Question

我遇到了一些问题。问题有两个表，即黑客和挑战。这是架构：

Hackers(hacker_id: int, name string)
Challenges(challenge_id: int, hacker_id int)

我正在尝试编写一个查询来打印hacker_id，名称以及每个学生创建的挑战总数。按降序排列挑战总数对结果进行排序。如果多个学生创建了相同数量的挑战，则按hacker_id对结果进行排序。如果多个学生创建了相同数量的挑战且计数小于创建的最大挑战数，则将结果中的学生排除在外。

这是我的问题：

select hackers.hacker_id , 
       hackers.name , 
       count(challenges.challenge_id) as challenges_created 
       from 
       hackers left join  challenges 
       on 
       hackers.hacker_id = challenges.hacker_id  
       having 
       count(challenges_created) >= max(challenges_created) 
       order by challenges_created desc, hackers.hacker_id asc

输错了！请告诉我哪里出错了！

示例输入：

黑客表：image 挑战者表：image

Sample Output:
21283  Angela  6
88255  Patrick  5
96196  Lisa  1

Answer 1

这只是在mysql中解决的疯狂要求。使用CTE（不支持mysql），您可以重用查询：

with tmp as (
    select h.hacker_id, h.name, count(1) as challenges_created
    from hackers h
    left join challenges c on c.hacker_id = h.hacker_id
    group by h.hacker_id, h.name
    order by challenges_created desc, h.hacker_id asc 
), max_challenges_created as (
    select max(challenges_created) as challenges_created
    from tmp
), count_per_challenges_created as (
    select challenges_created, count(1) as c
    from tmp
    group by challenges_created
)
select *
from tmp
cross join max_challenges_created m
left join count_per_challenges_created c on c.challenges_created = tmp.challenges_created
where tmp.challenges_created = m.challenges_created
   or c.c = 1

http://rextester.com/XGYQ11641

在MySQL中，您甚至无法使用临时表，因为您无法在一次查询中多次使用它。所以你必须复制并粘贴相同的查询三次（并希望mysql只使用缓存执行一次）：

select *
from (
    select h.*, count(1) as challenges_created
    from hackers h
    left join challenges c on c.hacker_id = h.hacker_id
    group by h.hacker_id
    order by challenges_created desc, h.hacker_id 
) tmp
cross join (
    select max(challenges_created) as challenges_created
    from (
        select h.*, count(1) as challenges_created
        from hackers h
        left join challenges c on c.hacker_id = h.hacker_id
        group by h.hacker_id
        order by challenges_created desc, h.hacker_id asc
    ) tmp
) m
left join (
    select challenges_created, count(1) as c
    from (
        select h.*, count(1) as challenges_created
        from hackers h
        left join challenges c on c.hacker_id = h.hacker_id
        group by h.hacker_id
        order by challenges_created desc, h.hacker_id asc
    ) tmp
    group by challenges_created
) c on c.challenges_created = tmp.challenges_created
where tmp.challenges_created = m.challenges_created
   or c.c = 1

http://rextester.com/IDUMRB49795

Answer 2

尝试开始解决查询的最小部分并继续加入它们。您可以尝试我的SQL（Oracle）代码

SELECT h.hacker_id, 
       h.name, 
       COUNT(c.challenge_id) AS challenge_count
FROM Hackers h
JOIN Challenges c ON c.hacker_id = h.hacker_id
GROUP BY h.hacker_id, h.name
HAVING challenge_count = 
    (SELECT COUNT(challenge_id) AS count_max
     FROM Challenges
     GROUP BY hacker_id 
     ORDER BY count_max DESC limit 1)
OR challenge_count IN 
    (SELECT DISTINCT c_compare AS c_unique
     FROM (SELECT h2.hacker_id, 
                  h2.name, 
                  COUNT(challenge_id) AS c_compare
           FROM Hackers h2
           JOIN Challenges c ON c.hacker_id = h2.hacker_id
           GROUP BY h2.hacker_id, h2.name) counts
     GROUP BY c_compare
     HAVING COUNT(c_compare) = 1)
ORDER BY challenge_count DESC, h.hacker_id;

Answer 3

尝试一下，这很简单。...

select
    h.hacker_id,
    h.name
from hackers h, difficulty d, challenges c, submissions s
where
    h.hacker_id=s.hacker_id
and s.challenge_id=c.challenge_id
and d.difficulty_level=c.difficulty_level
and d.score=s.score
group by h.hacker_id, h.name
having count(*)>1
ORDER BY COUNT(*) DESC, h.hacker_id;

Answer 4

这不是真正的答案，而是你的起点：

http://sqlfiddle.com/#!9/6910c/5

SELECT h.hacker_id , 
       h.name , 
       COUNT(challenges.challenge_id) as challenges_created 
FROM  hackers h 
LEFT JOIN challenges 
ON  h.hacker_id = challenges.hacker_id  
GROUP BY h.hacker_id
#HAVING challenges_created >= max(challenges_created) 
ORDER BY challenges_created desc, h.hacker_id asc

