Question

我有一个简单的MySQL表看起来：

    Table: Citizen

    Name    Country
    ----    ------
    Adam    USA
    Eva     Russia
    Arman   India
    Peter   USA
    Martin  Russia
    Jack    Finland
    Juno    Russia
    Fox     USA
    Maya    Finland

像

这样的简单查询

SELECT Country, COUNT(Name) People FROM Citizen  GROUP BY Country

会给我

Country     People
-------     ------
USA         3
Russia      3
Finland     2
India       1

我想找到人数最多的国家/地区。在平局的情况下（即两个国家的最大人数相同），列出所有这样的国家。结果表应如下所示

Country     MaxPeople
-------     ------
USA         3
Russia      3

我学习了一些关于派生表的知识并提出了这个问题：

SELECT Country, MAX(TotCit) 
FROM (SELECT Country, COUNT(People) TotCit 
      FROM Citizen GROUP BY Country) 
AS CitTable

结果是

Country   Max(TotCit)
-------   -----------
Finland   3

任何可能的解决方案？

Answer 1

如果您需要所有国家/地区，那么查询有点麻烦：

SELECT Country, COUNT(Name) as People
FROM Citizen 
GROUP BY Country
HAVING COUNT(Name) = (select max(cnt)
                      from (select COUNT(Name) as cnt
                            from Citizen
                            group by Country
                           ) c
                     );

查询可以满足您的需求。 HAVING子句首先计算每个国家/地区的计数，然后计算max()并将其与汇总值进行比较。

顺便说一下，这在大多数其他数据库中要容易得多，因为它们支持窗口/分析功能。

Answer 2

戈登的询问，简化了一下。

除非COUNT(Name)是可以为空的列，否则您不需要Nname。

可以使用LIMIT的简单子查询替换2嵌套子查询：

SELECT Country, COUNT(*) AS People
FROM Citizen 
GROUP BY Country
HAVING COUNT(*) = 
       ( SELECT COUNT(*) AS People
         FROM Citizen
         GROUP BY Country
         ORDER BY People DESC
         LIMIT 1
       ) ;

HAVING中的条件可能会导致查询效率非常低，因此您可以重写为：

SELECT 
    g.Country, g.People
FROM
  ( SELECT COUNT(*) AS People
    FROM Citizen
    GROUP BY Country
    ORDER BY cnt DESC
    LIMIT 1
  ) AS m
  JOIN
  ( SELECT Country, COUNT(*) AS People
    FROM Citizen  
    GROUP BY Country
  ) AS g
      ON g.People = m.People ;

选择从COUNT（）派生的所有类似MAX值

2 个答案: