Question

这是我的Sample table Structure，

 Create table #table(advId int identity(1,1),name nvarchar(100),ranks nvarchar(5),ReferId int ,ReferalRank nvarchar(5))

 insert into #table(name,ranks,ReferId,ReferalRank) values('King','MGR',0,'0')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Maceij','MGR',1,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Los','MGR',1,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Los1','ADV',1,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Griff','MGR',1,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('SA','MGR',2,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('CASSANDRA','MGR',2,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Jason','MGR',3,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Smith','MGR',3,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Akee','MGR',6,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Manasa','ADV',6,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Akee','MGR',10,'MGR')
 insert into #table (name,ranks,ReferId,ReferalRank) values('Manasa','ADV',10,'MGR')

 select *from #table

让我谈谈我的表结构这里， AdvId 1由admin推荐，（2,3,4,5）是1级的第1级（6,7,8,9）是1级的第2级（10,11）是1级的第3级（12,13）是1级的第4级每个顾问都有相同的逻辑 Like This Structure

如何为每个顾问选择最多3个级别的经理人数（代理人下有多少经理人）

 advId name     CountOfmanager
 1     king       8            --2,3,5,6,7,8,9,10
 2     Maceij     3            --6,7,10
 3      los       2            --8,9
 4     Los1       0            -- nobody 
 5     Griff      0        -- nobody 
 6     SA         2            -- 10,12   
 7     CASSANDRA  0            -- nobody 
 8     Jason      0
 9     Smith      0
10    Akee       1             --12
11    manasa     0
12    Akee       0
13    Manasa     0

这是我试过的。

with cte (advId,ReferId,Level)
as
(
 select AdvId,ReferId,1 as Level from table where ReferId=1
 union all
 select a.AdvId,a.ReferId ,Level+1 from table   
 as a inner join cte as b on b.AdvId=a.ReferId
 )
 select COUNT(b.AdvId) From cte as a inner join table 
 as b on a.advId=b.advId where a.level<=3 and b.ranks='MGR'

我希望它明确，协助我得到结果

Answer 1

我的结果与你的前两行有点不同，因为对于国王和Maceij你缺少12（Akee），如果你有12对10，你也必须有1和2个优点。

首先，我找到每一行的深层次。

其次我把父母放在每一排。

第三次计算孩子

最后我只是显示结果。

;WITH alignRef AS (

        SELECT advId, name, ranks, ReferId, ReferalRank, 0 AS DeepLevel 
        FROM #table
        WHERE ReferId = 0

        UNION ALL

        SELECT T.advId, T.name, T.ranks, T.ReferId, T.ReferalRank, R.DeepLevel+1 AS DeepLevel 
        FROM #table T
        INNER JOIN alignRef AS R
            ON T.ReferId = R.advId
            --AND T.ranks = R.ranks

     ), getParents AS ( 
         SELECT advId, name, ranks, ReferId, ReferalRank, DeepLevel, advId AS ParentID
         FROM alignRef
         WHERE ranks='MGR' AND DeepLevel <= 4

         UNION ALL

         SELECT R.advId, R.name, R.ranks, R.ReferId, R.ReferalRank, R.DeepLevel, CC.ParentID AS ParentID 
         FROM getParents AS CC
         INNER JOIN alignRef AS R
            ON CC.advId = R.ReferId
        WHERE R.ranks='MGR' AND R.DeepLevel <= 4

    ), CountChilds AS (
        SELECT ParentID, COUNT(*) AS Amount
        FROM getParents
        WHERE advId <> ParentID
        GROUP BY ParentID
    )
    SELECT T.advId, T.name, ISNULL(CC.Amount, 0) AS CountOfmanager
    FROM #table AS T
    LEFT JOIN CountChilds AS CC
        ON CC.ParentID = T.advId
    ORDER BY T.advId

结果：

advId   name        CountOfmanager
1       King        9
2       Maceij      4
3       Los         2
4       Los1        0
5       Griff       0
6       SA          2
7       CASSANDRA   0
8       Jason       0
9       Smith       0
10      Akee        1
11      Manasa      0
12      Akee        0
13      Manasa      0

我如何使用CTE

1 个答案: