Question

我正在尝试在Laravel / MySQL中实现某些目标，似乎无法为解决方案指明正确的方向。我可以通过子查询实现我想要的功能，但是有人告诉我它们不如联接有效。而且，我将不得不将解决方案转换为Eloquent / Query Builder，而使用子查询和联合的方法似乎并不容易转换。

我要根据行的created_at日期从两个可能的表中选择一行。我想将此created_at值与我的users表一起添加为名为started_at的新列。以下是一些示例数据，以及如何使用两个可能的表的子查询/联合来实现查询，我可以从中获取数据：

CREATE TABLE users (
  id INTEGER, 
  first_name TEXT,
  last_name TEXT
 );

INSERT INTO users (id, first_name, last_name)
VALUES 
(1, 'Craig', 'Smith'), 
(2, 'Bill', 'Nye'), 
(3, 'Bloop', 'Blop');

CREATE TABLE old_activity (
  id INTEGER,
  user_id INTEGER,
  firm_id INTEGER,
  amount INTEGER,
  created_at DATE
 );

INSERT INTO old_activity (id, user_id, firm_id, amount, created_at)
VALUES 
(1, 1, 3, 5.24, '2019-04-29'), 
(2, 2, 7, 4, '2019-03-28'), 
(3, 3, 4, 6.99, '2019-04-28');

CREATE TABLE new_activity (
  id INTEGER,
  user_id INTEGER,
  firm_id INTEGER,
  plays INTEGER,
  saves INTEGER,
  created_at DATE
);

INSERT INTO new_activity (id, user_id, firm_id, plays, saves, created_at)
VALUES 
(1, 1, 3, 10, 1, '2019-04-27'), 
(2, 2, 3, 12, 2, '2019-03-29'), 
(3, 3, 3, 6, 3, '2019-04-27');

CREATE TABLE firms (
  id INTEGER,
  name TEXT
);

INSERT INTO firms (id, name)
VALUES 
(1, 'apple'), 
(2, 'banana'), 
(3, 'orange');

select 
id,
first_name,
last_name,
(select created_at from old_activity 
 where user_id = users.id
 union
 select created_at from new_activity 
 where user_id = users.id
 order by created_at asc 
 limit 1) as started_at
from users

查询应仅返回两个created_at表之一中特定用户的最早的activity。

如何通过联接实现此目标？任何帮助，将不胜感激。

Answer 1

嗯。。。您可以随时使用窗口功能：

select u.*, a.*
from users u left join
     (select a.*,
             row_number() over (partition by a.user_id order by a.created_at desc) as seqnum
      from ((select oa.* from old_activity oa) union all
            (select na.* from new_activity na)
           ) a
     ) a
     on a.user_id = a.id and a.seqnum = 1

如何从两个表联接中选择一个值？

1 个答案: