如何创建SQL Server函数以将子查询中的多行“连接”到单个分隔字段?

时间:2008-08-09 20:11:18

标签: sql sql-server string-concatenation

为了说明,假设我有两个表如下:

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

我想写一个查询来返回以下结果:

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

但是,正如您所看到的,这需要大量代码。我想要的是一个通用函数,可以让我做这样的事情:

SELECT VehicleID
     , Name
     , JOIN(SELECT City FROM Locations WHERE VehicleID = Vehicles.VehicleID, ', ') AS Locations
FROM Vehicles

这可能吗?或类似的东西?

13 个答案:

答案 0 :(得分:246)

如果您使用的是SQL Server 2005,则可以使用FOR XML PATH命令。<​​/ p>

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]

它比使用光标容易得多,而且效果相当不错。

答案 1 :(得分:85)

请注意Matt's code会在字符串末尾产生额外的逗号;使用COALESCE(或ISNULL),如Lance帖子中的链接所示,使用类似的方法,但不会留下额外的逗号来删除。为了完整起见,这里是来自sqlteam.com上Lance链接的相关代码:

DECLARE @EmployeeList varchar(100)
SELECT @EmployeeList = COALESCE(@EmployeeList + ', ', '') + 
    CAST(EmpUniqueID AS varchar(5))
FROM SalesCallsEmployees
WHERE SalCal_UniqueID = 1

答案 2 :(得分:44)

我不相信有一种方法可以在一个查询中执行此操作,但您可以使用临时变量来播放这样的技巧:

declare @s varchar(max)
set @s = ''
select @s = @s + City + ',' from Locations

select @s

这肯定比走过光标的代码少,而且效率可能更高。

答案 3 :(得分:23)

在单个SQL查询中,不使用FOR XML子句 公用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

答案 4 :(得分:23)

从我所看到的FOR XML(如前所述)是唯一的方法,如果你想像OP那样选择其他列(我猜的最多)。 使用COALESCE(@var...不允许包含其他列。

更新: 感谢programmingsolutions.net,有一种方法可以删除“尾随”逗号。 通过将其设置为前导逗号并使用MSSQL的STUFF函数,您可以将第一个字符(前导逗号)替换为空字符串,如下所示:

stuff(
    (select ',' + Column 
     from Table
         inner where inner.Id = outer.Id 
     for xml path('')
), 1,1,'') as Values

答案 5 :(得分:14)

SQL Server 2005

SELECT Stuff(
  (SELECT N', ' + Name FROM Names FOR XML PATH(''),TYPE)
  .value('text()[1]','nvarchar(max)'),1,2,N'')

在SQL Server 2016中

您可以使用FOR JSON syntax

即。

SELECT per.ID,
Emails = JSON_VALUE(
   REPLACE(
     (SELECT _ = em.Email FROM Email em WHERE em.Person = per.ID FOR JSON PATH)
    ,'"},{"_":"',', '),'$[0]._'
) 
FROM Person per

结果将变为

Id  Emails
1   abc@gmail.com
2   NULL
3   def@gmail.com, xyz@gmail.com

这甚至会使您的数据包含无效的XML字符

''},{“”:“'是安全的,因为如果您的数据包含'”},{“”:“',它将被转义为”},{\“ _ \ “:\”

您可以将','替换为任何字符串分隔符


在SQL Server 2017中,Azure SQL数据库

您可以使用新的STRING_AGG function

答案 6 :(得分:12)

以下代码适用于Sql Server 2000/2005/2008

CREATE FUNCTION fnConcatVehicleCities(@VehicleId SMALLINT)
RETURNS VARCHAR(1000) AS
BEGIN
  DECLARE @csvCities VARCHAR(1000)
  SELECT @csvCities = COALESCE(@csvCities + ', ', '') + COALESCE(City,'')
  FROM Vehicles 
  WHERE VehicleId = @VehicleId 
  return @csvCities
END

-- //Once the User defined function is created then run the below sql

SELECT VehicleID
     , dbo.fnConcatVehicleCities(VehicleId) AS Locations
FROM Vehicles
GROUP BY VehicleID

答案 7 :(得分:7)

我通过创建以下功能找到了解决方案:

CREATE FUNCTION [dbo].[JoinTexts]
(
  @delimiter VARCHAR(20) ,
  @whereClause VARCHAR(1)
)
RETURNS VARCHAR(MAX)
AS 
BEGIN
    DECLARE @Texts VARCHAR(MAX)

