SQL Server中的SQL group_concat函数

Question

如果有一个名为employee的表

EmpID           EmpName
----------      -------------
1               Mary
1               John
1               Sam
2               Alaina
2               Edward

结果我需要这种格式:

EmpID           EmpName
----------      -------------
1               Mary, John, Sam
2               Alaina, Edward

问:这条记录在同Employee一张表中.我几乎没有使用UDF,存储过程的经验,我需要通过查询完成这个事情.这可能不使用UDF,SP.

Answer 1

1. FOR XML PATH 技巧和文章
2. CLR用户定义的聚合
3. 对于SQL Server早期版本2005 - 临时表

#1的一个例子

DECLARE @t TABLE (EmpId INT, EmpName VARCHAR(100))
INSERT @t VALUES
(1, 'Mary'),(1, 'John'),(1, 'Sam'),(2, 'Alaina'),(2, 'Edward')
SELECT distinct
    EmpId,
    (
        SELECT EmpName+','
        FROM @t t2
        WHERE t2.EmpId = t1.EmpId
        FOR XML PATH('')
    ) Concatenated
FROM @t t1

如何剥离最终的逗号 - 是你自己的

CLR聚合#2的c#代码

using System;
using System.Collections.Generic;
using System.Data.SqlTypes;
using System.Text;
using Microsoft.SqlServer.Server;
using System.IO;

namespace DatabaseAssembly
{
    [Serializable]
    [SqlUserDefinedAggregate(Format.UserDefined,
        IsInvariantToNulls = true,
        IsInvariantToDuplicates = true,
        IsInvariantToOrder = true,
        MaxByteSize = -1)]
    public struct StringJoin : IBinarySerialize
    {
        private Dictionary<string, string> AggregationList
        {
            get
            {
                if (_list == null)
                    _list = new Dictionary<string, string>();
                return _list;
            }
        }
        private Dictionary<string, string> _list;

        public void Init()
        {

        }

        public void Accumulate(SqlString Value)
        {
            if (!Value.IsNull)
                AggregationList[Value.Value.ToLowerInvariant()] = Value.Value;

        }

        public void Merge(StringJoin Group)
        {
            foreach (var key in Group.AggregationList.Keys)
                AggregationList[key] = Group.AggregationList[key];
        }

        public SqlChars Terminate()
        {
            var sb = new StringBuilder();
            foreach (var value in AggregationList.Values)
                sb.Append(value);
            return new SqlChars(sb.ToString());
        }

        #region IBinarySerialize Members

        public void Read(System.IO.BinaryReader r)
        {

            try
            {
                while (true)
                    AggregationList[r.ReadString()] = r.ReadString();
            }
            catch (EndOfStreamException)
            {

            }
        }

        public void Write(System.IO.BinaryWriter w)
        {
            foreach (var key in AggregationList.Keys)
            {
                w.Write(key);
                w.Write(AggregationList[key]);
            }
        }

        #endregion
    }
}

Answer 2

来自@ OlegDok的所选答案可能会返回正确的结果.但表现可能很糟糕.这个测试场景将说明它.

创建临时表:

CREATE table #temp (EmpId INT, EmpName VARCHAR(100))
;WITH N(N)AS 
(SELECT 1 FROM(VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1))M(N)),
tally(N)AS(SELECT ROW_NUMBER()OVER(ORDER BY N.N)FROM N,N a,N b,N c,N d,N e,N f)
INSERT #temp
SELECT EmpId, EmpName FROM (values(1, 'Mary'),(1, 'John'),(1, 'Sam')) x(EmpId, EmpName)
CROSS APPLY 
(SELECT top 2000 N FROM tally) y
UNION ALL
SELECT EmpId, EmpName FROM (values(2, 'Alaina'),(2, 'Edward')) x(EmpId, EmpName)
CROSS APPLY
(SELECT top 2000 N FROM tally) y

这只有10.000行.但是很多相同的EmpId.

Oleg的答案中的这个查询在我的数据库上花费了64秒.

SELECT distinct
    EmpId,
    (
        SELECT EmpName+','
        FROM #temp t2
        WHERE t2.EmpId = t1.EmpId
        FOR XML PATH('')
    ) Concatenated
FROM #temp t1

在这种情况下,区别不是清理行的正确方法.要避免此笛卡尔联接,请在加入之前减少ID的初始数量.

这是处理这个问题的正确方法:

;WITH CTE as
(
  SELECT distinct EmpId
  FROM #temp
)
SELECT 
    EmpId,
    STUFF((
        SELECT ','+EmpName
        FROM #temp t2
        WHERE t2.EmpId = t1.EmpId
        FOR XML PATH('')
    ), 1,1,'') Concatenated
FROM CTE t1

这需要不到1秒钟