SQL Server 2005 中的树形数据处理示例-2

王朝mssql·作者佚名  2006-01-09
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-- =====================================================

-- 直接查询的应用实例

-- =====================================================

-- 1. 每个叶子结点的 FullName

WITH stb([id],[FullName],[pid],[flag])

AS(

SELECT [id],CAST(RTRIM([name]) as nvarchar(4000)),[pid],1

FROM [tb] A

WHERE NOT EXISTS(

SELECT 1 FROM [tb]

WHERE [pid]=A.[id])

UNION ALL

SELECT A.[id],RTRIM(B.[name])+'/'+A.[FullName],B.[pid],A.flag+1

FROM stb A,[tb] B

WHERE A.[pid]=B.[id])

SELECT [id],[FullName] FROM stb A

WHERE NOT EXISTS(

SELECT * FROM stb

WHERE [id]=A.[id]

AND flag>A.flag)

ORDER BY [id]

GO

-- 2. 每个结点的 FullName

WITH stb([id],[FullName],[pid],[flag])

AS(

SELECT [id],CAST(RTRIM([name]) as nvarchar(4000)),[pid],1

FROM [tb]

UNION ALL

SELECT A.[id],RTRIM(B.[name])+'/'+A.[FullName],B.[pid],A.flag+1

FROM stb A,[tb] B

WHERE A.[pid]=B.[id])

SELECT [id],[FullName] FROM stb A

WHERE NOT EXISTS(

SELECT * FROM stb

WHERE [id]=A.[id]

AND flag>A.flag)

ORDER BY [id]

GO

-- 3. 树形显示数据

WITH stb([id],[level],[sid])

AS(

SELECT [id],1,CAST(RIGHT(10000+[id],4) as varchar(8000))

FROM [tb]

WHERE [pid]=0

UNION ALL

SELECT A.[id],B.[level]+1,B.sid+RIGHT(10000+A.[id],4)

FROM [tb] A,stb B

WHERE A.[pid]=B.[id])

SELECT N'|'+REPLICATE('-',B.[level]*4)+A.name

FROM [tb] A,stb B

WHERE a.[id]=b.[id]

ORDER BY b.sid

GO

-- 4. 检查不规范的数据

WITH chktb([id],[pid],[level],[Path],[Flag])

AS(

SELECT [id],[pid],1,

CAST([id] as varchar(8000)),

CASE WHEN [id]=[pid] THEN 1 ELSE 0 END

FROM [tb]

UNION ALL

SELECT A.[id],B.[pid],B.[level]+1,

CAST(B.[Path]+' > '+RTRIM(A.[id]) as varchar(8000)),

CASE WHEN A.[id]=B.[pid] THEN 1 ELSE 0 END

FROM [tb] A,chktb B

WHERE A.[pid]=B.[id]

AND B.[Flag]=0)

SELECT * FROM chktb

WHERE [Flag]=1

ORDER BY [Path]

GO

-- 5. 查询结点的所有子结点数

WITH sumtb([id],[level])

AS(

SELECT [pid],1

FROM [tb] A

WHERE [pid]<>0

UNION ALL

SELECT A.[pid],B.[level]+1

FROM [tb] A,sumtb B

WHERE A.[id]=B.[id]

AND A.[pid]<>0)

SELECT A.[id],ChildCounts=COUNT(b.[id])

FROM [tb] A

LEFT JOIN sumtb B

ON A.[id]=B.[id]

GROUP BY A.[id]

GO

-- 6. 查询结点的所有父结点数

WITH sumtb([id],[level],[ParentCounts])

AS(

SELECT [id],1,0

FROM [tb] A

WHERE [pid]=0

UNION ALL

SELECT A.[id],B.[level]+1,B.[ParentCounts]+1

FROM [tb] A,sumtb B

WHERE A.[pid]=B.[id])

SELECT * FROM sumtb

order by [ID]

GO

 
 
 
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