Question

编写查询以打印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。

我还附上了相应表格的图像

表黑客：

hacker_id name
     5077 Rose
    21283 Angela
    62743 Frank
    88255 Patrick
    96196 Lisa

表挑战：

challenge_id hacker_id
       61654      5077
       58302     21283
       40587     88255
       29477      5077
        1220     21283
       69514     21283
       46561     62743
       58077     62743
       18483     88255
       76766     21283
       52382      5077
       74467     21283
       33625     96196
       26053     88255
       42665     62743
       12859     62743
       70094     21283
       34599     88255
       54680     88255
       61881      5077

所以，我做到了这一点

SELECT c.hacker_id, h.name, COUNT(c.challenge_id) AS challenge_count
FROM Challenges c LEFT JOIN Hackers h on  c.hacker_id = h.hacker_id
GROUP by 1,c.hacker_id HAVING challenge_count >= 
MAX(challenge_count) ORDER BY challenge_count DESC ,c.hacker_id DESC;

但没有获得预期的产出。我的输出

我需要排除输出中的重复结果，例如应排除具有相同数量挑战的黑客。

Answer 1

这里有几个标准：

hacker_id，姓名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。
如果多名学生创建了相同数量的挑战，则将结果中的学生排除在外。
除非计数等于创建的最大挑战数，

以下涉及项目1,2和＆amp; 3 ...

SELECT h.*
     , COUNT(c.challenge_id) challenge_count 
  FROM hackers h 
  JOIN challenges c 
    ON c.hacker_id = h.hacker_id 
 GROUP 
    BY h.hacker_id
 ORDER 
    BY challenge_count DESC, hacker_id;

我们可以将此查询加入到自身一次，以解决标准4，然后再次解决第5项，如下所示：

SELECT DISTINCT a.*
           FROM 
              ( SELECT h.*
                     , COUNT(c.challenge_id) challenge_count 
                  FROM hackers h 
                  JOIN challenges c 
                    ON c.hacker_id = h.hacker_id 
                 GROUP 
                    BY h.hacker_id
              ) a
           LEFT
           JOIN
              ( SELECT h.*
                     , COUNT(c.challenge_id) challenge_count 
                  FROM hackers h 
                  JOIN challenges c 
                    ON c.hacker_id = h.hacker_id 
                 GROUP 
                    BY h.hacker_id
              ) b
             ON b.hacker_id <> a.hacker_id AND b.challenge_count = a.challenge_count
           LEFT
           JOIN
              ( SELECT h.*
                     , COUNT(c.challenge_id) challenge_count 
                  FROM hackers h 
                  JOIN challenges c 
                    ON c.hacker_id = h.hacker_id 
                 GROUP 
                    BY h.hacker_id
              ) c
             ON c.challenge_count > a.challenge_count
          WHERE b.hacker_id IS NULL  
             OR c.hacker_id IS NULL
          ORDER 
             BY challenge_count DESC, hacker_id;

Answer 2

SELECT t1.name,
       t1.hacker_id,
       COALESCE(t2.challengeCount, 0) AS challengeCount
FROM Hackers t1
LEFT JOIN
(
    SELECT hacker_id, COUNT(*) AS challengeCount
    FROM Challenges
    GROUP BY hacker_id
) t2
    ON t1.hacker_id = t2.hacker_id
WHERE COALESCE(t2.challengeCount, 0) IN
(
    SELECT t1.challengeCount
    FROM
    (
        SELECT t1.hacker_id,
               COALESCE(t2.challengeCount, 0) AS challengeCount
        FROM Hackers t1
        LEFT JOIN
        (
            SELECT hacker_id, COUNT(*) AS challengeCount
            FROM Challenges
            GROUP BY hacker_id
        ) t2
            ON t1.hacker_id = t2.hacker_id
    ) t1
    GROUP BY t1.challengeCount
    HAVING COUNT(*) = 1
) OR COALESCE(t2.challengeCount, 0) =
(
    SELECT MAX(t.challengeCount) FROM
    (
        SELECT COUNT(*) AS challengeCount
        FROM Challenges GROUP BY hacker_id
    ) t
)
ORDER BY COALESCE(t2.challengeCount, 0) DESC,
         t1.hacker_id

根据测试用例排除结果

2 个答案: