使用t-sql组合'LIKE'和'IN'

时间:2011-05-23 20:04:49

标签: sql sql-server tsql

我该如何做这种选择:

SELECT * 
FROM Street 
WHERE StreetName LIKE IN ('% Main Street', 'foo %')

请不要告诉我,我可以使用OR,因为这些实际上来自查询。

7 个答案:

答案 0 :(得分:35)

没有合并LIKEIN语法,但您可以LIKE JOIN使用;WITH Query(Result) As ( SELECT '% Main Street' UNION ALL SELECT 'foo %' ) SELECT DISTINCT s.* FROM Street s JOIN Query q ON StreetName LIKE q.Result 到您的查询中。

SELECT DISTINCT s.* 
FROM Street s
JOIN CarStreets cs ON s.StreetName LIKE cs.name + '%'
WHERE cs.Streets = 'offroad'

或者在评论中使用您的示例

{{1}}

答案 1 :(得分:4)

你在这里没有太多选择。

SELECT * FROM Street Where StreetName LIKE '% Main Street' OR StreetName LIKE 'foo %'

如果这是现有的,更复杂的查询(这是我得到的印象)的一部分,您可以创建一个表值函数来为您进行检查。

SELECT * FROM Street Where StreetName IN (dbo.FindStreetNameFunction('% Main Street|foo %'))

我建议使用最简单的解决方案(第一个)。如果这是嵌套在更大,更复杂的查询中,请发布它,我们将看一看。

答案 2 :(得分:1)

You can resort to Dynamic SQL and wrapping up all in a stored procedure.

If you get the LIKE IN param in a string as tokens with a certain separator, like

'% Main Street,foo %,Another%Street'

first you need to create a function that receives a list of LIKE "tokens" and returns a table of them.

CREATE FUNCTION [dbo].[SplitList]
(
  @list nvarchar(MAX),
  @delim nvarchar(5)
)  
RETURNS @splitTable table 
(       
  value nvarchar(50)
) 
AS BEGIN
  While (Charindex(@delim, @list)>0) Begin 
    Insert Into @splitTable (value)
      Select ltrim(rtrim(Substring(@list, 1, Charindex(@delim, @list)-1))) 
    Set @list = Substring(@list, Charindex(@delim, @list)+len(@delim), len(@list))
  End     
  Insert Into @splitTable (value) Select ltrim(rtrim(@list))
  Return
END 

Then in the SP you have the following code

declare 
  @sql nvarchar(MAX),
  @subWhere nvarchar(MAX)
  @params nvarchar(MAX)

-- prepare the where sub-clause to cover LIKE IN (...)
-- it will actually generate where sub clause StreetName Like option1 or StreetName Like option2 or ...   
set @subWhere = STUFF(
  (
    --(**)
    SELECT ' OR StreetName like ''' + value + '''' FROM SplitList('% Main Street,foo %,Another%Street', ',') 
      FOR XML PATH('')
  ), 1, 4, '')

-- create the dynamic SQL
set @sql ='select * from [Street]
  where 
    (' + @subWhere + ')
    -- and any additional query params here, if needed, like
    AND StreetMinHouseNumber = @minHouseNumber
    AND StreetNumberOfHouses between (@minNumberOfHouses and @maxNumberOfHouses)'

set @params = ' @minHouseNumber nvarchar(5),
  @minNumberOfHouses int,
  @minNumberOfHouses int'     

EXECUTE sp_executesql @sql, @params,    
  @minHouseNumber,
  @minNumberOfHouses,
  @minNumberOfHouses     

Of course, if you have your LIKE IN parameters in another table or you gather it through a query, you can replace that in line (**)

答案 3 :(得分:0)

我相信我可以澄清他在寻找什么,但我不知道答案。我会用我的情况来证明。我有一个表格,其中包含一个名为" Query"持有SQL查询。这些查询有时包含来自我的一个数据库的表名。我需要查找包含特定数据库中的表名的所有Query行。所以,我可以使用以下代码来获取表名:

SELECT TABLE_NAME FROM INFORMATION_SCHEMA.TABLES

我试图使用WHERE IN子句来标识包含我感兴趣的表名的查询行:

SELECT *
  FROM [DatasourceQuery]
 WHERE Query IN LIKE
(
    SELECT TABLE_NAME FROM INFORMATION_SCHEMA.TABLES
)

我相信OP正试图做那样的事情。

答案 4 :(得分:0)

这是我的方式:

首先创建一个表函数:

create function [splitDelimeter](@str nvarchar(max), @delimeter nvarchar(10)='*')
returns @r table(val nvarchar(max))
as
begin

    declare @x nvarchar(max)=@str
    set @x='<m>'+replace(@x, @delimeter, '</m><m>')+'</m>'

    declare @xx xml=cast(@x as xml)

    insert @r(val)
    SELECT Tbl.Col.value('.', 'nvarchar(max)') id
    FROM @xx.nodes('/m') Tbl(Col)

    return 
end

然后使用首选项分隔符分割搜索文本。之后,您可以使用左联接进行选择,如下所示:

  declare @s nvarchar(max)='% Main Street*foo %'

  select a.* from street a
     left join gen.splitDelimeter(@s, '*') b
        on a.streetname like b.val
  where val is not null

答案 5 :(得分:0)

我有一个类似的难题,但是由于只需要匹配字符串的开头,因此将“ like”更改为SUBSTRING

SELECT *  
FROM codes  
WHERE SUBSTRING(code, 1, 12) IN ('012316963429', '012315667849')  

答案 6 :(得分:0)

我在解决类似问题时所做的是:

SELECT DISTINCT S.* 
FROM Street AS S
JOIN (SELECT value FROM String_Split('% Main Street,foo %', N',')) T 
    ON S.StreetName LIKE T.value;

这在功能上类似于 Martin 的回答,但更直接地回答了问题。

注意:使用 DISTINCT 是因为您可能会在一行中获得多个匹配项。