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 

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+ '|%'

    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 ,',')])

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)

这很恶心，但如果你保证至少有一个，你可以这样做:

SELECT ...
       ...
 WHERE tag IN( @tag1, ISNULL( @tag2, @tag1 ), ISNULL( @tag3, @tag1 ), etc. )

有IN('tag1'， 'tag2'， 'tag1'， 'tag1'， 'tag1')将很容易被SQL Server优化掉。另外，你可以直接搜索索引