测试一个字符串包含至少2个字在SQL

Question

我有一个像SQL请求：

SELECT * 
FROM table 
WHERE lower(title) LIKE lower('%It's a beautiful string i think%') 

我需要检查，如果在我的字符串It's a beautiful string i think至少2个字包含在我的领域标题...我怎样才能做到这一点 ？

例如，如果在这个领域的头衔，我有串I think it's beautiful，该查询应该返回我这个对象...

谢谢！

Answer 1

则可以将字符串分割到一个临时表（比方说，使用这样的：http://ole.michelsen.dk/blog/split-string-to-table-using-transact-sql/），然后做一个连接，具有计数。

Answer 2

你可以动态生成下面的SQL语句：

SELECT title, count(*) 
FROM 
    (
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% It %') 
    UNION ALL 
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% s %') 
    UNION ALL 
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% a %') 
    UNION ALL 
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% beautiful %') 
    UNION ALL 
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% string %') 
    UNION ALL 
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% I %') 
    UNION ALL 
    SELECT title 
    FROM table1 
    WHERE (' ' + lower(title) + ' ') LIKE lower('% think %') 
    ) AS table2 
GROUP BY title 
HAVING COUNT(*) >= 2 

存储过程可能更有效，你可以对服务器端的整个工作完成。

Answer 3

你可以使用这样的函数

CREATE FUNCTION [dbo].[CheckSentece] (@mainSentence varchar(128), @checkSentence varchar(128)) 
RETURNS NUMERIC AS 
BEGIN 
    SET @mainSentence=LOWER(@mainSentence) 
    SET @checkSentence=LOWER(@checkSentence) 
    DECLARE @pos INT 
    DECLARE @word varchar(32) 
    DECLARE @count NUMERIC 
    SET @count=0 
    WHILE CHARINDEX(' ', @checkSentence) > 0 
    BEGIN 
    SELECT @pos = CHARINDEX(' ', @checkSentence) 
    SELECT @word = SUBSTRING(@checkSentence, 1, @pos-1)  
    DECLARE @LEN NUMERIC 

    //Simple containment check, better to use another charindex loop to check each word from @mainSentence 
    SET @LEN=(SELECT LEN(REPLACE(@mainSentence,@word,''))) 
    if (@LEN<LEN(@mainSentence)) SET @count=@count+1  
    SELECT @checkSentence = SUBSTRING(@checkSentence, @pos+1, LEN(@checkSentence)-@pos) 
    END 
    return @count 
END 

并从包含在第一个