我已经注释了你的HAVING条款，因为无法理解你想要实现的目标。

解释一下，为什么你期望结果只有3行？为什么应该从结果集中排除Rose和Frank？

Answer 5

在必要的表上使用多个连接时，解决方案很容易。为了清楚起见，我喜欢将结果分组为派生表，但也可以在没有这些派生表的情况下完成。

prod

Answer 6

select hacker_id, name, tmp.challenges_created
from (
    select h.*, count(1) as challenges_created
    from hackers h
    left join challenges c on c.hacker_id = h.hacker_id
    group by h.hacker_id, h.name
    order by challenges_created desc, h.hacker_id 
) tmp
cross join (
    select max(challenges_created) as challenges_created
    from (
        select h.*, count(1) as challenges_created
        from hackers h
        left join challenges c on c.hacker_id = h.hacker_id
        group by h.hacker_id, h.name
        order by challenges_created desc, h.hacker_id asc
    ) tmp
) m
left join (
    select challenges_created, count(1) as c
    from (
        select h.*, count(1) as challenges_created
        from hackers h
        left join challenges c on c.hacker_id = h.hacker_id
        group by h.hacker_id, h.name
        order by challenges_created desc, h.hacker_id asc
    ) tmp
    group by challenges_created
) c on c.challenges_created = tmp.challenges_created
where tmp.challenges_created = m.challenges_created
   or c.c = 1

Answer 7

SELECT a.hacker_id, a.name, b.id
  FROM Hackers a
 INNER JOIN (SELECT hacker_id, id
               FROM (SELECT hacker_id, count(challenge_id) AS id
                       FROM Challenges
                      GROUP BY hacker_id) x
               WHERE id = 50
             UNION ALL
             SELECT hacker_id,id
               FROM (SELECT hacker_id, count(challenge_id) AS id
                       FROM Challenges
                      GROUP BY hacker_id) y
              WHERE id IN (SELECT id
                             FROM (SELECT hacker_id, count(challenge_id) AS id
                                     FROM Challenges
                                    GROUP BY hacker_id) x
                            GROUP BY id
                           HAVING count(*) = 1)
            ) b ON a.hacker_id = b.hacker_id
  ORDER BY id DESC, hacker_id

Answer 8

这是hackerranck.com中的问题，您可以检查它here [link] 在此站点中，您可以通过许多SQL DBMS（oracale，maysql，sqlserver和DB2）传递答案

对于 SQL-SERVER

来说，效果很好

with t 
as(
Select  s.hacker_id ,
        s.score,d.Difficulty_level as Difficulty_level, 
        c.Challenge_id
From submissions s 
    inner join Challenges  c  on c.Challenge_id = s.Challenge_id
    inner join Difficulty d on d.Difficulty_level = c.Difficulty_level and d.score = s.score
) 

Select  t.hacker_id, h.name
From t inner join hackers h ON t.hacker_id=h.hacker_id
Where t.score in (Select score from Difficulty)
Group by t.hacker_id,h.name
Having count(t.Challenge_id) > 1
Order by count(t.Challenge_id) desc,t.hacker_id asc

Answer 9

我已经在SQL Server中完成我加入了2个不同的选择，第一个使用等级来确定具有等级1的选择，第二个使用重复的选择，因此在联接结果中，我将结果过滤为具有第一个等级或没有重复（cnt = 1）

select s.hacker_id, s.name, s.challenge_created from (
select * ,rank() over (order by challenge_created desc) rn from (
select h.hacker_id, name, count(challenge_id) challenge_created from hackers h 
inner join challenges c on h.hacker_id = c.hacker_id
group by h.hacker_id, name) as T) as S
inner join (select A.challenge_created , count(1) cnt from (select h.hacker_id, name, count(challenge_id) challenge_created from hackers h 
inner join challenges c on h.hacker_id = c.hacker_id
group by h.hacker_id, name) as A group by A.challenge_created) as O
on s.challenge_created = O.challenge_created
where rn = 1 or O.cnt = 1
order by s.challenge_created desc , hacker_id

Answer 10

以下查询在mysql中工作正常对于具有条款条件我已经包含了计数等于创建的最大挑战数或挑战的数量是独一无二的

select c.hacker_id,h.name,count(c.challenge_id) as cnt from
challenges c join hackers h
on c.hacker_id = h.hacker_id
group by c.hacker_id,h.name
having cnt = 
(select max(counts) from (select count(challenge_id) as counts from challenges group by hacker_id) as countstable1) or
cnt in (select counts from (select count(challenge_id) as counts from challenges group by hacker_id) as countstable2 group by counts having count(*)=1)
order by cnt desc,c.hacker_id

MySQL：根据指定条件排除某些结果

10 个答案: