Sql 将所有类似人员集中到一组_Sql_Sql Server_Tsql_Sql Server 2012

Sql 将所有类似人员集中到一组

sql sql-server tsql sql-server-2012

Sql 将所有类似人员集中到一组,sql,sql-server,tsql,sql-server-2012,Sql,Sql Server,Tsql,Sql Server 2012,我有一个人有几个身份证。其中一些在Id1列中，一些在Id2列中。我想将所有相同的个人Id收集到一个组中如果id1=10，则与id2=20在同一行中。这意味着id1=10的人和id2=20的人是同一个人输入和输出示例：输入输出安娜，这是个好例子吗？这是一个连接组件问题输入输出我怀疑这可以通过递归CTE实现，但这里有一个不那么简单的解决方案 -- CREATE Temps CREATE TABLE #Table (id1 INT, id2 INT) CREATE TABLE #

我有一个人有几个身份证。其中一些在Id1列中，一些在Id2列中。我想将所有相同的个人Id收集到一个组中

如果id1=10，则与id2=20在同一行中。这意味着id1=10的人和id2=20的人是同一个人

输入和输出示例：

输入输出

安娜，这是个好例子吗？这是一个连接组件问题

输入输出

我怀疑这可以通过递归CTE实现，但这里有一个不那么简单的解决方案

-- CREATE Temps
CREATE TABLE #Table (id1 INT, id2 INT)
CREATE TABLE #NewTable (NewID INT, OldID INT)
CREATE TABLE #AllIDs (ID INT)

-- Insert Test data
INSERT #Table
        ( id1, id2 )
VALUES  (  10, 20 ),
        (  10, 30 ),
        (  30, 20 ),
        (  10, 40 ),

        (  50, 70 ),
        (  60, 50 ),
        (  70, 70 ),
        (  110, 120 ),
        (  120, 130 ),
        (  140, 130 )

-- Assemble all possible OldIDs
INSERT INTO #AllIDs
    SELECT id1 FROM #Table
    UNION
    SELECT id2 FROM #Table

DECLARE @NewID INT = 1,
        @RowCnt int

-- Insert seed OldID
INSERT #NewTable
    SELECT TOP 1 @NewID, id
    FROM #AllIDs
    WHERE id NOT IN (SELECT OldID FROM #NewTable)
    ORDER BY 2

SET @RowCnt = @@ROWCOUNT

WHILE @RowCnt > 0
BEGIN   
    WHILE @RowCnt > 0
    BEGIN
        -- Check for id2 that match current OldID
        INSERT #NewTable
            SELECT DISTINCT @NewID, id2
            FROM #Table t
                INNER JOIN #NewTable nt ON t.id1 = nt.OldID
            WHERE nt.[NewID] = @NewID
                AND t.id2 NOT IN (SELECT OldID FROM #NewTable WHERE [NewID] = @NewID)

        SELECT @RowCnt = @@ROWCOUNT

        -- Check for id1 that match current OldID
        INSERT #NewTable
            SELECT DISTINCT @NewID, id1
            FROM #Table t
                INNER JOIN #NewTable nt ON t.id2 = nt.OldID
            WHERE nt.[NewID] = @NewID
                AND t.id1 NOT IN (SELECT OldID FROM #NewTable WHERE [NewID] = @NewID)

        SELECT @RowCnt = @RowCnt + @@ROWCOUNT
    END

    SET @NewID = @NewID + 1

    -- Add another seed OldID if any left
    INSERT #NewTable
        SELECT TOP 1 @NewID, id
        FROM #AllIDs
        WHERE id NOT IN (SELECT OldID FROM #NewTable)
        ORDER BY 2

    SELECT @RowCnt = @@ROWCOUNT
END

-- Get Results
SELECT * FROM #NewTable ORDER BY [NewID], OldID

CTE版本。注意，我还添加了一些数据点来模拟重复的ID和单独的ID

--create test data
declare @table table (Id1 int, Id2 int);
insert  @table values 
        (10, 20),
        (10, 30),
        (30, 30),
        (10, 40),
        (40, 45),
        (20, 40),
        (50, 70),
        (60, 50),
        (70, 70),
        (80, 80);
select  *
from    @table;

--join related IDs with recursive CTE
;with min_first_cte as (
        select  case when Id1 <= Id2 then Id1 else Id2 end Id1,
                case when Id1 <= Id2 then Id2 else Id1 end Id2
        from    @table
), related_ids_cte as (
        --anchor IDs
        select  distinct Id1 BaseId, Id1 ParentId, Id1 ChildId
        from    min_first_cte
        where   Id1 not in (    select  Id2
                                from    min_first_cte
                                where   Id2 <> Id1)
        union all
        --related recursive IDs
        select  r.BaseId, m.Id1 ParentId, M.Id2 ChildId
        from    min_first_cte   m
        join    related_ids_cte r
                on  r.ChildId = m.Id1
                and m.Id1 <> m.Id2
), distinct_ids_cte as (
        select  distinct r.BaseId, r.ChildId
        from    related_ids_cte r
)
select  dense_rank() over (order by d.BaseId) [NewId],
        d.ChildId OldId
from    distinct_ids_cte d
order   by BaseId, ChildId;

对于递归任务，应该使用递归CTE

with cq as 
    (
        select distinct Id2, Id1 from #Tmp -- get your table 
        union
        select distinct Id1, Id2 from #Tmp -- get your table (or sort output)
        union
        select distinct Id1, Id1 from #Tmp -- add root from Id1 
        union
        select distinct Id2, Id2 from #Tmp -- add root from Id2
    ), cte (Id1, Id2, lvl) 
    as (
        select t.Id1, t.Id2, 0 lvl 
        from cq t
        union all
        select t2.Id2, c.Id1, lvl + 1 lvl 
        from cq t2, cte c
        where t2.Id1 = c.Id2
            and t2.Id1 != c.Id1 
            and c.lvl < 5 -- maximum level of recursion
        )
    select 
        Id1, 
        min(Id2) FirstId1,
        dense_rank() over(order by min(Id2)) rn
    from cte
    group by Id1

最大lvl和有！=如果您的表排列得很好，则不必这样做。

从概念上讲，它是指在给定连接对列表的情况下查找连接的组件。然后，为每个组分配一个新id。以下实现工作：

CREATE TABLE #pairs (a int, b int)
CREATE TABLE #groups (a int, group_id int)

INSERT INTO #pairs
VALUES (1, 2), (3, 4), (5, 6), (5, 7), (3, 9), (8, 10), (11, 12), (1, 3)

-- starting stage - all items belong to their own group
INSERT INTO #groups(a, group_id)
SELECT a, a
  FROM #pairs
 UNION
SELECT b, b
  FROM #pairs

DECLARE @a INT
DECLARE @b INT
DECLARE @cGroup INT

SET ROWCOUNT 0
SELECT * INTO #mytemp FROM #pairs

SET ROWCOUNT 1

SELECT @a = a, @b = b FROM #mytemp

WHILE @@rowcount <> 0
BEGIN
    SET ROWCOUNT 0

    DECLARE @aGroup INT, @bGroup INT, @newGroup INT
    SELECT @aGroup = group_id FROM #groups WHERE a = @a
    SELECT @bGroup = group_id FROM #groups WHERE a = @b
    SELECT @newGroup = MIN(group_id) FROM #groups WHERE a IN (@a, @b)

    -- update the grouping table with the new group
    UPDATE #groups 
       SET group_id = @newGroup 
     WHERE group_id IN (@aGroup, @bGroup)

