请不要指向我读过很多关于如何在 SQL 中创建树结构或 CTE 的文章!!!我认为这对于内心深处的 t-sql 来说可能不是那么难，但对我来说绝对是难:)。
情况是这样，我必须创建一个如下所示的报告:
alt text http://img85.imageshack.us/img85/6372/70337249.png
当我的存储过程(SQL Server sproc)的参数设置为“全部”时，这非常有用，因为这只会抓取所有数据，最终用户可以展开/折叠项目以查看层次结构。例如，当我运行报告并选择一个名称(例如在本例中为“Kevin Bicking”)时，就会出现问题，请查看结果:
alt text http://img69.imageshack.us/img69/8398/46964880.png
问题是我只得到了 kevin 的直接下属，但我实际上需要查看所有的子直接下属。例如，在第一张图片中，我希望我的报告显示所有低于 kevin、低于 kelvin 和低于 Tim 的人，等等。
我了解这个问题，但我不知道如何在 T-SQL 中处理它。这是我的存储过程:
CREATE PROCEDURE [dbo].[rptContactsHierarchy]
@ContactID varchar(100)='All'
AS
BEGIN
SET NOCOUNT ON;
SELECT
c1.id AS EmployeeID,
c2.id as ManagerID,
c1.first_name + ' ' + c1.last_name AS [EmployeeName],
c1.title AS Title,
c2.first_name + ' ' + c2.last_name AS [ReportsTo]
FROM
Contacts c1
INNER JOIN
Contacts c2
ON
c1.reports_to_id = c2.id
WHERE
c1.deleted=0
AND (@ContactID='All' OR (c2.first_name + ' ' + c2.last_name = @ContactID OR (c1.first_name + ' ' + c1.last_name = @ContactID)))
END
sproc 工作正常，没有错误，但我的问题是使用我在此处列出的字段，如何更改它以获取其他名称下的直接报告，如上所述。基本上 EmployeeName 字段每次都是顶级(即报表参数)，ReportsTo 别名是您在图像中看到的报表上的字段。
我没有关于 SSRS 报告的问题，只是关于如何修改查询，以便在这种情况下，如果我选择 Kevin Bicking 并将其传递给我的存储过程。它目前只返回直接员工 Kelvin Squires。但我希望它返回的不仅仅是Kelvin，而是所有向Kelvin汇报的人，以及所有在Kelvin手下可能是老板但也有直接下属的人。
非常感谢任何帮助。感谢您的宝贵时间!
编辑部分
我正在使用 sql server 2005。有人要求提供表定义，请注意我没有创建此表，它是基于 CRM 的系统，自动生成:USE [sugarcrm]
GO
/****** Object: Table [dbo].[contacts] Script Date: 07/22/2010 10:44:31 ******/
SET ANSI_NULLS OFF
GO
SET QUOTED_IDENTIFIER ON
GO
SET ANSI_PADDING OFF
GO
CREATE TABLE [dbo].[contacts](
[id] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NOT NULL,
[date_entered] [datetime] NULL,
[date_modified] [datetime] NULL,
[modified_user_id] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[created_by] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[description] [text] COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[deleted] [bit] NULL DEFAULT ('0'),
[assigned_user_id] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[team_id] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[salutation] [varchar](5) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[first_name] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[last_name] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[title] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[department] [varchar](255) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[do_not_call] [bit] NULL DEFAULT ('0'),
[phone_home] [varchar](25) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[phone_mobile] [varchar](25) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[phone_work] [varchar](25) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[phone_other] [varchar](25) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[phone_fax] [varchar](25) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[primary_address_street] [varchar](150) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[primary_address_city] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[primary_address_state] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[primary_address_postalcode] [varchar](20) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[primary_address_country] [varchar](255) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[alt_address_street] [varchar](150) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[alt_address_city] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[alt_address_state] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[alt_address_postalcode] [varchar](20) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[alt_address_country] [varchar](255) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[assistant] [varchar](75) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[assistant_phone] [varchar](25) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[lead_source] [varchar](100) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[reports_to_id] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[birthdate] [datetime] NULL,
[portal_name] [varchar](255) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[portal_active] [bit] NOT NULL DEFAULT ('0'),
