create FUNCTION [dbo].[ConvertStringToList]


      (@str VARCHAR (MAX), @delimeter CHAR (1))
        RETURNS 
        @result TABLE (
            [ID] INT NULL)
    AS
    BEG

IN

    DECLARE @x XML 
    SET @x = '<t>' + REPLACE(@str, @delimeter, '</t><t>') + '</t>'

    INSERT INTO @result
    SELECT DISTINCT x.i.value('.', 'int') AS token
    FROM @x.nodes('//t') x(i)
    ORDER BY 1

RETURN
END

——你的查询

select * from table where id in ([dbo].[ConvertStringToList(YOUR comma separated string ,',')])

下面是我用过的一个快速而又复杂的技巧:

SELECT * FROM Tags
WHERE '|ruby|rails|scruffy|rubyonrails|'
LIKE '%|' + Name + '|%'

下面是c#代码:

string[] tags = new string[] { "ruby", "rails", "scruffy", "rubyonrails" };
const string cmdText = "select * from tags where '|' + @tags + '|' like '%|' + Name + '|%'";

using (SqlCommand cmd = new SqlCommand(cmdText)) {
   cmd.Parameters.AddWithValue("@tags", string.Join("|", tags);
}

两个问题:

演出糟透了。像“%……%"查询没有索引。 请确保没有任何|、blank或null标记，否则将无法工作

有些人可能认为还有其他更清洁的方法可以做到这一点，所以请继续阅读。

唯一的制胜方法就是不玩。

对你来说没有无限的可变性。只有有限的可变性。

在SQL中，你有这样一个子句:

and ( {1}==0 or b.CompanyId in ({2},{3},{4},{5},{6}) )

在c#代码中，你可以这样做:

  int origCount = idList.Count;
  if (origCount > 5) {
    throw new Exception("You may only specify up to five originators to filter on.");
  }
  while (idList.Count < 5) { idList.Add(-1); }  // -1 is an impossible value
  return ExecuteQuery<PublishDate>(getValuesInListSQL, 
               origCount,   
               idList[0], idList[1], idList[2], idList[3], idList[4]);

基本上，如果count为0，那么就没有过滤器，所有东西都会通过。如果计数高于0，则该值必须在列表中，但列表已被填充为5个不可能的值(因此SQL仍然有意义)

有时蹩脚的解决方案是唯一真正有效的。

如果你有SQL Server 2008或更高版本，我会使用表值参数。

如果你不幸被困在SQL Server 2005上，你可以添加这样一个CLR函数，

[SqlFunction(
    DataAccessKind.None,
    IsDeterministic = true,
    SystemDataAccess = SystemDataAccessKind.None,
    IsPrecise = true,
    FillRowMethodName = "SplitFillRow",
    TableDefinintion = "s NVARCHAR(MAX)"]
public static IEnumerable Split(SqlChars seperator, SqlString s)
{
    if (s.IsNull)
        return new string[0];

    return s.ToString().Split(seperator.Buffer);
}

public static void SplitFillRow(object row, out SqlString s)
{
    s = new SqlString(row.ToString());
}

你可以这样用，

declare @desiredTags nvarchar(MAX);
set @desiredTags = 'ruby,rails,scruffy,rubyonrails';

select * from Tags
where Name in [dbo].[Split] (',', @desiredTags)
order by Count desc

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)

    create FUNCTION [dbo].[ConvertStringToList]


      (@str VARCHAR (MAX), @delimeter CHAR (1))
        RETURNS 
        @result TABLE (
            [ID] INT NULL)
    AS
    BEG

IN

    DECLARE @x XML 
    SET @x = '<t>' + REPLACE(@str, @delimeter, '</t><t>') + '</t>'

    INSERT INTO @result
    SELECT DISTINCT x.i.value('.', 'int') AS token
    FROM @x.nodes('//t') x(i)
    ORDER BY 1

RETURN
END

——你的查询

select * from table where id in ([dbo].[ConvertStringToList(YOUR comma separated string ,',')])