将名称/值对列表传递到存储过程

Question

我在List<T>中有一个名称/值对，并且需要找到将这些传递给存储过程的最佳方法。

Id Name 
1 abc 
2 bbc 
3 cnn 
.... 
... 

完成此操作的最佳方法是什么？

Answer 1

我endup使用foreach <insert>

Answer 2

在Arrays and Lists in SQL Server 2008看看得到一些想法

SQL Server 2008还支持这种多行值语法在SQL Server之前，表的可用性解决这个2005年（

create table #bla (id int, somename varchar(50)) 

insert #bla values(1,'test1'),(2,'Test2') 

select * from #bla 

Answer 3

的一种方式有价值的参数）是通过一个分隔列表并使用Split功能。如果你使用的是两列的阵列，您可能需要使用两个不同的分隔符：

Declare @Values varchar(max) 
Set @Values = '1,abc|2,bbc|3,cnn' 

With SplitItems As 
    (
    Select S.Value As [Key] 
     , S2.Value 
     , Row_Number() Over (Partition By S.Position Order By S2.Position) As ElementNum 
    From dbo.Split(@Values,'|') As S 
     Outer Apply dbo.Split(S.Value, ',') As S2 
    ) 
Select [Key] 
    , Min(Case When S.ElementNum = 1 Then S.Value End) As ListKey 
    , Min(Case When S.ElementNum = 2 Then S.Value End) As ListValue 
From SplitItems As S 
Group By [Key] 

Create Function [dbo].[Split] 
( 
    @DelimitedList nvarchar(max) 
    , @Delimiter nvarchar(2) = ',' 
) 
RETURNS TABLE 
AS 
RETURN 
    (
    With CorrectedList As 
     (
     Select Case When Left(@DelimitedList, Len(@Delimiter)) <> @Delimiter Then @Delimiter Else '' End 
      + @DelimitedList 
      + Case When Right(@DelimitedList, Len(@Delimiter)) <> @Delimiter Then @Delimiter Else '' End 
      As List 
      , Len(@Delimiter) As DelimiterLen 
     ) 
     , Numbers As 
     (
     Select Row_Number() Over (Order By c1.object_id) As Value 
     From sys.columns As c1 
      Cross Join sys.columns As c2 
     ) 
    Select CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen As Position 
     , Substring (
        CL.List 
        , CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen  
        , CharIndex(@Delimiter, CL.list, N.Value + 1)       
         - (CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen) 
        ) As Value 
    From CorrectedList As CL 
     Cross Join Numbers As N 
    Where N.Value < Len(CL.List) 
     And Substring(CL.List, N.Value, CL.DelimiterLen) = @Delimiter 
    ) 

另一种方式来处理这个不表值参数是通过XML作为一个nvarchar（最大）：

Declare @Values nvarchar(max) 
Set @Values = '<root><Item Key="1" Value="abc"/> 
<Item Key="2" Value="bbc"/> 
<Item Key="3" Value="cnn"/></root>' 

Declare @docHandle int 
exec sp_xml_preparedocument @docHandle output, @Values 

Select * 
From OpenXml(@docHandle, N'/root/Item', 1) 
    With([Key] int, Value varchar(10))