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如何对“row_number()over(partition by[Col]order by[Col][按[Col]分区)”

row-number sql-server-2005 sql-server sql

9

Scott Chamberlain · 技术社区 · 13 年前

我正在尝试将数据表中重复的条目组合起来,并给它们一个新的编号。

以下是一个示例数据集( runnable copy )

declare @tmpTable table
    (ID Varchar(1), 
     First varchar(4), 
     Last varchar(5), 
     Phone varchar(13),
     NonKeyField varchar(4))

insert into @tmpTable select 'A', 'John', 'Smith', '(555)555-1234', 'ASDF'
insert into @tmpTable select 'B', 'John', 'Smith', '(555)555-1234', 'GHJK'
insert into @tmpTable select 'C', 'Jane', 'Smith', '(555)555-1234', 'QWER'
insert into @tmpTable select 'D', 'John', 'Smith', '(555)555-1234', 'RTYU'
insert into @tmpTable select 'E', 'Bill', 'Blake', '(555)555-0000', 'BVNM'
insert into @tmpTable select 'F', 'Bill', 'Blake', '(555)555-0000', '%^&*'
insert into @tmpTable select 'G', 'John', 'Smith', '(555)555-1234', '!#RF'

select row_number() over (partition by First, Last, Phone order by ID) NewIDNum, *  
from @tmpTable order by ID

现在它给了我结果

NewIDNum             ID   First Last  Phone         NonKeyField
-------------------- ---- ----- ----- ------------- -----------
1                    A    John  Smith (555)555-1234 ASDF
2                    B    John  Smith (555)555-1234 GHJK
1                    C    Jane  Smith (555)555-1234 QWER
3                    D    John  Smith (555)555-1234 RTYU
1                    E    Bill  Blake (555)555-0000 BVNM
2                    F    Bill  Blake (555)555-0000 %^&*
4                    G    John  Smith (555)555-1234 !#RF

然而,这与我想要的相反 NewIDNum 每当它找到一个新的密钥组合时,就会重置它的计数器。我希望所有相同的组合都有相同的ID。因此,如果它按照我想要的方式运行,我会得到以下结果

NewIDNum             ID   First Last  Phone         NonKeyField
-------------------- ---- ----- ----- ------------- -----------
1                    A    John  Smith (555)555-1234 ASDF
1                    B    John  Smith (555)555-1234 GHJK
2                    C    Jane  Smith (555)555-1234 QWER
1                    D    John  Smith (555)555-1234 RTYU
3                    E    Bill  Blake (555)555-0000 BVNM
3                    F    Bill  Blake (555)555-0000 %^&*
1                    G    John  Smith (555)555-1234 !#RF

得到我想要的结果的正确方法是什么?

我在最初的帖子中没有包括这一要求 :我需要 新IDNum 如果添加了更多的行,则在后续运行该查询时为现有行生成相同的数字(假设如果对ID列执行了排序,则所有新行都将具有更高的ID“值”)

因此,如果在后一天完成以下操作

insert into @tmpTable select 'H', 'John', 'Smith', '(555)555-1234', '4321'
insert into @tmpTable select 'I', 'Jake', 'Jons', '(555)555-1234', '1234'
insert into @tmpTable select 'J', 'John', 'Smith', '(555)555-1234', '2345'

再次运行正确的查询会给出

NewIDNum             ID   First Last  Phone         NonKeyField
-------------------- ---- ----- ----- ------------- -----------
1                    A    John  Smith (555)555-1234 ASDF
1                    B    John  Smith (555)555-1234 GHJK
2                    C    Jane  Smith (555)555-1234 QWER
1                    D    John  Smith (555)555-1234 RTYU
3                    E    Bill  Blake (555)555-0000 BVNM
3                    F    Bill  Blake (555)555-0000 %^&*
1                    G    John  Smith (555)555-1234 !#RF
1                    H    John  Smith (555)555-1234 4321
4                    I    Jake  Jons  (555)555-1234 1234
1                    J    John  Smith (555)555-1234 2345

4 回复 | 直到 13 年前

1

8

Andomar 13 年前

你可以使用 dense_rank() 以下为:

dense_rank() over (order by First, Last, Phone) as NewIDNum

作为对你评论的回应,你可以对旧的 Id 具有相同行的每组列 (First, Last, Phone) 组合:

select  *
from    (
        select  dense_rank() over (order by min_id) as new_id
        ,       *
        from    (
                select  min(id) over (
                            partition by First, Last, Phone) as min_id
                ,       *
                from    @tmpTable 
                ) as sub1
        ) as sub3
order by
        new_id

2

1

etliens 13 年前

基于@Andomar的原始答案——这将适用于您更新的需求(尽管这不太可能很好地扩展)

select
    DENSE_RANK() over (ORDER BY IdRank, First, Last, Phone) AS NewIDNum,
    ID,
    First,
    Last,
    Phone,
    NonKeyField
from
(
    select
        MIN(ID) OVER (PARTITION BY First, Last, Phone) as IdRank,
        *
    from
        @tmpTable
) as x
order by
    ID;

3

0

Community Mohan Dere 8 年前

幸亏 Andomar's answer 作为起点,我自己解决了

select sub1.rn, tt.*
from @tmpTable tt
inner join (
    select row_number() over (order by min(ID)) as rn, first, last, phone
    from @tmpTable
    group by first, last, phone
    ) as sub1 on tt.first = sub1.first and tt.last = sub1.last and tt.phone = sub1.phone

这会产生

rn                   ID   First Last  Phone         NonKeyField
-------------------- ---- ----- ----- ------------- -----------
1                    A    John  Smith (555)555-1234 ASDF
1                    B    John  Smith (555)555-1234 GHJK
1                    D    John  Smith (555)555-1234 RTYU
1                    G    John  Smith (555)555-1234 !#RF
1                    H    John  Smith (555)555-1234 4321
1                    J    John  Smith (555)555-1234 2345
2                    C    Jane  Smith (555)555-1234 QWER
3                    E    Bill  Blake (555)555-0000 BVNM
3                    F    Bill  Blake (555)555-0000 %^&*
4                    I    Jake  Jons  (555)555-1234 1234

从SQL执行计划来看,Adnomar的答案对于更大的数据集会比我的答案运行得更快。(53%的执行时间VS相邻运行时的47%执行时间,并选中“包括实际执行计划”。

4

-1

iruvar 13 年前

这应该有效

select dense_rank() over (order by First, Last, Phone) NewIDNum, *  
from @tmpTable order by ID