在我看来，解决这个问题的最佳来源是这个网站上发布的内容:

Syscomments。Dinakar Nethi

CREATE FUNCTION dbo.fnParseArray (@Array VARCHAR(1000),@separator CHAR(1))
RETURNS @T Table (col1 varchar(50))
AS 
BEGIN
 --DECLARE @T Table (col1 varchar(50))  
 -- @Array is the array we wish to parse
 -- @Separator is the separator charactor such as a comma
 DECLARE @separator_position INT -- This is used to locate each separator character
 DECLARE @array_value VARCHAR(1000) -- this holds each array value as it is returned
 -- For my loop to work I need an extra separator at the end. I always look to the
 -- left of the separator character for each array value

 SET @array = @array + @separator

 -- Loop through the string searching for separtor characters
 WHILE PATINDEX('%' + @separator + '%', @array) <> 0 
 BEGIN
    -- patindex matches the a pattern against a string
    SELECT @separator_position = PATINDEX('%' + @separator + '%',@array)
    SELECT @array_value = LEFT(@array, @separator_position - 1)
    -- This is where you process the values passed.
    INSERT into @T VALUES (@array_value)    
    -- Replace this select statement with your processing
    -- @array_value holds the value of this element of the array
    -- This replaces what we just processed with and empty string
    SELECT @array = STUFF(@array, 1, @separator_position, '')
 END
 RETURN 
END

Use:

SELECT * FROM dbo.fnParseArray('a,b,c,d,e,f', ',')

致谢:Dinakar Nethi

我将传递一个表类型参数(因为它是SQL Server 2008)，并做一个where exists，或内部连接。您也可以使用XML，使用sp_xml_preparedocument，然后甚至可以索引临时表。

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)

对于这样数量可变的参数，我所知道的唯一方法是显式地生成SQL，或者做一些涉及用所需项填充临时表并与临时表连接的事情。

如果你从。net调用，你可以使用Dapper dot net:

string[] names = new string[] {"ruby","rails","scruffy","rubyonrails"};
var tags = dataContext.Query<Tags>(@"
select * from Tags 
where Name in @names
order by Count desc", new {names});

这里是达普在思考，所以你不用思考。当然，类似的事情也可能发生在LINQ to SQL中:

string[] names = new string[] {"ruby","rails","scruffy","rubyonrails"};
var tags = from tag in dataContext.Tags
           where names.Contains(tag.Name)
           orderby tag.Count descending
           select tag;

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)