    SELECT  @Texts = COALESCE(@Texts + @delimiter, '') + T.Texto
    FROM    SomeTable AS T
    WHERE   T.SomeOtherColumn = @whereClause

    RETURN @Texts
END
GO

用法:

SELECT dbo.JoinTexts(' , ', 'Y')

答案 8 :(得分:3)

Mun的答案对我没有用,所以我对这个答案做了一些修改,让它发挥作用。希望这有助于某人。 使用SQL Server 2012:

SELECT [VehicleID]
     , [Name]
     , STUFF((SELECT DISTINCT ',' + CONVERT(VARCHAR,City) 
         FROM [Location] 
         WHERE (VehicleID = Vehicle.VehicleID) 
         FOR XML PATH ('')), 1, 2, '') AS Locations
FROM [Vehicle]

答案 9 :(得分:3)

版本注意:对于此解决方案,您必须使用SQL Server 2005或更高版本,且兼容级别设置为90或更高。

有关创建用户定义聚合函数的第一个示例,请参阅此MSDN article,该聚合函数连接从表中的列获取的一组字符串值。

我的谦虚建议是省略附加的逗号,以便您可以使用自己的临时分隔符(如果有的话)。

参考示例1的C#版本:

change:  this.intermediateResult.Append(value.Value).Append(',');
    to:  this.intermediateResult.Append(value.Value);

change:  output = this.intermediateResult.ToString(0, this.intermediateResult.Length - 1);
    to:  output = this.intermediateResult.ToString();

这样,当您使用自定义聚合时,您可以选择使用自己的分隔符,或者根本不使用分隔符,例如:

SELECT dbo.CONCATENATE(column1 + '|') from table1

注意:请注意您尝试在汇总中处理的数据量。如果您尝试连接数千个行或许多非常大的数据类型,您可能会收到一个.NET Framework错误,指出“缓冲区不足。”

答案 10 :(得分:3)

通过其他答案,阅读答案的人必须了解车辆表并创建车辆表和数据以测试解决方案。

下面是一个使用SQL Server“Information_Schema.Columns”表的示例。通过使用此解决方案,无需创建表或添加数据。此示例为数据库中的所有表创建逗号分隔的列名列表。

SELECT
    Table_Name
    ,STUFF((
        SELECT ',' + Column_Name
        FROM INFORMATION_SCHEMA.Columns Columns
        WHERE Tables.Table_Name = Columns.Table_Name
        ORDER BY Column_Name
        FOR XML PATH ('')), 1, 1, ''
    )Columns
FROM INFORMATION_SCHEMA.Columns Tables
GROUP BY TABLE_NAME 

答案 11 :(得分:1)

如果您正在运行SQL Server 2005,则可以编写custom CLR aggregate function来处理此问题。

C#版本:

using System;
using System.Data;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using System.Text;
using Microsoft.SqlServer.Server;
[Serializable]
[Microsoft.SqlServer.Server.SqlUserDefinedAggregate(Format.UserDefined,MaxByteSize=8000)]
public class CSV:IBinarySerialize
{
    private StringBuilder Result;
    public void Init() {
        this.Result = new StringBuilder();
    }

    public void Accumulate(SqlString Value) {
        if (Value.IsNull) return;
        this.Result.Append(Value.Value).Append(",");
    }
    public void Merge(CSV Group) {
        this.Result.Append(Group.Result);
    }
    public SqlString Terminate() {
        return new SqlString(this.Result.ToString());
    }
    public void Read(System.IO.BinaryReader r) {
        this.Result = new StringBuilder(r.ReadString());
    }
    public void Write(System.IO.BinaryWriter w) {
        w.Write(this.Result.ToString());
    }
}

答案 12 :(得分:1)

尝试此查询

SELECT v.VehicleId, v.Name, ll.LocationList
FROM Vehicles v 
LEFT JOIN 
    (SELECT 
     DISTINCT
        VehicleId,
        REPLACE(
            REPLACE(
                REPLACE(
                    (
                        SELECT City as c 
                        FROM Locations x 
                        WHERE x.VehicleID = l.VehicleID FOR XML PATH('')
                    ),    
                    '</c><c>',', '
                 ),
             '<c>',''
            ),
        '</c>', ''
        ) AS LocationList
    FROM Locations l
) ll ON ll.VehicleId = v.VehicleId