有没有更快的方法来查找 COLUMN 的顺序？

Question

我的 SQL 服务器 table 看起来像这样

ID   a_Toyota   a_Mazda   a_Nissan   a_Kia   a_Honda   a_Subaru     SoldCar   CarOrder
1    8000       7000      6200       8500    6500      7000         Mazda     NULL
2    4000       5000      4500       3500    3500      5000         Mazda     NULL
3    5400       5000      4500       5500    5500      4600         Mazda     NULL
4    5600       6300      7500       8200    6500      7300         Mazda     NULL
5    8500       7400      7400       6500    9500      9000         Mazda     NULL
6    9900       8000      9900       7300    8100      8000         Mazda     NULL

我想更新 CarOrder 字段，所以它具有已售出汽车的价格与其他汽车价格相比的顺序。

因此，对于 ID 1 的汽车价格，如 a_Kia (8500) 是 1st 和 a_Toyota (8000) 是 2nd 和 a_Mazda & a_Subaru (7000) 是3rd and a_Honda (6500) is 5th and a_Nissan (6200) 是 第 6 售出的汽车是马自达，排名第三，所以 table 应该如下

ID   a_Toyota   a_Mazda   a_Nissan   a_Kia   a_Honda   a_Subaru     SoldCar   CarOrder
1    8000       7000      6200       8500    6500      7000         Mazda     3
2    4000       5000      4500       3500    3500      5000         Subaru    1
3    5400       5000      4500       5500    5500      4600         Toyota    3
4    5600       6300      7500       8200    6500      7300         Honda     4
5    8500       7400      7400       6500    9500      9000         Honda     1
6    9900       8000      9900       7300    8100      8000         Honda     3

我可以找到带有大 CASE 语句的订单

UPDATE mytable
SET CarOrder =
CASE WHEN SoldCar = 'Toyota' AND a_Toyota>=a_Mazda AND a_Toyota>=a_Nissan AND ... AND a_Toyota>=a_Subaru THEN 1 
CASE WHEN SoldCar = 'Toyota' AND a_Toyota<a_Mazda AND a_Toyota>=a_Nissan AND ... AND a_Toyota>=a_Subaru THEN 2 
CASE WHEN SoldCar = 'Toyota' AND a_Toyota>=a_Mazda AND a_Toyota<a_Nissan AND ... AND a_Toyota>=a_Subaru THEN 2 
.
.
.
CASE WHEN SoldCar = 'Toyota' AND a_Toyota>=a_Mazda AND a_Toyota>=a_Nissan AND ... AND a_Toyota<a_Subaru THEN 2 
.
.
.
CASE WHEN SoldCar = 'Toyota' AND a_Toyota<a_Mazda AND a_Toyota<a_Nissan AND ... AND a_Toyota>=a_Subaru THEN 3
..
..
..

但这将是一个巨大的案例陈述。

不知道有人有更简单的方法吗？

Answer 1

假设 table 结构类似于以下内容：

CREATE TABLE tempdb..cars
(
    ID INT NOT NULL,
    a_Toyota INT NOT NULL,
    a_Mazda INT NOT NULL,
    a_Nissan INT NOT NULL,
    a_Kia INT NOT NULL,
    a_Honda INT NOT NULL,
    a_Subaru INT NOT NULL,
    SoldCar VARCHAR(100) NOT NULL,
    CarOrder INT NULL
);

一种方法是利用 APPLY operator. Something like the following should give you a resultset from a table structure like above, assuming the use of CROSS APPLY and a non-dense RANK (vs a DENSE RANK) 以及降序来确定您的顺序：

SELECT  c.ID, c.SoldCar, o.ord AS CarOrder
FROM    tempdb..cars c
CROSS APPLY
        (
        SELECT  t.ord
        FROM    (
                SELECT  r.car, RANK() OVER (ORDER BY r.qty DESC) AS ord
                FROM    (
                        SELECT  c.a_Toyota AS qty, 'Toyota' AS car
                        UNION ALL
                        SELECT  c.a_Mazda AS qty, 'Mazda' AS car
                        UNION ALL
                        SELECT  c.a_Nissan AS qty, 'Nissan' AS car
                        UNION ALL
                        SELECT  c.a_Kia AS qty, 'Kia' AS car
                        UNION ALL
                        SELECT  c.a_Honda AS qty, 'Honda' AS car
                        UNION ALL
                        SELECT  c.a_Subaru AS qty, 'Subaru' AS car
                        ) r
                ) t
        WHERE   t.car = c.SoldCar
        ) o

Answer 2

table 结构需要从一开始就转向它应该有的东西，这将有助于轻松确定正确的顺序并满足任何数量的品牌。

您可以使用交叉应用和 row_number 将每个值与其序号位置相匹配，使用 updatable CTE 更新基础 table。

with cars as (
    select * from t
    cross apply (
        select case soldcar
          when 'Toyota' then a_Toyota
          when 'Mazda'  then a_Mazda
          when 'Nissan' then a_Nissan
          when 'Honda'  then a_Honda
          when 'Subaru' then a_Subaru
        end
    )s(SoldValue)
    cross apply (
      select rank() over (order by v desc) co, v
      from (values(a_toyota),(a_mazda),(a_nissan),(a_kia),(a_honda),(a_subaru))v(v)
    )c
    where SoldValue=v
)
update cars set carOrder=co

