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求一SQL语句

时间:2011-12-13

来源:互联网

Table:
id name pid
1 a 0
2 b 0
3 a1 1
4 a2 1
5 b1 2

一个分类表,pid就是父类id,现在想获得如下结果:
id name pid
1 a 0
3 a1 1
4 a2 1
2 b 0
5 b1 2

也就是分类显示,父类和子类一块显示,请问SQL语句怎么写?

作者: mr_tanglin   发布时间: 2011-12-13

BOM按节点排序。

作者: fredrickhu   发布时间: 2011-12-13

引用 1 楼 fredrickhu 的回复:

BOM按节点排序。

不懂,能讲解下吗?

作者: mr_tanglin   发布时间: 2011-12-13

去网上找下 物料排序。 很多实例的

作者: szstephenzhou   发布时间: 2011-12-13

SQL code
--> 测试数据:[tb]
if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([id] int,[name] varchar(2),[pid] int)
insert [tb]
select 1,'a',0 union all
select 2,'b',0 union all
select 3,'a1',1 union all
select 4,'a2',1 union all
select 5,'b1',2

;with t as
(
    select*,cast(id as varbinary(max)) as px from tb as a
    where not exists(select * from tb where id=a.pid)
    union all 
    select a.*, cast(b.px+cast(a.id as varbinary) as varbinary(max))
    from tb as a 
    join t as b on a.pid=b.id

)

select id,name, pid from t order by px

/*
id          name pid
----------- ---- -----------
1           a    0
3           a1   1
4           a2   1
2           b    0
5           b1   2

*/

作者: Beirut   发布时间: 2011-12-13

SQL code
BOM按节点排序应用实例 
--------------------------------------------------------------------------

--  Author : htl258(Tony)

--  Date   : 2010-04-23 02:37:28

--  Version:Microsoft SQL Server 2008 (RTM) - 10.0.1600.22 (Intel X86) 

--          Jul  9 2008 14:43:34 

--          Copyright (c) 1988-2008 Microsoft Corporation

--          Developer Edition on Windows NT 5.1 <X86> (Build 2600: Service Pack 3)

--  Subject: BOM按节点排序应用实例

--------------------------------------------------------------------------

 

--实例1:

--> 生成测试数据表:tb

 

IF NOT OBJECT_ID('[tb]') IS NULL

    DROP TABLE [tb]

GO

CREATE TABLE [tb]([id] INT,[code] NVARCHAR(10),[pid] INT,[name] NVARCHAR(10))

INSERT [tb]

SELECT 1,'01',0,N'服装' UNION ALL

SELECT 2,'01',1,N'男装' UNION ALL

SELECT 3,'01',2,N'西装' UNION ALL

SELECT 4,'01',3,N'全毛' UNION ALL

SELECT 5,'02',3,N'化纤' UNION ALL

SELECT 6,'02',2,N'休闲装' UNION ALL

SELECT 7,'02',1,N'女装' UNION ALL

SELECT 8,'01',7,N'套装' UNION ALL

SELECT 9,'02',7,N'职业装' UNION ALL

SELECT 10,'03',7,N'休闲装' UNION ALL

SELECT 11,'04',7,N'西装' UNION ALL

SELECT 12,'01',11,N'全毛' UNION ALL

SELECT 13,'02',11,N'化纤' UNION ALL

SELECT 14,'05',7,N'休闲装'

GO

--SELECT * FROM [tb]

 

-->SQL查询如下:

 

;WITH T AS

(

    SELECT CAST(CODE AS VARCHAR(20)) AS CODE,*,

        CAST(ID AS VARBINARY(MAX)) AS px 

    FROM tb AS A

    WHERE NOT EXISTS(SELECT * FROM tb WHERE id=A.pid)

    UNION ALL 

    SELECT CAST(B.CODE+A.CODE AS VARCHAR(20)),A.*,

         CAST(B.px+CAST(A.ID AS VARBINARY) AS VARBINARY(MAX))    

    FROM tb AS A

        JOIN T AS B

           ON A.pid=B.id

)

SELECT Code,Name FROM T 

ORDER BY px

/*

Code                 Name

-------------------- ----------

01                   服装

0101                 男装

010101               西装

01010101             全毛

01010102             化纤

010102               休闲装

0102                 女装

010201               套装

010202               职业装

010203               休闲装

010204               西装

01020401             全毛

01020402             化纤

010205               休闲装

 

(14 行受影响)

*/

 

--实例2:

--> 生成测试数据表:tb

IF NOT OBJECT_ID('[tb]') IS NULL
    DROP TABLE [tb]
GO
CREATE TABLE [tb]([id] INT,[parentid] INT,[categoryname] NVARCHAR(10))
INSERT [tb]
SELECT 1,0,'test1' UNION ALL
SELECT 2,0,'test2' UNION ALL
SELECT 3,1,'test1.1' UNION ALL
SELECT 4,2,'test2.1' UNION ALL
SELECT 5,3,'test1.1.1' UNION ALL
SELECT 6,1,'test1.2'
GO
--SELECT * FROM [tb]

-->SQL查询如下:
;WITH T AS
(
    SELECT *,CAST(ID AS VARBINARY(MAX)) AS px 
    FROM tb AS A
    WHERE NOT EXISTS(SELECT * FROM tb WHERE id=A.[parentid])
    UNION ALL 
    SELECT A.*,CAST(B.px+CAST(A.ID AS VARBINARY) AS VARBINARY(MAX))  
    FROM tb AS A
        JOIN T AS B
           ON A.[parentid]=B.id
)
SELECT [id],[parentid],[categoryname] FROM T 
ORDER BY px
/*
id          parentid    categoryname
----------- ----------- ------------
1           0           test1
3           1           test1.1
5           3           test1.1.1
6           1           test1.2
2           0           test2
4           2           test2.1

(6 行受影响)
*/



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作者: fredrickhu   发布时间: 2011-12-13