Question

表格：候选人

+-----+---------+
| id  | Name    |
+-----+---------+
| 1   | A       |
| 2   | B       |
| 3   | C       |
| 4   | D       |
| 5   | E       |
+-----+---------+

表格：投票

+-----+--------------+
| id  | CandidateId  |
+-----+--------------+
| 1   |     2        |
| 2   |     4        |
| 3   |     3        |
| 4   |     2        |
| 5   |     5        |
+-----+--------------+
id is the auto-increment primary key, CandidateId is the id appeared in Candidate table.

编写一个sql来查找获胜者的姓名，上面的示例将返回获胜者B。

+------+
| Name |
+------+
| B    |
+------+

注意：您可能会假设没有平局，换句话说，最多将有一名获胜者。

为什么此代码无法正常工作？尝试无限使用

SELECT c.Name AS Name
FROM Candidate AS c 
               JOIN 
                   (SELECT r.CandidateId AS can, MAX(r.Total_vote) AS big
                    FROM (SELECT CandidateId, COUNT(id) AS Total_vote
                          FROM Vote
                          GROUP BY CandidateId) AS r) AS v
ON c.id = v.can;

Answer 1

在您的查询中，这里：Single<String>

您使用SELECT r.CandidateId AS can, MAX(r.Total_vote) AS big聚合函数，而没有MAX，这是不正确的SQL。

尝试：

group by

Answer 2

这是一个join / group by查询，带有order by：

select c.name
from candidate c join
     vote v
     on v.candidateid = c.id
group by c.id, c.name
order by count(*) desc
limit 1;

Answer 3

SELECT c.Name AS Name
FROM Candidate AS c JOIN (SELECT r.CandidateId AS can
                      FROM
                       (SELECT CandidateId, COUNT(id) AS Total_vote
                        FROM Vote
                        GROUP BY CandidateId) AS r
                      WHERE r.Total_vote = (SELECT MAX(r.Total_vote) FROM (SELECT 
CandidateId, COUNT(id) AS Total_vote
                        FROM Vote
                        GROUP BY CandidateId) r)) AS v
ON c.id = v.can;

这是更新的代码

我的代码有两个错误。第一个是“如果选择列表中有任何未聚合的列，则使用像Max这样的聚合需要使用Group By子句”，但是不确定为什么我的先前代码仍然可以运行并且没有显示错误。也许系统在运行时自动按功能添加分组。

第二个是max不能与这种格式的Group by一起使用。

leetcode 574获奖候选人查询

3 个答案: