我有一个不需要UDF的答案，XML 因为IN接受一个选择语句 例如:SELECT * FROM Test where Data IN (SELECT Value FROM TABLE)

您实际上只需要一种将字符串转换为表的方法。

这可以通过递归CTE或使用数字表(或Master..spt_value)的查询来完成。

这是CTE的版本。

DECLARE @InputString varchar(8000) = 'ruby,rails,scruffy,rubyonrails'

SELECT @InputString = @InputString + ','

;WITH RecursiveCSV(x,y) 
AS 
(
    SELECT 
        x = SUBSTRING(@InputString,0,CHARINDEX(',',@InputString,0)),
        y = SUBSTRING(@InputString,CHARINDEX(',',@InputString,0)+1,LEN(@InputString))
    UNION ALL
    SELECT 
        x = SUBSTRING(y,0,CHARINDEX(',',y,0)),
        y = SUBSTRING(y,CHARINDEX(',',y,0)+1,LEN(y))
    FROM 
        RecursiveCSV 
    WHERE
        SUBSTRING(y,CHARINDEX(',',y,0)+1,LEN(y)) <> '' OR 
        SUBSTRING(y,0,CHARINDEX(',',y,0)) <> ''
)
SELECT
    * 
FROM 
    Tags
WHERE 
    Name IN (select x FROM RecursiveCSV)
OPTION (MAXRECURSION 32767);

创建一个存储名称的临时表，然后使用以下查询:

select * from Tags 
where Name in (select distinct name from temp)
order by Count desc

我有一个不需要UDF的答案，XML 因为IN接受一个选择语句 例如:SELECT * FROM Test where Data IN (SELECT Value FROM TABLE)

您实际上只需要一种将字符串转换为表的方法。

这可以通过递归CTE或使用数字表(或Master..spt_value)的查询来完成。

这是CTE的版本。

DECLARE @InputString varchar(8000) = 'ruby,rails,scruffy,rubyonrails'

SELECT @InputString = @InputString + ','

;WITH RecursiveCSV(x,y) 
AS 
(
    SELECT 
        x = SUBSTRING(@InputString,0,CHARINDEX(',',@InputString,0)),
        y = SUBSTRING(@InputString,CHARINDEX(',',@InputString,0)+1,LEN(@InputString))
    UNION ALL
    SELECT 
        x = SUBSTRING(y,0,CHARINDEX(',',y,0)),
        y = SUBSTRING(y,CHARINDEX(',',y,0)+1,LEN(y))
    FROM 
        RecursiveCSV 
    WHERE
        SUBSTRING(y,CHARINDEX(',',y,0)+1,LEN(y)) <> '' OR 
        SUBSTRING(y,0,CHARINDEX(',',y,0)) <> ''
)
SELECT
    * 
FROM 
    Tags
WHERE 
    Name IN (select x FROM RecursiveCSV)
OPTION (MAXRECURSION 32767);

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)

我使用了一个更简洁的投票结果:

List<SqlParameter> parameters = tags.Select((s, i) => new SqlParameter("@tag" + i.ToString(), SqlDbType.NVarChar(50)) { Value = s}).ToList();

var whereCondition = string.Format("tags in ({0})", String.Join(",",parameters.Select(s => s.ParameterName)));

它循环两次标记参数;但大多数时候这并不重要(它不会成为你的瓶颈;如果是，展开循环)。

如果你真的对性能感兴趣，不想重复循环两次，这里有一个不太漂亮的版本:

var parameters = new List<SqlParameter>();
var paramNames = new List<string>();
for (var i = 0; i < tags.Length; i++)  
{
    var paramName = "@tag" + i;

    //Include size and set value explicitly (not AddWithValue)
    //Because SQL Server may use an implicit conversion if it doesn't know
    //the actual size.
    var p = new SqlParameter(paramName, SqlDbType.NVarChar(50) { Value = tags[i]; } 
    paramNames.Add(paramName);
    parameters.Add(p);
}

var inClause = string.Join(",", paramNames);

在默认情况下，我将通过向IN条件传递一个表值函数(从字符串返回一个表)来实现这一点。

下面是UDF的代码(我从Stack Overflow的某个地方得到了它，我现在找不到源代码)

CREATE FUNCTION [dbo].[Split] (@sep char(1), @s varchar(8000))
RETURNS table
AS
RETURN (
    WITH Pieces(pn, start, stop) AS (
      SELECT 1, 1, CHARINDEX(@sep, @s)
      UNION ALL
      SELECT pn + 1, stop + 1, CHARINDEX(@sep, @s, stop + 1)
      FROM Pieces
      WHERE stop > 0
    )
    SELECT 
      SUBSTRING(@s, start, CASE WHEN stop > 0 THEN stop-start ELSE 512 END) AS s
    FROM Pieces
  )

一旦你得到了这个，你的代码就会像这样简单:

select * from Tags 
where Name in (select s from dbo.split(';','ruby;rails;scruffy;rubyonrails'))
order by Count desc

除非你有一个长得离谱的字符串，否则这应该与表索引一起工作得很好。

如果需要，你可以把它插入一个临时表，索引它，然后运行一个连接…