1

我有一张桌子

create table posts (
   text varchar(20),
   created_at datetime,
   user_id integer
);

create table users (
   id integer,
   name varchar(20),
   important bolean
);

create table categories (
   id integer,
   name varchar(20),
);
create table categories_posts (
   category_id integer,
    post_id integer
);

想要计算特定时间之间每个类别的帖子数量,结果如下

START_TIME_PERIOD, category, count_all, count_important, count_normal
'2013-03-06 23:40', 'dogs',    20,       5,               15
'2013-03-06 23:40', 'cats',    22,       6,               16
'2013-03-06 23:40', 'birds',   24,       7,               17

其中重要性基于用户的重要性。

得到一个计数,例如重要的

select '2013-03-06 23:40', categories.name, count(*)
from posts, users, categories
where posts.id = categories_posts.post_id
and categories.id = categories_posts.category_id
and posts.user_id = users.id
and users.important = true
and posts.created_at BETWEEN '2013-03-06 23:40' AND  '2013-03-06 23:20'
group by 1, 2

在单个结果集中获得三个计数的最佳方法是什么?

谢谢

4

2 回答 2

2
SELECT ..... , count(*), SUM(IF(user.important,1,0)) as count_important, COUNT(*) - SUM(IF(user.important,1,0)) as count_normal
于 2013-03-06T23:52:42.557 回答
1

除了 using COUNT,您还可以使用SUMwith CASE,例如

SELECT COUNT(*) AS count_all, 
    SUM(CASE WHEN important = 1 THEN 1 ELSE 0 END) AS count_important
于 2013-03-06T23:51:15.227 回答