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

请使用以下存储过程。它使用了一个自定义分割函数，可以在这里找到。

 create stored procedure GetSearchMachingTagNames 
    @PipeDelimitedTagNames varchar(max), 
    @delimiter char(1) 
    as  
    begin
         select * from Tags 
         where Name in (select data from [dbo].[Split](@PipeDelimitedTagNames,@delimiter) 
    end

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SQL Server 2016+中，另一种可能是使用OPENJSON函数。

OPENJSON中有关于这种方法的博客——按id列表选择行的最佳方法之一。

下面是一个完整的示例

CREATE TABLE dbo.Tags
  (
     Name  VARCHAR(50),
     Count INT
  )

INSERT INTO dbo.Tags
VALUES      ('VB',982), ('ruby',1306), ('rails',1478), ('scruffy',1), ('C#',1784)

GO

CREATE PROC dbo.SomeProc
@Tags VARCHAR(MAX)
AS
SELECT T.*
FROM   dbo.Tags T
WHERE  T.Name IN (SELECT J.Value COLLATE Latin1_General_CI_AS
                  FROM   OPENJSON(CONCAT('[', @Tags, ']')) J)
ORDER  BY T.Count DESC

GO

EXEC dbo.SomeProc @Tags = '"ruby","rails","scruffy","rubyonrails"'

DROP TABLE dbo.Tags 

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

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