Question

我在mySql中有以下表格。

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)

user_profile_analytics

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

用户

Field        Type        
-----------  ------------
id           int(11)
email        varchar(255)

我想要一个查询，它可以从blog表中为每个用户提供博客共享，从user_profile表中获得每个用户的总个人资料份额，从昨天开始的总博客视图，即user_blog_analytics表，来自user_profile_analytics表格的所有时间观点。

我创建了一个查询，但没有给出我期望的结果，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;

当我运行此查询时，它只给我一个结果但是当我手动检查表时，它应该给我至少2个结果。告诉我我错在哪里，如何通过修改此查询来获得结果。

Answer 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

Answer 2

不是答案，而是要考虑的事情......

DROP TABLE IF EXISTS my_table;

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

INSERT INTO my_table VALUES (123456789012345,123456789012345);

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

所以，括号中的数字对你来说没有多大作用！

sql没有显示正确的结果

2 个答案: