如何使用CTE这个

Question

这是我的样本表结构，

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 

让我对我的表结构 这里， ADVID 1个词语是由管理员提到， （2,3,4 ，5）是第一级1 （6,7,8,9）是第二级1 （10,11）是第三级1 （12,13）是第4级1 同样逻辑For每个顾问 Like This Structure

如何选择计数管理器（有多少管理者的代理下）对于每个顾问达3级

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 

这是我试过了。

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' 

我希望它清除，协助我得到的结果

Answer 1

我结果的位前两排跟你不一样，因为国王和Maceij你失踪12（Akee），如果你有12个10，你也必须有1和2 advIDs。

首先，我发现每一行的深层次。

第二我把父母放到每一行。

第三计数孩子

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：从您预期的答案

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