力扣中SQL刷题

news2024/11/24 20:33:16

小知识点

取余数：mod(数,2)
取第一个字母：left(name,1) 或者name like ‘M%’

196. 删除重复的电子邮箱

题型：删除列A中重复的记录，指保留列B最小的那一行记录
答案：delete t1 from 表名 t1,表名 t2 where t1.列A=t2.列A and t1.列B<t2.列B
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DELETE p1
from Person p1 , Person p2 
where p1. email=p2.email and p1.id>p2.id

删除语句： DELETE 表 from表 where 条件

627. 变更性别

题型：列A满足条件1时，列B的记录改为记录1，否则为记录2
解答：update 表名 set 列B= case 列A when 条件1 then 记录1 else 记录2 end;
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UPDATE salary
SET
    sex = CASE sex
        WHEN 'm' THEN 'f'
        ELSE 'm'
    END;

更新语句：UPDATE 表名 SET 更改内容

1070. 产品销售分析 III

题型：根据列A分组，取各组中列B记录最小的那一行
解答1：select * from 表名 where (列A，列B) in (select 列A，min(列B) from 表名 group by 列A)
解答2：窗口函数，先排序，再删选rank=1的列
select * from (select *,rank()over(partition by 列A order by 列B) as rr from 表名) as t
where rr=1
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select product_id,year as first_year,quantity,price
from Sales
where (product_id,year) in 
(select product_id,min(year) from Sales group by product_id)

select product_id,year as first_year,quantity,price
from (select *,rank()over(partition by product_id order by year ) as rr from Sales
) as t
where rr=1

534. 游戏玩法分析 III

题型：根据ID列A分组，然后对于日期列B，进行列C值的累计求和
解答：select *, sum(列c值)over(partition by 列A order by 列B) from table
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select  player_id,event_date,
sum(games_played)over(partition by player_id order by event_date asc) as games_played_so_far
from Activity

550. 游戏玩法分析 IV

题型：计算比率–count(distinct 列A)
读题一定要仔细，这里不是连续两日登录，是首次登录后的第二天
函数：data()、ifnull(值,0)、round(值,2)、
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select 
    ifnull(
        round(
            (count(distinct player_id))/(select count(distinct player_id) from activity)
            ,2)
        ,0) as fraction
from activity
where (player_id,event_date) in
(
    select player_id,date(min(event_date)+1) from activity group by player_id
)