Mysql 来自hackerrank的15天Sql_Mysql_Sql Server

Mysql 来自hackerrank的15天Sql

mysql sql-server

Mysql 来自hackerrank的15天Sql,mysql,sql-server,Mysql,Sql Server,我无法理解这一行在代码中的用法。有人能解释一下这一点吗？或者给出一些不同的方法来解决这个问题吗问题的链接：代码：无法理解他们为什么使用这部分代码中的这一行： (s3.submission_date < s1.submission_date) = dateDIFF(s1.submission_date , '2016-03-01')) 要理解这句话 ( SELECT COUNT(distinct s3.submission_date) FROM

我无法理解这一行在代码中的用法。有人能解释一下这一点吗？或者给出一些不同的方法来解决这个问题吗

问题的链接：

代码：

无法理解他们为什么使用这部分代码中的这一行：

 (s3.submission_date < s1.submission_date) = dateDIFF(s1.submission_date , '2016-03-01'))

要理解这句话

(  SELECT COUNT(distinct s3.submission_date) 
          FROM      Submissions s3 
          WHERE 
          s3.hacker_id = s2.hacker_id 
          AND s3.submission_date < s1.submission_date)
                  = dateDIFF(s1.submission_date , '2016-03-01')

首先了解左手边：

此行统计每个黑客id在当前日期之前的唯一提交日期，因此，如果一行的日期为2016-03-05，则将计算黑客id的唯一提交。在此日期注释中，将一个黑客在一天内提交的多份提交计算为1份

换句话说，这需要一个黑客id，并开始检查从第一天起每天是否有该黑客id的提交。直到今天，它将在每个提交日期执行此操作

然后了解右手边：日期差异1.提交日期'2016-03-01' 这将取当前日期2016-03-05与第一天2016-03-01的差值

现在理解整个声明：

因此，如果黑客在2016-03-05至2016-03-01期间每天至少提交一份申请，那么上述代码的双方都是平等的，

也就是说，从5号到1号的日期差将是5号右边，对于从1号到5号每天至少提交一次的黑客来说，不同的提交日期也将是5号左边，这不是整行。它是从提交s3中选择COUNTdistinct s3.submission\u date，其中s3.hacker\u id=s2.hacker\u id和s3.submission\u date

CREATE TABLE #max_submissions (
    submission_date date,
    hacker_id integer,
    submission_count integer,
    ordering_row integer
)

insert into #max_submissions 
select 
    submission_date, 
    hacker_id, 
    submission_count, 
    row_number() over(partition by submission_date order by submission_count desc, hacker_id) as ordering_row
from (
    select submission_date, 
            hacker_id,
            count(hacker_id) as submission_count
    from submissions
    group by submission_date, hacker_id
) tbl_submission_count 

CREATE TABLE #hacker_counts (
    submission_date date,
    hacker_count integer
)

insert into #hacker_counts
select tbl.submission_date, 
        COUNT(distinct tbl.hacker_id) as cc 
from (
    select *,
            (case when (
                            (select count(*) 
                                from (select distinct * 
                                        from (select s1.hacker_id, 
                                                    s1.submission_date 
                                                from Submissions s1 
                                                where s1.hacker_id = s.hacker_id and
                                                (s1.submission_date >= '2016-03-01' and 
                                                s1.submission_date <= s.submission_date)) t1 
                                    ) t2 
                            ) >= (DATEDIFF(day, '2016-03-01', s.submission_date) + 1) )
                    then 1
                    else 0
            end) as logic
    from Submissions s
) tbl
where tbl.logic = 1
group by tbl.submission_date



select max_submissions.submission_date, 
    hacker_counts.hacker_count, 
    max_submissions.hacker_id, 
    h.name 
from #max_submissions max_submissions
inner join hackers h on max_submissions.hacker_id = h.hacker_id
left join #hacker_counts hacker_counts on max_submissions.submission_date = hacker_counts.submission_date
where max_submissions.ordering_row = 1
order by max_submissions.submission_date

drop table #max_submissions
drop table #hacker_counts

(  SELECT COUNT(distinct s3.submission_date) 
          FROM      Submissions s3 
          WHERE 
          s3.hacker_id = s2.hacker_id 
          AND s3.submission_date < s1.submission_date)
                  = dateDIFF(s1.submission_date , '2016-03-01')

(SELECT COUNT(distinct s3.submission_date) FROM Submissions s3 WHERE s3.hacker_id = s2.hacker_id AND s3.submission_date < s1.submission_date)