示例

标题

id | title

1 | Cars
2 | Computers

输入

id | entry    | title_id | user_id | time
1  | bmw      | 1        | 1       | 1418595811
2  | mercedes | 2        | 2       | 1418595812
3  | ibm      | 2        | 2       | 1418595813
4  | hp       | 2        | 3       | 1418595814
5  | dell     | 2        | 1       | 1418595815
6  | acer     | 2        | 3       | 1418595816
7  | asus     | 2        | 4       | 1418595817
8  | toshiba  | 2        | 4       | 1418595818
9  | ferrari  | 1        | 3       | 1418595819
10 | toyota   | 1        | 4       | 1418595820
11 | honda    | 1        | 2       | 1418595821

预期结果：

如果user_id = 1 - ＆gt;

Cars (3)
Computers (3)

如果user_id = 2 - ＆gt;

Cars (0)
Computers (5)

如果user_id = 3 - ＆gt;

Cars (2)
Computers (2)

如上所示，[user_id = 1]提交汽车（宝马）。之后，其他用户提交了该标题的条目。用户上次提交后，我需要查询每个标题的新提交。（使用时间字段）唯一需要的是，如果有新的提交。如果是这样，有多少？

我需要这样的东西：

select count(entry) 
from entry where time>last_entry_time[user_id] 
left join titles 
where title.id=entry.title_id

Answer 1

这适用于所有用户：

SELECT e1.user_id, t.title, COUNT(e2.id) AS count
FROM titles AS t
LEFT JOIN (SELECT user_id, title_id, MAX(time) AS last_entry
           FROM entry
           GROUP BY user_id, title_id) AS e1
ON t.id = e1.title_id
LEFT JOIN entry AS e2 ON e2.title_id = e1.title_id AND e2.user_id != e1.user_id AND e2.time > e1.last_entry
GROUP BY e1.user_id, t.id
ORDER BY e1.user_id, t.title

DEMO

为特定用户执行此操作：

SET @user_id = 1;

SELECT t.title, COUNT(e2.id) AS count
FROM titles AS t
LEFT JOIN (SELECT user_id, title_id, MAX(time) AS last_entry
           FROM entry
           WHERE user_id = @user_id
           GROUP BY title_id) AS e1
ON t.id = e1.title_id
LEFT JOIN entry AS e2 ON e2.title_id = e1.title_id AND e2.user_id != @user_id AND e2.time > e1.last_entry
GROUP BY t.id
ORDER BY t.title

DEMO

Answer 2

我不确定我是否正确地关注了您的问题。试试这个

select e.id, e.tittle_id, e.user_id from entry e 
       where e.id not in(
                        select e.id from entry e, titles t where t.id = e.id
                        )

mysql过滤结果与子查询结果

示例

2 个答案: