【发布时间】:2020-02-04 04:02:44
【问题描述】:
假设我们有一个名为Activities的表
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
我想找出 player_id 并且它是 first event_date 作为 first_date。
SQL:
SELECT Activities.player_id, min(Activities.event_date) as first_date
FROM `activities`
GROUP BY `activities`.`player_id`
结果表:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
+-----------+-------------+
如何在 Rails 中做到这一点?
我已经尝试过这个,但检索到一个仅包含 player_id 的 Activity 集合。
Activity.select('Activities.player_id, min(Activities.event_date) as first_date')
.group(:player_id)
这样
[#<Activity:0x00007f94923bb888 player_id: 1>, #<Activity:0x00007f94923bb608 player_id: 2>, #<Activity:0x00007f94923b9ba0 player_id: 3>]
【问题讨论】:
-
当您在 select 语句中指定自定义名称时,rails 会自动将其创建为对象上的方法名称。例如
activities = [#<Activity:0x00007f94923bb888 player_id: 1>]; activities.first.first_date -
试试
to_json,你也能看到first_date。 -
你到底是怎么得到那个 SQL 输出的?您选择
SELECT Activities.player_id, min(Activities.event_date) as first_date并在结果中得到player_id和device_id完全没有意义。这不会发生。 -
@max,对不起,我输入了完全错误的列名。表格已更新
标签: sql ruby-on-rails activerecord