[portal_password] [varchar](32) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[portal_app] [varchar](255) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
[campaign_id] [varchar](36) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
CONSTRAINT [pk_contacts] PRIMARY KEY CLUSTERED
(
[id] ASC
)WITH (IGNORE_DUP_KEY = OFF) ON [PRIMARY]
) ON [PRIMARY] TEXTIMAGE_ON [PRIMARY]
GO
SET ANSI_PADDING OFF
解决方案
在你们的帮助下，这是我的解决方案
set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
go
-- =============================================
-- Author: <Author,,Name>
-- Create date: <Create Date,,>
-- Description: <Description,,>
-- =============================================
ALTER PROCEDURE [dbo].[rptContactsHierarchy]
@ContactID varchar(100)='All'
AS
BEGIN
SET NOCOUNT ON;
--grab id of @contactid
DECLARE @Test varchar(36)
SELECT @Test = (SELECT id FROM contacts c1 WHERE c1.first_name + ' ' + c1.last_name = @ContactID)
;WITH StaffTree AS
(
SELECT
c.id,
c.Title,
c.first_name,
c.last_name,
c.reports_to_id,
c.reports_to_id as Manager_id,
cc.first_name AS Manager_first_name,
cc.last_name as Manager_last_name,
cc.first_name + ' ' + cc.last_name AS [ReportsTo],
c.first_name + ' ' + c.last_name as EmployeeName,
1 AS LevelOf
FROM Contacts c
LEFT OUTER JOIN Contacts cc ON c.reports_to_id=cc.id
WHERE c.id=@Test OR (@Test IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id,
s.Title,
s.first_name,
s.last_name,
s.reports_to_id,
t.id,
t.first_name,
t.last_name,
t.first_name + ' ' + t.last_name,
s.first_name + ' ' + s.last_name,
t.LevelOf+1
FROM StaffTree t
INNER JOIN Contacts s ON t.id=s.reports_to_id
WHERE s.reports_to_id=@Test OR @Test IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
END
最佳答案
EDIT 基于 OP 的表格:
这是一个使用 OP 表定义中的列的示例，我的示例数据如下:
1-Jerome
|
2-Joe
/ \
3-Paul 6-David
/ \ / \
4-Jack 5-Daniel 7-Ian 8-Helen
--I only included the needed columns from the OP's table here
DECLARE @Contacts table (id varchar(36), first_name varchar(100), reports_to_id varchar(36))
INSERT @Contacts VALUES ('1','Jerome', NULL )
INSERT @Contacts VALUES ('2','Joe' ,'1')
INSERT @Contacts VALUES ('3','Paul' ,'2')
INSERT @Contacts VALUES ('4','Jack' ,'3')
INSERT @Contacts VALUES ('5','Daniel','3')
INSERT @Contacts VALUES ('6','David' ,'2')
INSERT @Contacts VALUES ('7','Ian' ,'6')
INSERT @Contacts VALUES ('8','Helen' ,'6')
DECLARE @Root_id char(4)
--get complete tree---------------------------------------------------
SET @Root_id=null
PRINT '@Root_id='+COALESCE(''''+@Root_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE c.id=@Root_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE s.reports_to_id=@Root_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
--get all below 2---------------------------------------------------
SET @Root_id=2
PRINT '@Root_id='+COALESCE(''''+@Root_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE c.id=@Root_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE s.reports_to_id=@Root_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
--get all below 6---------------------------------------------------
SET @Root_id=6
PRINT '@Root_id='+COALESCE(''''+@Root_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE c.id=@Root_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE s.reports_to_id=@Root_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
输出:
@Root_id=null
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
1 Jerome NULL NULL NULL 1
2 Joe 1 1 Jerome 2
3 Paul 2 2 Joe 3
6 David 2 2 Joe 3
7 Ian 6 6 David 4
8 Helen 6 6 David 4
4 Jack 3 3 Paul 4
5 Daniel 3 3 Paul 4
(8 row(s) affected)
@Root_id='2 '
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
2 Joe 1 1 Jerome 1
3 Paul 2 2 Joe 2
6 David 2 2 Joe 2
7 Ian 6 6 David 3
8 Helen 6 6 David 3
4 Jack 3 3 Paul 3
5 Daniel 3 3 Paul 3
(7 row(s) affected)
@Root_id='6 '
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
6 David 2 2 Joe 1
7 Ian 6 6 David 2
8 Helen 6 6 David 2
(3 row(s) affected)
关于sql-server - 难以显示组织结构图的 T-SQL(层次结构/递归)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3309405/