Question

I m trying to find the number of rows which contain a value greater than 7 and less than 13.99. However, the following query always returns 0, and that s just not possible.

任何人能否在我的询问中看到问题?

SELECT COUNT(*) FROM lead_status AS ls 
INNER JOIN leads AS l ON l.id = ls.lead_id 
INNER JOIN contacts AS c ON c.lead_id = l.id 
WHERE ls.discriminator =  AUTO_POST  AND l.affiliate_id=1003 
AND ls.winner =1 AND l.test =0 AND l.create_date BETWEEN  2011-10-03  AND  2011-10-17  
AND ls.amount >= 7 AND ls.amount <= 13.99;

Answer 1

未成年人的ed、一些格式和怀疑:

SELECT COUNT(*)
FROM   lead_status ls 
INNER  JOIN leads l ON l.id = ls.lead_id    -- ls.id .. l.lead_id ??
INNER  JOIN contacts c ON c.lead_id = l.id 
WHERE  ls.discriminator =  AUTO_POST 
AND    ls.winner = 1
AND    ls.amount >= 7
AND    ls.amount <  14
AND    l.affiliate_id = 1003 
AND    l.test = 0
AND    l.create_date BETWEEN  2011-10-03  AND  2011-10-17 ;

You have ls.lead_id and c.lead_id, but l.id. That seems inconsistent. Typo?

页: 1

页: 1

你的法典说:

不超过或等于13.99

我把它改为(我怀疑你真的希望是什么):

大于或等于7或14

Answer 2

你的问询似乎是罚款。但是,只有确定先去除所有电离层过滤器,然后由一位您添加,过滤器将造成零。

Answer 3

你没有把你的表象张贴在后面,但你可能需要把你的指示日期提高到实际价值:

...AND l.create_date BETWEEN STR_TO_DATE( 2011-10-03 ,  %Y-%M-%d ) AND 
                             STR_TO_DATE( 2011-10-17 ,  %Y-%M-%d )

友情链接