AND om.createdAt <= paramdateto) result_b,
            (SELECT 
                    COUNT(DISTINCT om.createdby) AS count
                FROM
                    order_match om
                WHERE
                    om.order_status_id IN (4 , 5, 6, 8)
                        AND om.createdAt <= paramdateto
                        AND om.createdAt >= paramdatefrom
                        AND EXISTS( SELECT 
                            1
                        FROM
                            order_match om2
                        WHERE
                            om.createdby = om2.createdby
                                AND om2.createdAt < paramdatefrom
                                AND om2.order_status_id IN (4 , 5, 6, 8))) result_c
    ) a
于 2020-03-28T12:49:25.877 回答