识别 SQL Server 中的链接 ID 集

发布于 2024-11-26 22:31:13 字数 880 浏览 1 评论 0原文

我有一个简单的表,如下所示:

ClientID    ItemID
1           1
1           2
1           3
2           1
2           2
3           3
4           3
5           1
5           2
5           4
5           5

其中两列组合为主键。我现在的任务是识别分配给 ClientID 的所有唯一的 ItemID 集。因此,在我的示例中,集合将是:

ItemIDs 1,2,3 (used by ClientID 1)
ItemIDs 1,2 (used by ClientID 2)
ItemIDs 3 (used by ClientIDs 3 and 4)
ItemIDs 1,2,4,5 (used by ClientID 5)

理想情况下,输出将是两个表:

SetID    ItemID
1        1
1        2
1        3
2        1
2        2
3        3
4        1
4        2
4        4
4        5

ClientID    SetID
1           1
2           2
3           3
4           3
5           4

其中 SetID 将是在其他地方使用的新字段。

目前,我识别唯一集的方法包括使用游标为每个 ClientID 构建有序 ItemID 的字符串,然后比较输出以获取唯一字符串,最后将其解析回来。写得很快,但感觉很糟糕。

我确信一定有比这更好的方法。有什么想法吗?

I have a simple table that looks like this:

ClientID    ItemID
1           1
1           2
1           3
2           1
2           2
3           3
4           3
5           1
5           2
5           4
5           5

where both columns combine to be the primary key. I am now tasked with identifying all the unique sets of ItemIDs assigned to ClientIDs. So in my example, the sets would be:

ItemIDs 1,2,3 (used by ClientID 1)
ItemIDs 1,2 (used by ClientID 2)
ItemIDs 3 (used by ClientIDs 3 and 4)
ItemIDs 1,2,4,5 (used by ClientID 5)

Ideally the output would be two tables:

SetID    ItemID
1        1
1        2
1        3
2        1
2        2
3        3
4        1
4        2
4        4
4        5

ClientID    SetID
1           1
2           2
3           3
4           3
5           4

where SetID would be a new field for use elsewhere.

Currently the way I have of identifying the unique sets involves using a cursor to build a string of the ordered ItemIDs for each ClientID, then comparing the output to get the unique strings, and finally parsing it back. It was quick enough to write but feels horrible.

I'm sure there must be a better way than this. Any ideas?

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も星光 2024-12-03 22:31:13
-- Table to hold test data
declare @T table
(
  ClientID int,
  ItemID int
)

insert into @T values
(1, 1),(1, 2),(1, 3),
(2, 1),(2, 2),
(3, 3),(4, 3),
(5, 1),(5, 2),(5, 4),(5, 5)


-- Temp table that will hold the generated set's
declare @Tmp table
(
  ClientID int,
  ItemIDSet varchar(max),
  SetID int
)

-- Query the sets using rank() over a comma separated ItemIDSet
insert into @Tmp
select ClientID,
       ItemIDSet,
       rank() over(order by ItemIDSet) as SetID
from (
      select T1.ClientID,
             stuff((select ','+cast(T2.ItemID as varchar(10))
                    from @T as T2
                    where T1.ClientID = T2.ClientID
                    order by T2.ItemID
                    for xml path('')), 1, 1, '') as ItemIDSet
      from @T as T1
      group by T1.ClientID
     ) as T

-- Get ClientID and SetID from @Tmp
select ClientID, 
       SetID
from @Tmp
order by ClientID

-- Get SetID and ItemID from @Tmp
select SetID,
       T3.N.value('.', 'int') as ItemID
from ( 
       select distinct
              SetID,
              '<i>'+replace(ItemIDSet, ',', '</i><i>')+'</i>' as ItemIDSet
       from @Tmp
     ) as T1
  cross apply 
     ( 
       select cast(T1.ItemIDSet as xml) as ItemIDSet
     ) as T2
  cross apply T2.ItemIDSet.nodes('i') as T3(N)

结果:

ClientID    SetID
----------- -----------
1           2
2           1
3           4
4           4
5           3

SetID       ItemID
----------- -----------
1           1
1           2
2           1
2           2
2           3
3           1
3           2
3           4
3           5
4           3

SetID 的值与您提供的输出中的值不完全相同,但我认为这不是一个大问题。 SetID 是根据按 ItemIDSet 排序的排名函数 rank() over(order by ItemIDSet) 生成的。

拿它来旋转

-- Table to hold test data
declare @T table
(
  ClientID int,
  ItemID int
)

insert into @T values
(1, 1),(1, 2),(1, 3),
(2, 1),(2, 2),
(3, 3),(4, 3),
(5, 1),(5, 2),(5, 4),(5, 5)


-- Temp table that will hold the generated set's
declare @Tmp table
(
  ClientID int,
  ItemIDSet varchar(max),
  SetID int
)

-- Query the sets using rank() over a comma separated ItemIDSet
insert into @Tmp
select ClientID,
       ItemIDSet,
       rank() over(order by ItemIDSet) as SetID
from (
      select T1.ClientID,
             stuff((select ','+cast(T2.ItemID as varchar(10))
                    from @T as T2
                    where T1.ClientID = T2.ClientID
                    order by T2.ItemID
                    for xml path('')), 1, 1, '') as ItemIDSet
      from @T as T1
      group by T1.ClientID
     ) as T

-- Get ClientID and SetID from @Tmp
select ClientID, 
       SetID
from @Tmp
order by ClientID

-- Get SetID and ItemID from @Tmp
select SetID,
       T3.N.value('.', 'int') as ItemID
from ( 
       select distinct
              SetID,
              '<i>'+replace(ItemIDSet, ',', '</i><i>')+'</i>' as ItemIDSet
       from @Tmp
     ) as T1
  cross apply 
     ( 
       select cast(T1.ItemIDSet as xml) as ItemIDSet
     ) as T2
  cross apply T2.ItemIDSet.nodes('i') as T3(N)

Result:

ClientID    SetID
----------- -----------
1           2
2           1
3           4
4           4
5           3

SetID       ItemID
----------- -----------
1           1
1           2
2           1
2           2
2           3
3           1
3           2
3           4
3           5
4           3

The values of SetID's is not exactly the same as in the output you have provided but I don't think that would be a big issue. The SetID's are generated from the rank function rank() over(order by ItemIDSet) ordered by ItemIDSet.

Take it for a spin.

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