7

我有一张这样的桌子：

Products
(
   ID int not null primary key,
   Type int not null,
   Route varchar(20) null
)

我在客户端上有一个这种格式的列表：

Type=1, Percent=0.4, Route=A
Type=1, Percent=0.4, Route=B
Type=1, Percent=0.2, Route=C
Type=2, Percent=0.5, Route=A
Type=2, Percent=0.5, Route=B
Type=3, Percent=1.0, Route=C
...etc

完成后，我想将 40% 的 1 类产品分配给 Route A，40% 分配给 Route B，20% 分配给 Route C。然后 50% 的 2 类产品分配给 Route A，50% 的 2 类产品分配给 Route乙等。

有没有办法在单个更新语句中做到这一点？

如果不是在一个巨大的语句中，是否可以在每个类型一个语句或每个路由一个语句中完成？目前我们正在为每种类型+路线做一个，以上任何一项都会是一种改进。

4

4 回答 4

1

这是我在您发布您正在使用 SQL-Server 之前准备的 Oracle 声明，但它可能会给您一些想法，尽管您必须使用 CTE 和自联接来滚动您自己的 ratio_to_report 分析函数。我们计算产品和客户端路由表中每种类型的累积比例，并对匹配的比例带进行非等连接。我使用的样本数据有一些四舍五入，但对于较大的数据集，这些数据会减少。

这是设置：

create table products (id int not null primary key, "type" int not null, route varchar (20) null);
create table clienttable ( "type" int not null, percent number (10, 2) not null, route varchar (20) not null);
insert into clienttable ("type", percent, route) values (1, 0.4, 'A');
insert into clienttable ("type", percent, route) values (1, 0.4, 'B');
insert into clienttable ("type", percent, route) values (1, 0.2, 'C');
insert into clienttable ("type", percent, route) values (2, 0.5, 'A');
insert into clienttable ("type", percent, route) values (2, 0.5, 'B');
insert into clienttable ("type", percent, route) values (3, 1.0, 'C');

insert into products (id, "type", route) values (1, 1, null);
insert into products (id, "type", route) values (2, 1, null);
insert into products (id, "type", route) values (3, 1, null);
insert into products (id, "type", route) values (4, 1, null);
insert into products (id, "type", route) values (5, 1, null);
insert into products (id, "type", route) values (6, 1, null);
insert into products (id, "type", route) values (7, 1, null);
-- 7 rows for product type 1 so we will expect 3 of route A, 3 of route B, 1 of route C (rounded)

insert into products (id, "type", route) values (8, 2, null);
insert into products (id, "type", route) values (9, 2, null);
insert into products (id, "type", route) values (10, 2, null);
insert into products (id, "type", route) values (11, 2, null);
insert into products (id, "type", route) values (12, 2, null);
-- 5 rows for product type 2 so we will expect 3 of route A and 2 of route B (rounded)

insert into products (id, "type", route) values (13, 3, null);
insert into products (id, "type", route) values (14, 3, null);
-- 2 rows for product type 3 so we will expect 2 of route C

这是声明

select prods.id, prods."type", client.route cr from
(
select
p.id, 
p."type", 
row_number () over (partition by p."type" order by p.id) / count (*) over (partition by p."type") cum_ratio
from
products p
) prods
inner join 
(
select "type", route, nvl (lag (cum_ratio, 1) over (partition by "type" order by route), 0) ratio_start, cum_ratio ratio_end from 
(select "type", route, sum (rr) over (partition by "type" order by route) cum_ratio
from (select c."type", c.route, ratio_to_report (c.percent) over (partition by "type") rr from clienttable c))) client 
on prods."type" = client."type" 
and prods.cum_ratio >= client.ratio_start and prods.cum_ratio < client.ratio_end

这给出了以下结果：-

+----+------+----+
| ID | type | CR |
+----+------+----+
|  1 |    1 | A  |
|  2 |    1 | A  |
|  3 |    1 | B  |
|  4 |    1 | B  |
|  5 |    1 | B  |
|  6 |    1 | C  |
|  8 |    2 | A  |
|  9 |    2 | A  |
| 10 |    2 | B  |
| 11 |    2 | B  |
| 13 |    3 | C  |
+----+------+----+
于 2013-01-22T22:00:18.750 回答
0

像这样的东西怎么样

--For updating type 1, set every route for type 1 as null.

UPDATE MyTable
SET [Route] = null
WHERE [Type] = '1'

--Update Route A(40%)
DECLARE @myVal int;
SET @myVal  =CAST(0.4*(SELECT COUNT(*) FROM myTable WHERE [Type]='1') AS INT);
WITH    tab AS
    (
    SELECT  TOP (@myVal) *
    FROM myTable
    )
UPDATE  tab
SET     [Route] = 'A'
WHERE [Route] is null

--Update Route B (40%)
DECLARE @myVal int;
SET @myVal  =CAST(0.4*(SELECT COUNT(*) FROM myTable WHERE [Type]='1') AS INT);
WITH    tab AS
    (
    SELECT  TOP (@myVal) *
    FROM myTable
    )
UPDATE  tab
SET     [Route] = 'B'
WHERE [Route] is null


--Update Route C (20%)
DECLARE @myVal int;
SET @myVal  =CAST(0.2*(SELECT COUNT(*) FROM myTable WHERE [Type]='1') AS INT);
WITH    tab AS
    (
    SELECT  TOP (@myVal) *
    FROM myTable
    )
UPDATE  tab
SET     [Route] = 'C'
WHERE [Route] is null
于 2013-01-22T21:17:04.587 回答
0

我不知道 SQL Server 中是否存在类似的功能。在 Oracle 中有 SAMPLE 子句。下面的查询从表中选择 10% 的行：

SELECT empno
  FROM scott.emp
SAMPLE (10)
/

那么你的更新会很容易......也许SQL Server中存在类似的东西。您还可以计算行数或数据，然后计算百分比然后更新...

于 2013-01-22T21:26:16.720 回答
0 
WITH po AS
  ( SELECT 
        ID,
        Type,
        ROW_NUMBER() OVER ( PARTITION BY Type
                            ORDER BY ID
                          ) AS Rn,
        COUNT(*) OVER (PARTITION BY Type) AS CntType
    FROM
          Products
  )    
, ro AS
  ( SELECT 
        Type,
        Route,
        ( SELECT SUM(rr.Percent) 
          FROM Route AS rr 
          WHERE rr.Type = r.Type 
            AND rr.Route <= r.Route
        ) AS SumPercent 
    FROM
          Routes AS r
  )
UPDATE p
SET p.Route =
            ( SELECT MIN(ro.Route) 
              FROM ro 
              WHERE ro.Type = po.Type 
                AND ro.SumPercent >= po.Rn / po.CntType
            )
FROM    Products AS p
    JOIN
        po   ON po.ID = p.ID ;
于 2013-01-22T22:17:33.853 回答