问一道SQL 题，求解答

问大家一道leetcode变形题，求解答??

given table A:
+---------+------------+----------+--------+
| user_id | date       | platform | amount |
+---------+-------

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7-01 | desktop  | 2           |
| 2019-07-01 | mobile   | 3           |
| 2019-07-01 | both     | 1           |

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我这里用了两个CTE
第一个里面先 left join 来判断 platform
第二个里面用了 union 来合并所有的可能性

根据 test case 提供的数据可以在 MySQL 里面得到如下结果：

1. WITH CTE1 as   (
   
2. SELECT DISTINCT
   
3. a.user_id,
   
4. a.date,
   
5. case
   
6. when a.platform = 'mobile' and b.platform is null then 'mobile'
   
7. when a.platform = 'desktop' and b.platform is null then 'desktop'
   .1point3acres
8. when a.platform is not null and b.platform is not null then 'both'
   
9. else null end as 'platform'
   
10. 
    
11. FROM platform a LEFT JOIN platform b
    
12. ON a.user_id = b.user_id
    
13. AND a.date = b.date
    
14. AND a.platform <> b.platform
    
15. ),
    . Χ
16. 
    
17. CTE2 as   (
    
18. select user_id, date, platform from CTE1
    
19. union
    
20. select user_id, date, 'mobile' as platform from CTE1 where platform = 'both'
    
21. union
    
22. select user_id, date, 'desktop' as platform from CTE1 where platform = 'both'
    
23. )
    
24. 
    
25. 
    
26. SELECT
    
27. date,
    
28. platform,
    
29. count(user_id) as total_users
    
30. 
    
31. FROM CTE2
    
32. GROUP BY date, platform
    
33. ORDER BY date, platform desc
    
34. ;

复制代码

..

方法比较笨，但是应该有效

1. with a as (
   
2. select date, user_id, sum(case when platform='mobile' then amount else 0 end ) as ma,sum(case when platform='desktop' then amount else 0 end) as da from A group by date, user),
   
3. ref as (select distinct date, 'mobile' as platform from A union (select distinct date,'desktop' as platform from A) union select distinct date,'both' as platform from A),
   
4. b as (select date, count(case when ma>0 and da>0 then userid else null end) as bothn, count(case when ma=0 and da>0 then userid else null end) as dn, count(case when da=0 and ma>0 then userid else null end) as mn from a group by date)
   
5. select date, platform, coalesce(total_users, 0) as total_users from (
   
6. select date, 'both' as p , bothn as total_users from b where bothn >0 union select date, 'mobile' as p, mn as total_users from b where mn>0 union select date,  'desktop' as p ,dn as total_users from b where dn>0) sub right join ref on sub.date=ref.date,sub.p=ref.platform

复制代码

combine rows，要用for, 再用一个case when

Group by date, count user id. Having date = 2019-7-1

cloverlyy 发表于 2020-11-25 01:36
Group by date, count user id. Having date = 2019-7-1

这题最终结果应该不只2019-07-01的

踩我的朋友，要是有5个platform 你会scale吗？不写for你来做一个？count(userid) = 2你能知道是哪2个platform？莫名其妙

Qomo 发表于 2020-11-25 02:19
这题最终结果应该不只2019-07-01的

嗯嗯嗯，我的答案不对。 积分不够，看不了全文。不知道答案中both是什么。。。

cloverlyy 发表于 2020-11-25 04:02
嗯嗯嗯，我的答案不对。 积分不够，看不了全文。不知道答案中both是什么。。。

both 应该是指 each user_id 既有 mobile transaction 也有 desktop transaction.

能想到的最简单的方法是用union。desktop和mobile的用group by就可以，后面both的可以self join一下date+user_id并且左边取desktop右边取mobile，两个都不是null就是1，然后sum一下，最后两个table union all起来。