    DELETE FROM #mytemp 
     WHERE a = @a 
       AND b = @b

    SET ROWCOUNT 1
    SELECT @a = a, @b = b FROM #mytemp
END
SET ROWCOUNT 0

SELECT * FROM #groups

DROP TABLE #mytemp
DROP TABLE #pairs
DROP TABLE #groups

解释如下：

最初，为每个数字分配一组它自己的值在这些对上迭代每对查找最小值并将其设置为新组id 将组id设置为当前组id与当前对中的数字相同的所有数字

就一个过程而言，这是两次迭代，不断地将组ID更新到组-On2中的最小值。

预期结果plzd您想要两列相同的select语句还是在id1和id2相同时插入id3？您的问题很难理解。请重新编辑。此外，不同DBMS的解决方案可能不同。指定您使用的是哪一个。将根据什么标准选择NewId？它来自哪里？为什么在你的例子中前三条记录是一样的？在你的例子中不是。你所说的与你的例子不一致。由于第一个屏幕截图中的所有ID在ID1和ID2中都是相等的，所以所有行的NewId都应该是1，或者所有行的NewId都应该是不同的。我再次问：为什么前三行是1，而后面几行不是1？是的，这正是我想要的！这可能有一个缺陷。尝试在表格中插入这些值10,20,10,30,30,10,40,40,45,20,40,50,70,60,50,70,80,80,120,90,120@你说得对。递归非常好，可以清楚地识别基本成员。当我有更多的时间时，我将不得不重新讨论这个问题。这与EarlOfEnnui下面提出的附加案例不符：值10,20,10,30,30,30,10,40,40,45,20,40,50,70,60,50,70,80,80,120,90，120@SMM，如果您感兴趣，可以使用递归查询来完成。这个问题也是在dba.se上提出的，我写了详细的。固定值Id2>Id1。您可以控制层次结构路径，因此不需要固定级别和c.lvl<5-递归的最大级别可以使用递归查询完成。这个问题也是在dba.se上提出的，我写了详细的问题。

-- CREATE Temps
CREATE TABLE #Table (id1 INT, id2 INT)
CREATE TABLE #NewTable (NewID INT, OldID INT)
CREATE TABLE #AllIDs (ID INT)

-- Insert Test data
INSERT #Table
        ( id1, id2 )
VALUES  (  10, 20 ),
        (  10, 30 ),
        (  30, 20 ),
        (  10, 40 ),

        (  50, 70 ),
        (  60, 50 ),
        (  70, 70 ),
        (  110, 120 ),
        (  120, 130 ),
        (  140, 130 )

-- Assemble all possible OldIDs
INSERT INTO #AllIDs
    SELECT id1 FROM #Table
    UNION
    SELECT id2 FROM #Table

DECLARE @NewID INT = 1,
        @RowCnt int

-- Insert seed OldID
INSERT #NewTable
    SELECT TOP 1 @NewID, id
    FROM #AllIDs
    WHERE id NOT IN (SELECT OldID FROM #NewTable)
    ORDER BY 2

SET @RowCnt = @@ROWCOUNT

WHILE @RowCnt > 0
BEGIN   
    WHILE @RowCnt > 0
    BEGIN
        -- Check for id2 that match current OldID
        INSERT #NewTable
            SELECT DISTINCT @NewID, id2
            FROM #Table t
                INNER JOIN #NewTable nt ON t.id1 = nt.OldID
            WHERE nt.[NewID] = @NewID
                AND t.id2 NOT IN (SELECT OldID FROM #NewTable WHERE [NewID] = @NewID)

        SELECT @RowCnt = @@ROWCOUNT

        -- Check for id1 that match current OldID
        INSERT #NewTable
            SELECT DISTINCT @NewID, id1
            FROM #Table t
                INNER JOIN #NewTable nt ON t.id2 = nt.OldID
            WHERE nt.[NewID] = @NewID
                AND t.id1 NOT IN (SELECT OldID FROM #NewTable WHERE [NewID] = @NewID)

        SELECT @RowCnt = @RowCnt + @@ROWCOUNT
    END

    SET @NewID = @NewID + 1

    -- Add another seed OldID if any left
    INSERT #NewTable
        SELECT TOP 1 @NewID, id
        FROM #AllIDs
        WHERE id NOT IN (SELECT OldID FROM #NewTable)
        ORDER BY 2

