如何创建一个SQL Server函数“连接”多行从一个子查询到一个单独的分隔字段?

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?

当前回答

SQL Server 2005

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

SQL Server 2016

你可以使用FOR JSON语法

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

结果就会变成

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

即使数据包含无效的XML字符，这也可以工作

的 '"},{"":"' 你是安全的,因为如果数据包含吗 '"},{"":"', 它将逃到 "},{\"_\":\"

你可以用任何字符串分隔符替换'，'

在SQL Server 2017, Azure SQL数据库

您可以使用新的STRING_AGG函数

2010-09-09 00:15:49

其他回答

我通过创建以下函数找到了一个解决方案:

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')

2011-05-30 15:15:06

如果你正在运行SQL Server 2005，你可以编写一个自定义CLR聚合函数来处理这个问题。

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());
    }
}

2008-08-10 13:36:33

下面的代码将适用于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

2009-06-18 12:41:27