根据某些标准 (MYSQL) 选择每个重复元组的第一行

Question

我有一个这样的例子table：

Month City Person
8 LHR ABC
10 BEIJING BCS
11 NY JJJ
11 VENICE hghg
11 VENICE KKK
12 NY aa
12 ORL abc
12 ORL bbc

所以我想实现的是在特定月份看到访问人数最多的城市

就像输出应该是：

12 ORL  
11 VENINCE  
10 BEIJING  
8 LHR

我试过像

那样分组

SELECT month, city , count(*) AS 'no of people visiting'     
FROM table
GROUP BY month, city

这个table确实告诉我哪个城市和月份位置有多少人
访问但我无法提取最上面的月份和城市组合
关于某个月。

更新查询（有错误）

SELECT *  
FROM   
( SELECT monthname(reservation.PickupDate), location.LocationName, COUNT(*) AS count
FROM reservation  NATURAL JOIN location
WHERE reservation.pickupdate >= DATE_ADD(NOW(), INTERVAL - 3 MONTH)
GROUP BY month(reservation.PickupDate), location.LocationName) AS t1
WHERE NOT EXISTS (SELECT 1
FROM reservation R1 NATURAL JOIN location L1
WHERE  monthname(R1.PickupDate) = monthname(t1.PickupDate)
GROUP BY month(R1.PickupDate), L1.LocationName)

Answer 1

从您的查询开始，您只需要消除那些在该月有更多访客的城市的行：

SELECT * 
FROM
  (SELECT `month`, city, count(*) AS cnt 
  FROM `table` 
  GROUP BY `month`, city) t1
WHERE NOT EXISTS(SELECT 1 
  FROM `table` t2 
  WHERE t2.`month` = t1.`month`
  GROUP BY `month`, city 
  HAVING count(*) > t1.cnt)

根据某些标准 (MYSQL) 选择每个重复元组的第一行

selecting top row of each repeating tuple according to some criteria (MYSQL)

mysql

group-by