    SELECT @RowCnt = @@ROWCOUNT
END

-- Get Results
SELECT * FROM #NewTable ORDER BY [NewID], OldID

--create test data
declare @table table (Id1 int, Id2 int);
insert  @table values 
        (10, 20),
        (10, 30),
        (30, 30),
        (10, 40),
        (40, 45),
        (20, 40),
        (50, 70),
        (60, 50),
        (70, 70),
        (80, 80);
select  *
from    @table;

--join related IDs with recursive CTE
;with min_first_cte as (
        select  case when Id1 <= Id2 then Id1 else Id2 end Id1,
                case when Id1 <= Id2 then Id2 else Id1 end Id2
        from    @table
), related_ids_cte as (
        --anchor IDs
        select  distinct Id1 BaseId, Id1 ParentId, Id1 ChildId
        from    min_first_cte
        where   Id1 not in (    select  Id2
                                from    min_first_cte
                                where   Id2 <> Id1)
        union all
        --related recursive IDs
        select  r.BaseId, m.Id1 ParentId, M.Id2 ChildId
        from    min_first_cte   m
        join    related_ids_cte r
                on  r.ChildId = m.Id1
                and m.Id1 <> m.Id2
), distinct_ids_cte as (
        select  distinct r.BaseId, r.ChildId
        from    related_ids_cte r
)
select  dense_rank() over (order by d.BaseId) [NewId],
        d.ChildId OldId
from    distinct_ids_cte d
order   by BaseId, ChildId;

with cq as 
    (
        select distinct Id2, Id1 from #Tmp -- get your table 
        union
        select distinct Id1, Id2 from #Tmp -- get your table (or sort output)
        union
        select distinct Id1, Id1 from #Tmp -- add root from Id1 
        union
        select distinct Id2, Id2 from #Tmp -- add root from Id2
    ), cte (Id1, Id2, lvl) 
    as (
        select t.Id1, t.Id2, 0 lvl 
        from cq t
        union all
        select t2.Id2, c.Id1, lvl + 1 lvl 
        from cq t2, cte c
        where t2.Id1 = c.Id2
            and t2.Id1 != c.Id1 
            and c.lvl < 5 -- maximum level of recursion
        )
    select 
        Id1, 
        min(Id2) FirstId1,
        dense_rank() over(order by min(Id2)) rn
    from cte
    group by Id1

CREATE TABLE #pairs (a int, b int)
CREATE TABLE #groups (a int, group_id int)

INSERT INTO #pairs
VALUES (1, 2), (3, 4), (5, 6), (5, 7), (3, 9), (8, 10), (11, 12), (1, 3)

-- starting stage - all items belong to their own group
INSERT INTO #groups(a, group_id)
SELECT a, a
  FROM #pairs
 UNION
SELECT b, b
  FROM #pairs

DECLARE @a INT
DECLARE @b INT
DECLARE @cGroup INT

SET ROWCOUNT 0
SELECT * INTO #mytemp FROM #pairs

SET ROWCOUNT 1

SELECT @a = a, @b = b FROM #mytemp

WHILE @@rowcount <> 0
BEGIN
    SET ROWCOUNT 0

    DECLARE @aGroup INT, @bGroup INT, @newGroup INT
    SELECT @aGroup = group_id FROM #groups WHERE a = @a
    SELECT @bGroup = group_id FROM #groups WHERE a = @b
    SELECT @newGroup = MIN(group_id) FROM #groups WHERE a IN (@a, @b)

    -- update the grouping table with the new group
    UPDATE #groups 
       SET group_id = @newGroup 
     WHERE group_id IN (@aGroup, @bGroup)

    DELETE FROM #mytemp 
     WHERE a = @a 
       AND b = @b

    SET ROWCOUNT 1
    SELECT @a = a, @b = b FROM #mytemp
END
SET ROWCOUNT 0

SELECT * FROM #groups

DROP TABLE #mytemp
DROP TABLE #pairs
DROP TABLE #groups