为了说明这一点，假设我有如下两个表:
VehicleID Name
1 Chuck
2 Larry
LocationID VehicleID City
1 1 New York
2 1 Seattle
3 1 Vancouver
4 2 Los Angeles
5 2 Houston
我想写一个查询返回以下结果:
VehicleID Name Locations
1 Chuck New York, Seattle, Vancouver
2 Larry Los Angeles, Houston
我知道这可以使用服务器端游标完成，即:
DECLARE @VehicleID int
DECLARE @VehicleName varchar(100)
DECLARE @LocationCity varchar(100)
DECLARE @Locations varchar(4000)
DECLARE @Results TABLE
(
VehicleID int
Name varchar(100)
Locations varchar(4000)
)
DECLARE VehiclesCursor CURSOR FOR
SELECT
[VehicleID]
, [Name]
FROM [Vehicles]
OPEN VehiclesCursor
FETCH NEXT FROM VehiclesCursor INTO
@VehicleID
, @VehicleName
WHILE @@FETCH_STATUS = 0
BEGIN
SET @Locations = ''
DECLARE LocationsCursor CURSOR FOR
SELECT
[City]
FROM [Locations]
WHERE [VehicleID] = @VehicleID
OPEN LocationsCursor
FETCH NEXT FROM LocationsCursor INTO
@LocationCity
WHILE @@FETCH_STATUS = 0
BEGIN
SET @Locations = @Locations + @LocationCity
FETCH NEXT FROM LocationsCursor INTO
@LocationCity
END
CLOSE LocationsCursor
DEALLOCATE LocationsCursor
INSERT INTO @Results (VehicleID, Name, Locations) SELECT @VehicleID, @Name, @Locations
END
CLOSE VehiclesCursor
DEALLOCATE VehiclesCursor
SELECT * FROM @Results
然而，正如您所看到的，这需要大量的代码。我想要的是一个泛型函数，允许我做这样的事情:
SELECT VehicleID
, Name
, JOIN(SELECT City FROM Locations WHERE VehicleID = Vehicles.VehicleID, ', ') AS Locations
FROM Vehicles
这可能吗?或者类似的东西?
在单个SQL查询中，不使用FOR XML子句。
公共表表达式用于递归地连接结果。
-- rank locations by incrementing lexicographical order
WITH RankedLocations AS (
SELECT
VehicleID,
City,
ROW_NUMBER() OVER (
PARTITION BY VehicleID
ORDER BY City
) Rank
FROM
Locations
),
-- concatenate locations using a recursive query
-- (Common Table Expression)
Concatenations AS (
-- for each vehicle, select the first location
SELECT
VehicleID,
CONVERT(nvarchar(MAX), City) Cities,
Rank
FROM
RankedLocations
WHERE
Rank = 1
-- then incrementally concatenate with the next location
-- this will return intermediate concatenations that will be
-- filtered out later on
UNION ALL
SELECT
c.VehicleID,
(c.Cities + ', ' + l.City) Cities,
l.Rank
FROM
Concatenations c -- this is a recursion!
INNER JOIN RankedLocations l ON
l.VehicleID = c.VehicleID
AND l.Rank = c.Rank + 1
),
-- rank concatenation results by decrementing length
-- (rank 1 will always be for the longest concatenation)
RankedConcatenations AS (
SELECT
VehicleID,
Cities,
ROW_NUMBER() OVER (
PARTITION BY VehicleID
ORDER BY Rank DESC
) Rank
FROM
Concatenations
)
-- main query
SELECT
v.VehicleID,
v.Name,
c.Cities
FROM
Vehicles v
INNER JOIN RankedConcatenations c ON
c.VehicleID = v.VehicleID
AND c.Rank = 1
如果你使用的是SQL Server 2005，你可以使用FOR XML PATH命令。
SELECT [VehicleID]
, [Name]
, (STUFF((SELECT CAST(', ' + [City] AS VARCHAR(MAX))
FROM [Location]
WHERE (VehicleID = Vehicle.VehicleID)
FOR XML PATH ('')), 1, 2, '')) AS Locations
FROM [Vehicle]
这比使用光标容易得多，而且似乎工作得相当好。
更新
对于那些在新版本的SQL Server中仍然使用这种方法的人来说，还有另一种方法，它更简单，性能更好
自SQL Server 2017以来已经可用的STRING_AGG方法。
SELECT [VehicleID]
,[Name]
,(SELECT STRING_AGG([City], ', ')
FROM [Location]
WHERE VehicleID = V.VehicleID) AS Locations
FROM [Vehicle] V
这还允许将不同的分隔符指定为第二个参数，比前一种方法提供了更多的灵活性。
在单个SQL查询中，不使用FOR XML子句。
公共表表达式用于递归地连接结果。
-- rank locations by incrementing lexicographical order
WITH RankedLocations AS (
SELECT
VehicleID,
City,
ROW_NUMBER() OVER (
PARTITION BY VehicleID
ORDER BY City
) Rank
FROM
Locations
),
-- concatenate locations using a recursive query
-- (Common Table Expression)
Concatenations AS (
-- for each vehicle, select the first location
SELECT
VehicleID,
CONVERT(nvarchar(MAX), City) Cities,
Rank
FROM
RankedLocations
WHERE
Rank = 1
-- then incrementally concatenate with the next location
-- this will return intermediate concatenations that will be
-- filtered out later on
UNION ALL
SELECT
c.VehicleID,
(c.Cities + ', ' + l.City) Cities,
l.Rank
FROM
Concatenations c -- this is a recursion!
INNER JOIN RankedLocations l ON
l.VehicleID = c.VehicleID
AND l.Rank = c.Rank + 1
),
-- rank concatenation results by decrementing length
-- (rank 1 will always be for the longest concatenation)
RankedConcatenations AS (
SELECT
VehicleID,
Cities,
ROW_NUMBER() OVER (
PARTITION BY VehicleID
ORDER BY Rank DESC
) Rank
FROM
Concatenations
)
-- main query
SELECT
v.VehicleID,
v.Name,
c.Cities
FROM
Vehicles v
INNER JOIN RankedConcatenations c ON
c.VehicleID = v.VehicleID
AND c.Rank = 1
