如果你需要一个数组，用逗号分隔数组列:

SELECT * FROM (VALUES('WOMENS'),('MENS'),('CHILDRENS')) as X([Attribute])
,(VALUES(742),(318)) AS z([StoreID])

我在大多数SQL DB上创建了一个函数来实现这一点。

CREATE OR ALTER FUNCTION [dbo].[UTIL_SplitList](@parList Varchar(MAX),@splitChar Varchar(1)=',') 
  Returns @t table (Column_Value varchar(MAX))
  as
  Begin
    Declare @pos integer 
    set @pos = CharIndex(@splitChar, @parList)
    while @pos > 0
    Begin
      Insert Into @t (Column_Value) VALUES (Left(@parList, @pos-1))
      set @parList = Right(@parList, Len(@parList) - @pos)
      set @pos = CharIndex(@splitChar, @parList)
    End
    Insert Into @t (Column_Value) VALUES (@parList)
    Return
  End

一旦函数存在，它就非常简单

SELECT DISTINCT 
    *
FROM 
    [dbo].[UTIL_SplitList]('1,1,1,2,5,1,6',',') 

如果它是一个来自现有SQL表的参数列表，例如来自现有Table1的ID列表，那么你可以尝试这样做:

select distinct ID
      FROM Table1
      where 
      ID in (1, 1, 1, 2, 5, 1, 6)
ORDER BY ID;

或者，如果你需要参数列表作为一个SQL表常量(变量)，试试这个:

WITH Id_list AS (
     select ID
      FROM Table1
      where 
      ID in (1, 1, 1, 2, 5, 1, 6)
)
SELECT distinct * FROM Id_list
ORDER BY ID;

PostgreSQL提供了两种方法:

SELECT DISTINCT * FROM (VALUES('a'),('b'),('a'),('v')) AS tbl(col1)

or

SELECT DISTINCT * FROM (select unnest(array['a','b', 'a','v'])) AS tbl(col1)

使用数组方法，你也可以这样做:

SELECT DISTINCT * FROM (select unnest(string_to_array('a;b;c;d;e;f;a;b;d', ';'))) AS tbl(col1)

你可以使用的另一种方式是这样的查询:

SELECT DISTINCT
    LTRIM(m.n.value('.[1]','varchar(8000)')) as columnName
FROM 
    (SELECT CAST('<XMLRoot><RowData>' + REPLACE(t.val,',','</RowData><RowData>') + '</RowData></XMLRoot>' AS XML) AS x
     FROM (SELECT '1, 1, 1, 2, 5, 1, 6') AS t(val)
    ) dt
  CROSS APPLY 
    x.nodes('/XMLRoot/RowData') m(n);

这样的行构造函数只能在SQL Server 2008及以上版本上使用: 你可以用

SELECT DISTINCT *
FROM (
  VALUES (1), (1), (1), (2), (5), (1), (6)
) AS X(a)

有关更多信息，请参阅:

官方的女士 http://www.sql-server-helper.com/sql-server-2008/row-value-constructor-as-derived-table.aspx