5

我见过多个将单列卷成逗号分隔列表的示例，但我需要更多。

这是我需要的数据和结果的示例。

DECLARE @SalesPerson table (SalesPersonID int, SalesPersonName varchar(10))
DECLARE @Region table (RegionID int, RegionName varchar(15))
DECLARE @SalesPersonRegion table (SalesPersonID int, RegionID int)

INSERT INTO @SalesPerson (SalesPersonID, SalesPersonName) VALUES (1,'Jeff') 
INSERT INTO @SalesPerson (SalesPersonID, SalesPersonName) VALUES (2,'Pat') 
INSERT INTO @SalesPerson (SalesPersonID, SalesPersonName) VALUES (3,'Joe') 

INSERT INTO @Region (RegionID, RegionName) VALUES (1,'North') 
INSERT INTO @Region (RegionID, RegionName) VALUES (2,'South') 
INSERT INTO @Region (RegionID, RegionName) VALUES (3,'East') 
INSERT INTO @Region (RegionID, RegionName) VALUES (4,'West') 

INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (1,1)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (1,2)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (1,3)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (2,2)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (2,3)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (2,4)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (3,1)
INSERT INTO @SalesPersonRegion (SalesPersonID,RegionID) VALUES (3,4)

一个简单的选择将使我得到每个销售人员，每个销售人员所在的区域。

SELECT 
    sp.SalesPersonID,
    sp.SalesPersonName,
    r.RegionName
FROM @SalesPersonRegion spr
    JOIN @SalesPerson sp
        ON spr.SalesPersonID = sp.SalesPersonID
    JOIN @Region r
        ON spr.RegionID = r.RegionID

在这种情况下，它将返回 9 行。

我想得到这样的结果：

SalesPersonID    SalesPersonName      Regions
1                Jeff                 North,South,East
2                Pat                  South,East,West
3                Joe                  North,West
4

3 回答 3

9 
SELECT 
  sp.SalesPersonID, 
  sp.SalesPersonName, 
  Regions = STUFF
  (
    (
      SELECT ',' + r.RegionName
       FROM @Region AS r
       INNER JOIN @SalesPersonRegion AS spr
       ON r.RegionID = spr.RegionID
       WHERE spr.SalesPersonID = sp.SalesPersonID
       ORDER BY r.RegionID
       FOR XML PATH(''), TYPE
    ).value('.[1]','nvarchar(max)'),
    1,1,''
  )
FROM @SalesPerson AS sp
ORDER BY sp.SalesPersonID;
于 2013-08-09T20:48:37.240 回答
2

试试这个查询：

SELECT 
    sp.SalesPersonID,
    sp.SalesPersonName,
    reg.Regions
FROM @SalesPerson sp
    CROSS APPLY( -- or OUTER APPLY
        SELECT STUFF(
        (SELECT ','+r.RegionName
        FROM    @Region r INNER JOIN @SalesPersonRegion spr ON r.RegionID = spr.RegionID
        WHERE   spr.SalesPersonID = sp.SalesPersonID
        FOR XML PATH('')),1,1,'') AS Regions
    )reg;

结果：

SalesPersonID SalesPersonName Regions
------------- --------------- ----------------
1             Jeff            North,South,East
2             Pat             South,East,West
3             Joe             North,West
于 2013-08-09T20:48:50.350 回答
2 
select
    sp.SalesPersonID,
    sp.SalesPersonName,
    stuff(
       (
           select ',' + r.RegionName
           from @SalesPersonRegion as spr 
               inner join @Region as r on r.RegionID = spr.RegionID
           where spr.SalesPersonID = sp.SalesPersonID
           for xml path(''), type
       ).value('.', 'nvarchar(max)')
       , 1, 1, '')
from @SalesPerson as sp

参见sql fiddle示例

于 2013-08-09T20:49:13.240 回答