我想自己想出一个解决办法，但我的大脑绊倒了。我觉得很管用，但别让我早上解释。: P

DECLARE @table AS TABLE
(
    Number int not null
);

insert into @table select 2;
insert into @table select 4;
insert into @table select 9;
insert into @table select 15;
insert into @table select 22;
insert into @table select 26;
insert into @table select 37;
insert into @table select 49;

DECLARE @Count AS INT
SELECT @Count = COUNT(*) FROM @table;

WITH MyResults(RowNo, Number) AS
(
    SELECT RowNo, Number FROM
        (SELECT ROW_NUMBER() OVER (ORDER BY Number) AS RowNo, Number FROM @table) AS Foo
)
SELECT AVG(Number) FROM MyResults WHERE RowNo = (@Count+1)/2 OR RowNo = ((@Count+1)%2) * ((@Count+2)/2)

我想自己想出一个解决办法，但我的大脑绊倒了。我觉得很管用，但别让我早上解释。: P

DECLARE @table AS TABLE
(
    Number int not null
);

insert into @table select 2;
insert into @table select 4;
insert into @table select 9;
insert into @table select 15;
insert into @table select 22;
insert into @table select 26;
insert into @table select 37;
insert into @table select 49;

DECLARE @Count AS INT
SELECT @Count = COUNT(*) FROM @table;

WITH MyResults(RowNo, Number) AS
(
    SELECT RowNo, Number FROM
        (SELECT ROW_NUMBER() OVER (ORDER BY Number) AS RowNo, Number FROM @table) AS Foo
)
SELECT AVG(Number) FROM MyResults WHERE RowNo = (@Count+1)/2 OR RowNo = ((@Count+1)%2) * ((@Count+2)/2)

如果你想在SQL Server中使用创建聚合函数，这是如何做到的。这样做的好处是能够编写干净的查询。注意，这个过程可以很容易地计算一个百分位值。

创建一个新的Visual Studio项目，并将目标框架设置为。net 3.5(这是针对SQL 2008的，在SQL 2012中可能有所不同)。然后创建一个类文件，并放入以下代码或c#等效代码:

Imports Microsoft.SqlServer.Server
Imports System.Data.SqlTypes
Imports System.IO

<Serializable>
<SqlUserDefinedAggregate(Format.UserDefined, IsInvariantToNulls:=True, IsInvariantToDuplicates:=False, _
  IsInvariantToOrder:=True, MaxByteSize:=-1, IsNullIfEmpty:=True)>
Public Class Median
  Implements IBinarySerialize
  Private _items As List(Of Decimal)

  Public Sub Init()
    _items = New List(Of Decimal)()
  End Sub

  Public Sub Accumulate(value As SqlDecimal)
    If Not value.IsNull Then
      _items.Add(value.Value)
    End If
  End Sub

  Public Sub Merge(other As Median)
    If other._items IsNot Nothing Then
      _items.AddRange(other._items)
    End If
  End Sub

  Public Function Terminate() As SqlDecimal
    If _items.Count <> 0 Then
      Dim result As Decimal
      _items = _items.OrderBy(Function(i) i).ToList()
      If _items.Count Mod 2 = 0 Then
        result = ((_items((_items.Count / 2) - 1)) + (_items(_items.Count / 2))) / 2@
      Else
        result = _items((_items.Count - 1) / 2)
      End If

      Return New SqlDecimal(result)
    Else
      Return New SqlDecimal()
    End If
  End Function

  Public Sub Read(r As BinaryReader) Implements IBinarySerialize.Read
    'deserialize it from a string
    Dim list = r.ReadString()
    _items = New List(Of Decimal)

    For Each value In list.Split(","c)
      Dim number As Decimal
      If Decimal.TryParse(value, number) Then
        _items.Add(number)
      End If
    Next

  End Sub

  Public Sub Write(w As BinaryWriter) Implements IBinarySerialize.Write
    'serialize the list to a string
    Dim list = ""

    For Each item In _items
      If list <> "" Then
        list += ","
      End If      
      list += item.ToString()
    Next
    w.Write(list)
  End Sub
End Class

然后编译它，并将DLL和PDB文件复制到您的SQL Server机器，在SQL Server中运行以下命令:

CREATE ASSEMBLY CustomAggregate FROM '{path to your DLL}'
WITH PERMISSION_SET=SAFE;
GO

CREATE AGGREGATE Median(@value decimal(9, 3))
RETURNS decimal(9, 3) 
EXTERNAL NAME [CustomAggregate].[{namespace of your DLL}.Median];
GO

然后你可以写一个查询来计算中位数，就像这样: SELECT dbo.Median(Field) FROM Table

with t1 as (select *, row_number() over(order by ordqty) as rn,
count(*) over() as rc from ord_line)
select rn,* from t1 where rn in((rc+1)/2, (rc+2)/2);

它将计算偶数和奇数的中位数。

Ord_line是一个表 Ordqty是一个列

对于连续变量/测量'table1'中的'col1'

select col1  
from
    (select top 50 percent col1, 
    ROW_NUMBER() OVER(ORDER BY col1 ASC) AS Rowa,
    ROW_NUMBER() OVER(ORDER BY col1 DESC) AS Rowd
    from table1 ) tmp
where tmp.Rowa = tmp.Rowd

通常情况下，我们不仅需要为整个表计算Median，还需要为与某个ID相关的聚合计算Median。换句话说，计算表中每个ID的中位数，其中每个ID有许多记录。(基于@gdoron编辑的解决方案:性能良好，适用于许多SQL)

SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val, 
  COUNT(*) OVER (PARTITION BY our_id) AS cnt,
  ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rnk
  FROM our_table
) AS x
WHERE rnk IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;

希望能有所帮助。