见Demo Fiddle

Answer 3

又一个基于XQuery的方法。

对于ID=2的行，Subary和Mazda打成平手。他们都有5000的价值。

SQL

-- DDL and sample data population, start
DECLARE @tbl TABLE (
    ID INT IDENTITY(1,1) PRIMARY KEY, 
    a_Toyota INT,
    a_Mazda INT,
    a_Nissan INT,
    a_Kia INT,
    a_Honda INT,
    a_Subaru INT,
    SoldCar VARCHAR(20)
);
INSERT INTO @tbl
(
    a_Toyota,
    a_Mazda,
    a_Nissan,
    a_Kia,
    a_Honda,
    a_Subaru,
    SoldCar
) VALUES
(8000, 7000, 6200, 8500, 6500, 7000, 'Mazda'),
(4000, 5000, 4500, 3500, 3500, 5000, 'Subaru'),
(5400, 5000, 4500, 5500, 5500, 4600, 'Toyota'),
(5600, 6300, 7500, 8200, 6500, 7300, 'Honda'),
(8500, 7400, 7400, 6500, 9500, 9000, 'Honda'),
(9900, 8000, 9900, 7300, 8100, 8000, 'Honda');
-- DDL and sample data population, end

SELECT t.*, CarOrder 
FROM @tbl AS t
    CROSS APPLY (SELECT a_Toyota, a_Mazda, a_Nissan,
    a_Kia, a_Honda, a_Subaru, SoldCar
    FOR XML PATH(''), TYPE, ROOT('root')) AS t1(c)
CROSS APPLY (SELECT c.query('<root>
{
for $r in /root/*
order by data($r) descending
return <r>
        <make>{local-name($r)}</make>
        <salePrice>{data($r)}</salePrice>
    </r>
}
</root>').query('
    let $soldcar := sql:column("SoldCar")
    for $r in /root/r[contains((make/text())[1], $soldcar)]
    let $pos := count(root/*[. << $r])
    return $pos').value('.','INT')
) AS t2(CarOrder);

输出

+----+----------+---------+----------+-------+---------+----------+---------+----------+
| ID | a_Toyota | a_Mazda | a_Nissan | a_Kia | a_Honda | a_Subaru | SoldCar | CarOrder |
+----+----------+---------+----------+-------+---------+----------+---------+----------+
|  1 |     8000 |    7000 |     6200 |  8500 |    6500 |     7000 | Mazda   |        3 |
|  2 |     4000 |    5000 |     4500 |  3500 |    3500 |     5000 | Subaru  |        2 |
|  3 |     5400 |    5000 |     4500 |  5500 |    5500 |     4600 | Toyota  |        3 |
|  4 |     5600 |    6300 |     7500 |  8200 |    6500 |     7300 | Honda   |        4 |
|  5 |     8500 |    7400 |     7400 |  6500 |    9500 |     9000 | Honda   |        1 |
|  6 |     9900 |    8000 |     9900 |  7300 |    8100 |     8000 | Honda   |        3 |
+----+----------+---------+----------+-------+---------+----------+---------+----------+

Answer 4

这是一个选项，您不必枚举要逆透视的列。

这还假设列名的前缀为 a_

示例或dbFiddle

with cte as (
Select * 
 From  YourTable A
 Cross Apply ( 
               Select *
                     ,rn=row_number() over (order by convert(decimal(12,2),value) desc)
                 From OpenJson((Select A.* For JSON Path,Without_Array_Wrapper  )) 
                 Where [Key] not in ('ID','SoldCar','CarOrder')
             ) B
 Where [key] ='a_'+SoldCar  collate SQL_Latin1_General_CP1_CI_AS 
)
Update cte set CarOrder = RN

更新后Table

Answer 5

这个很简单。您只需将前 6 列中当前 SoldCar 的排名附加到相应的记录。在 SQL 中执行此操作很复杂。该语言需要 XQuery 或 OpenJson 和 window 函数辅助的交叉应用。该声明冗长且难以阅读。另一种方法是从数据库中导出数据并在 Python 或 SPL 中处理它。 SPL，open-source Java 包，更容易集成到 Java 程序中并生成更简单的代码。它只用三行代码就完成了：

	A
1	=MSSQL.query("select a_Toyota,a_Mazda,a_Nissan,a_Kia,a_Honda,a_Subaru,'a_'+SoldCar as SoldCar from cars")
2	=A1.fname().m(:-2)
3	=A1.derive([${A2.concat@c()}].ranks@z()(A2.pselect(~==A1.~.#7)):CarOrder)

有没有更快的方法来查找 COLUMN 的顺序？

Is there a faster way to find the order of a COLUMN?

sql

tsql

sql-server