这里有另一种选择。只需将一个以逗号分隔的列表作为字符串参数传递给存储过程，然后:

CREATE PROCEDURE [dbo].[sp_myproc]
    @UnitList varchar(MAX) = '1,2,3'
AS
select column from table
where ph.UnitID in (select * from CsvToInt(@UnitList))

函数:

CREATE Function [dbo].[CsvToInt] ( @Array varchar(MAX))
returns @IntTable table
(IntValue int)
AS
begin
    declare @separator char(1)
    set @separator = ','
    declare @separator_position int
    declare @array_value varchar(MAX)

    set @array = @array + ','

    while patindex('%,%' , @array) <> 0
    begin

        select @separator_position = patindex('%,%' , @array)
        select @array_value = left(@array, @separator_position - 1)

        Insert @IntTable
        Values (Cast(@array_value as int))
        select @array = stuff(@array, 1, @separator_position, '')
    end
    return
end

我听到Jeff/Joel今天在播客上谈论这个(第34集，2008-12-16 (MP3, 31 MB)， 1小时03分38秒-1小时06分45秒)，我想我记得Stack Overflow使用LINQ到SQL，但也许它被抛弃了。在LINQ to SQL中也是如此。

var inValues = new [] { "ruby","rails","scruffy","rubyonrails" };

var results = from tag in Tags
              where inValues.Contains(tag.Name)
              select tag;

就是这样。而且，是的，LINQ已经足够向后看了，但是Contains子句对我来说似乎太向后看了。当我不得不在工作中为一个项目做类似的查询时，我自然地试图用错误的方式来做这个，在本地数组和SQL Server表之间做一个连接，认为LINQ to SQL翻译器将足够聪明，以某种方式处理翻译。它没有这样做，但它确实提供了一个描述性的错误消息，并指示我使用Contains。

无论如何，如果您在强烈推荐的LINQPad中运行此命令，并运行此查询，您可以查看SQL LINQ提供程序生成的实际SQL。它将向您显示每个参数化为IN子句的值。

也许我们可以在这里使用XML:

    declare @x xml
    set @x='<items>
    <item myvalue="29790" />
    <item myvalue="31250" />
    </items>
    ';
    With CTE AS (
         SELECT 
            x.item.value('@myvalue[1]', 'decimal') AS myvalue
        FROM @x.nodes('//items/item') AS x(item) )

    select * from YourTable where tableColumnName in (select myvalue from cte)

步骤1:-

string[] Ids = new string[] { "3", "6", "14" };
string IdsSP = string.Format("'|{0}|'", string.Join("|", Ids));

步骤2:-

@CurrentShipmentStatusIdArray [nvarchar](255) = NULL

步骤3:-

Where @CurrentShipmentStatusIdArray is null or @CurrentShipmentStatusIdArray LIKE '%|' + convert(nvarchar,Shipments.CurrentShipmentStatusId) + '|%'

or

Where @CurrentShipmentStatusIdArray is null or @CurrentShipmentStatusIdArray LIKE '%|' + Shipments.CurrentShipmentStatusId+ '|%'

可以将参数作为字符串传递

这是弦

DECLARE @tags

SET @tags = ‘ruby|rails|scruffy|rubyonrails’

select * from Tags 
where Name in (SELECT item from fnSplit(@tags, ‘|’))
order by Count desc

然后你所要做的就是将字符串作为1参数传递。

这是我使用的分裂函数。

CREATE FUNCTION [dbo].[fnSplit](
    @sInputList VARCHAR(8000) -- List of delimited items
  , @sDelimiter VARCHAR(8000) = ',' -- delimiter that separates items
) RETURNS @List TABLE (item VARCHAR(8000))

BEGIN
DECLARE @sItem VARCHAR(8000)
WHILE CHARINDEX(@sDelimiter,@sInputList,0) <> 0
 BEGIN
 SELECT
  @sItem=RTRIM(LTRIM(SUBSTRING(@sInputList,1,CHARINDEX(@sDelimiter,@sInputList,0)-1))),
  @sInputList=RTRIM(LTRIM(SUBSTRING(@sInputList,CHARINDEX(@sDelimiter,@sInputList,0)+LEN(@sDelimiter),LEN(@sInputList))))

 IF LEN(@sItem) > 0
  INSERT INTO @List SELECT @sItem
 END

IF LEN(@sInputList) > 0
 INSERT INTO @List SELECT @sInputList -- Put the last item in
RETURN
END

在我看来，正确的方法是将列表存储在字符串中(长度受DBMS支持的限制);唯一的技巧是(为了简化处理)我在字符串的开头和结尾都有一个分隔符(在我的例子中是一个逗号)。其思想是“动态规范化”，将列表转换为单列表，每个值包含一行。这让你可以转弯

In (ct1,ct2, ct3…卡通)

成一个

在(选择…)

或者(我可能更喜欢的解决方案)常规连接，如果你只是添加一个“distinct”来避免列表中重复值的问题。

不幸的是，分割字符串的技术是相当特定于产品的。 下面是SQL Server的版本:

 with qry(n, names) as
       (select len(list.names) - len(replace(list.names, ',', '')) - 1 as n,
               substring(list.names, 2, len(list.names)) as names
        from (select ',Doc,Grumpy,Happy,Sneezy,Bashful,Sleepy,Dopey,' names) as list
        union all
        select (n - 1) as n,
               substring(names, 1 + charindex(',', names), len(names)) as names
        from qry
        where n > 1)
 select n, substring(names, 1, charindex(',', names) - 1) dwarf
 from qry;

Oracle版本:

 select n, substr(name, 1, instr(name, ',') - 1) dwarf
 from (select n,
             substr(val, 1 + instr(val, ',', 1, n)) name
      from (select rownum as n,
                   list.val
            from  (select ',Doc,Grumpy,Happy,Sneezy,Bashful,Sleepy,Dopey,' val
                   from dual) list
            connect by level < length(list.val) -
                               length(replace(list.val, ',', ''))));

MySQL版本:

select pivot.n,
      substring_index(substring_index(list.val, ',', 1 + pivot.n), ',', -1) from (select 1 as n
     union all
     select 2 as n
     union all
     select 3 as n
     union all
     select 4 as n
     union all
     select 5 as n
     union all
     select 6 as n
     union all
     select 7 as n
     union all
     select 8 as n
     union all
     select 9 as n
     union all
     select 10 as n) pivot,    (select ',Doc,Grumpy,Happy,Sneezy,Bashful,Sleepy,Dopey,' val) as list where pivot.n <  length(list.val) -
                   length(replace(list.val, ',', ''));

(当然，“pivot”必须返回与的最大行数相同的行数 我们可以在列表中找到的项目)