如何獲得列值和聚合值（MAX）

Question

我使用postgresql作爲我的存儲引擎。

我有一個員工與經理的自我外鍵參考。

這裏是我的employee表的外觀：

employee_id firstname lastname manager_id 
11;   "Chuck"; "Norris"; NULL 
1;   "Sergey"; "Brin"; 11 
3;   "Larry"; "Page"; 11 
5;   "Bill"; "Gates"; 4 
4;   "Father"; "Bill"; NULL 

現在我想查詢員工誰擁有最多的員工在他/她的顯示（經理標識和計數）。

這是上面的表格，我希望

id max_count 
    11  2 

此查詢工作;

select MAX(y.count) as max_count FROM 
(select m.manager_id as id, count(m.manager_id) as count from employee m GROUP BY (id)) y; 

max_count 
    2 

但這不起作用 - 我只是包括了id列。

select y.id, MAX(y.count) as max_count FROM 
(select m.manager_id as id, count(m.manager_id) as count from employee m GROUP BY (id)) y; 

我得到以下錯誤：

ERROR: column "y.id" must appear in the GROUP BY clause or be used in an aggregate function 
LINE 1: select y.id, MAX(y.count) as max_count FROM 
      ^

********** Error ********** 

ERROR: column "y.id" must appear in the GROUP BY clause or be used in an aggregate function 
SQL state: 42803 

Character: 8 

爲什麼這不起作用？

Answer 1

SELECT列表中的任何非聚合字段必須包含在GROUP BY中。您可以通過使用LIMIT或利用窗/解析函數得到你想要的東西：

select y.id, y.count as max_count 
FROM (select m.manager_id as id 
      , count(m.manager_id) as count 
     from employee m 
     GROUP BY id 
    ) y 
order by count DESC 
limit 1; 

正如評論指出的那樣，甚至更直截了當：

select m.manager_id as id 
    , count(m.manager_id) as count 
from employee m 
GROUP BY id 
order by count(m.manager_id) DESC 
limit 1;