學習SQL：需要幫助使用HAVING從聚合獲得最小結果

Question

希望得到一些幫助。 SQL新手，並試圖保持頭腦清醒。

我一直在考慮這個任務：

「使用HAVING，確定其直接下屬有最低工資平均。」

我的邏輯是在衍生表上加入一個連接，以計算每個具有直接報告的經理的工作人員的AVG工資。這個結果然後通過HAVING子句被過濾到MIN值。

SELECT Flash.firstname, Flash.lastname, Wally.AVGSalary, 
     Wally.[Direct Reports] FROM CrewMembers Flash 
    JOIN 
    (
     SELECT Barry.crewMemberId, AVG(Zoom.salary) AS 'AVGSalary', COUNT(Zoom.firstname) AS 'Direct Reports' 
        FROM CrewMembers Barry 
      JOIN CrewMembers Zoom 
          ON Zoom.managerId = Barry.crewMemberId 
     GROUP BY Barry.crewMemberId 
    ) 
    AS Wally ON Wally.crewMemberId = Flash.crewMemberId 

GROUP BY Flash.firstname, Flash.lastname, Wally.AVGSalary, Wally.[Direct Reports] 

HAVING MIN(Wally.AVGSalary) in (Wally.AVGSalary) 

ORDER BY Wally.AVGSalary asc 

我得到這個結果：

firstname lastname AVGSalary Direct Reports 
Mike  Patton  33666.500000 2 
Kurt  Corgan  37300.000000 2 
Amber  Bruckner 45851.666666 3 
Doug  Adams  86250.000000 2 
Montgomery Scott  92500.000000 2 
James  Kirk  132666.750000 3 

我難倒。我需要它給我帶最低價值的那一排，它給我所有的價值。我盯着這個，無法弄清楚我做錯了什麼。

我知道，通過在線觀察線索，使用TOP是一種過濾項目的方式，但尚未引入，我希望保留在迄今爲止已建立的參數範圍內。

任何批評我寫的或任何幫助將會很棒。就像我說的，我是新的，只是想跟上。

感謝，

Ĵ

Answer 1

你需要另一個子查詢（如果你不想使用上/窗口功能/等）：

select Flash.firstname, 
    Flash.lastname, 
    Wally.AVGSalary, 
    Wally.[Direct Reports] 
from CrewMembers Flash 
join (
    select Barry.crewMemberId, 
     AVG(Zoom.salary) as 'AVGSalary', 
     COUNT(Zoom.firstname) as 'Direct Reports' 
    from CrewMembers Barry 
    join CrewMembers Zoom on Zoom.managerId = Barry.crewMemberId 
    group by Barry.crewMemberId 
    having avg(Zoom.salary) = (
      select min(salary) 
      from (
       select AVG(Zoom.salary) as salary 
       from CrewMembers Barry 
       join CrewMembers Zoom on Zoom.managerId = Barry.crewMemberId 
       group by Barry.crewMemberId 
       ) t 
      ) 
    ) as Wally on Wally.crewMemberId = Flash.crewMemberId 

您可以使用top：

select top 1 Flash.firstname, 
    Flash.lastname, 
    Wally.AVGSalary, 
    Wally.[Direct Reports] 
from CrewMembers Flash 
join (
    select Barry.crewMemberId, 
     AVG(Zoom.salary) as 'AVGSalary', 
     COUNT(Zoom.firstname) as 'Direct Reports' 
    from CrewMembers Barry 
    join CrewMembers Zoom on Zoom.managerId = Barry.crewMemberId 
    group by Barry.crewMemberId 
    ) as Wally on Wally.crewMemberId = Flash.crewMemberId 
order by Wally.AVGSalary 

另一種方法是使用窗口功能rank：

select Flash.firstname, 
    Flash.lastname, 
    Wally.AVGSalary, 
    Wally.[Direct Reports] 
from CrewMembers Flash 
join (
    select Barry.crewMemberId, 
     AVG(Zoom.salary) as 'AVGSalary', 
     COUNT(Zoom.firstname) as 'Direct Reports', 
     rank() over (order by AVG(Zoom.salary)) as rnk 
    from CrewMembers Barry 
    join CrewMembers Zoom on Zoom.managerId = Barry.crewMemberId 
    group by Barry.crewMemberId 
    ) as Wally on Wally.crewMemberId = Flash.crewMemberId 
where Wally.rnk = 1; 

Answer 2

你應該使用HAVING與平等的，而不是IN所以像：

HAVING Wally.AVGSalary = MIN(Wally.AVGSalary)

你的情況與IN將返回所有行，你想只希望與工資等於最低的一個。

Answer 3

嘗試：

Select crewMemberId, avgSalary 
From (Select s.crewMemberId, 
     AVG(d.salary) avgSalary 
     From CrewMembers d join CrewMembers s 
     on s.crewMemberId = d.managerId 
     group by s.crewMemberId) x 
Where avgSalary = 
    (Select Min(avgSalary) 
    from (Select s.crewMemberId, 
      AVG(d.salary) avgSalary 
      From CrewMembers d join CrewMembers s 
       on s.crewMemberId = d.managerId 
      group by s.crewMemberId)x)  

- 在臨時表變量使用的示例數據：

declare @e table(crewmemberId int primary key not null, 
       managerId int null, salary decimal) 
Insert @e(crewmemberId, managerId, salary) 
values (1, null, 23), 
     (2,1,45), (3,1,33), (4,1,80), 
     (5,2,14), (6,2,8), (7,2,12), 
     (8,3,11), (9,3,5), (10,3,2), 
     (11,4,51), (12,4,38), (13,4,17) 

Select crewMemberId, avgSalary 
From (Select s.crewMemberId, 
     AVG(d.salary) avgSalary 
     From @e d join @e s 
     on s.crewMemberId = d.managerId 
     group by s.crewMemberId) x 
Where avgSalary = 
    (Select Min(avgSalary) 
    From (Select s.crewMemberId, 
      AVG(d.salary) avgSalary 
      From @e d join @e s 
      on s.crewMemberId = d.managerId 
      group by s.crewMemberId)x)   

這給出crewmemberId = 3，平均工資= 6.0000;

如果兩個或多個經理的直接聘用具有相同的平均工資，則它具有生成多個結果的優勢。