2014-12-09 159 views
3

我在mySql中有以下表格。sql没有显示正确的结果

博客

Field  Type   
---------- ------------ 
id   int(11) 
name  varchar(255) 
user_id  int(11) 
share  int(14) 

user_blog_analytics

Field  Type   
----------- ------------ 
id   int(11) 
blog_id  int(11) 
ip   varchar(255) 
impressions int(11) 
date   date 

user_profile

Field  Type   
----------- ------------ 
id   int(11) 
user_id  int(11) 
description text 
share  int(14) 

个user_profile_analytics

Field  Type   
----------- ------------ 
id   int(11) 
user_id  int(11) 
ip   varchar(255) 
impressions int(11) 
date   date 

用户

​​

我想要一个查询,让我每个用户的总博客股份从blog表,从user_profile表中每个用户的总轮廓股,占总博客的观点从昨天即从user_blog_analytics表,所有时间观点从user_profile_analytics表中的配置文件。

我创建了一个查询,但没有给我我期望的结果,它只给了我很少的结果。

SELECT a.user_id, COUNT(DISTINCT b.ip) AS blog_view_count, a.share AS blog_share_count, c.share AS profile_share_count, COUNT(DISTINCT d.ip) AS user_profile_view 
FROM blog AS a 
JOIN user_blog_analytics AS b ON b.blog_id=a.id 
JOIN user_profile AS c ON c.user_id=a.user_id 
JOIN user_profile_analytics AS d ON d.user_id=c.user_id 
JOIN users AS e ON e.id=a.user_id 
WHERE DATE_SUB(CURDATE(), INTERVAL 1 DAY) = b.date AND e.role_id=2 
GROUP BY a.id; 

当我运行此查询它给了我只有一个结果,但是当我手工检查表那么就应该给我至少2分的结果。告诉我我错在哪里,如何通过修改此查询来获得结果。

+0

这很难说没有数据,但你有几个聚合件在那里。你能分享你得到的结果和你期望的结果吗?这可能是足够的数据来说明发生了什么。 – 2014-12-09 11:31:16

+0

那么用户可以有多个配置文件?什么是INT(14)? !?!?! – Strawberry 2014-12-09 11:32:22

回答

1

试试这个:

SELECT u.id, u.email, b.blog_share_count, b.blog_view_count, 
     up.profile_share_count, upa.user_profile_view 
FROM users u 
LEFT JOIN (SELECT b.user_id, SUM(b.share) AS blog_share_count, COUNT(DISTINCT b.ip) AS blog_view_count 
      FROM blog b 
      LEFT JOIN user_blog_analytics AS uba ON uba.blog_id = b.id AND DATE_SUB(CURDATE(), INTERVAL 1 DAY) = uba.date 
      GROUP BY b.user_id 
     ) b ON u.id = b.user_id 
LEFT JOIN (SELECT up.user_id, SUM(up.share) AS profile_share_count 
      FROM user_profile up 
      GROUP BY up.user_id 
     ) up ON u.id = up.user_id 
LEFT JOIN (SELECT up.user_id, COUNT(DISTINCT up.ip) AS user_profile_view 
      FROM user_profile_analytics up 
      GROUP BY up.user_id 
     ) upa ON u.id = upa.user_id 
1

不是一个答案,但值得思考......

DROP TABLE IF EXISTS my_table; 

CREATE TABLE my_table(i14 INT(14),i4 INT(4)); 

INSERT INTO my_table VALUES (123456789,123456789); 

SELECT * FROM my_table; 
+------------+------------+ 
| i14  | i4   | 
+------------+------------+ 
| 2147483647 | 2147483647 | 
+------------+------------+ 

因此,括号内的数字是不是做多的哦!