Question

我需要以下查询的帮助。

我有一个表M，它包含membre_number，first_name，Last_name，Date_birth 我需要知道meny成员的年龄大于所有成员first_name的平均值。请注意，许多成员都被命名为保罗。

我试过以下

SELECT date_nais
  FROM membre
 WHERE prenom_membre = 'paul'

SELECT AVG(convert(GETDATE()-membre.date_nais)/365.242199)  
  FROM membre;

Answer 1

试试吧，

 DECLARE @M table
 (
 membre_number int identity, 
 first_name varchar(10), 
 Last_name varchar(10), 
 Date_birth date
 )

 INSERT @M (first_name, Last_name, Date_birth) Values 
 ('Paul', 'Surname 1' , '20170101')
 ,('dadada', 'Surname 1' , '20130101')
 ,('Paul', 'Surname 2' , '20090301')
 ,('Paul', 'Surname 3' , '19890401')
 ,('Paul', 'Surname 4' , '19300521')
  ,('Alana', 'XX' , '19750521')

 SELECT * 
 ,DATEDIFF(Year,Date_birth,Getdate()) Age
 ,Avg(Cast(DATEDIFF(Year,Date_birth,Getdate()) as decimal(5,2))) Over (Partition by first_name)  Agv_ByName
 ,Avg(Cast(DATEDIFF(Year,Date_birth,Getdate()) as decimal(5,2))) Over (Partition by 1) AllUsers
 ,Case when first_name = 'Paul' THEN 
            Avg(Cast(DATEDIFF(Year,Date_birth,Getdate()) as decimal(5,2))) Over (Partition by first_name) 
 Else NULL end Agv_Pauls_age
 FROM 
 @M

结果

membre_number first_name Last_name  Date_birth Age         Agv_ByName                              AllUsers                                Agv_Pauls_age
------------- ---------- ---------- ---------- ----------- --------------------------------------- --------------------------------------- ---------------------------------------
6             Alana      XX         1975-05-21 42          42.000000                               28.166666                               NULL
2             dadada     Surname 1  2013-01-01 4           4.000000                                28.166666                               NULL
3             Paul       Surname 2  2009-03-01 8           30.750000                               28.166666                               30.750000
4             Paul       Surname 3  1989-04-01 28          30.750000                               28.166666                               30.750000
5             Paul       Surname 4  1930-05-21 87          30.750000                               28.166666                               30.750000
1             Paul       Surname 1  2017-01-01 0           30.750000                               28.166666                               30.750000

(6 rows affected)

或者让Paul AVg参与所有行

;with cte as (

 SELECT * 
 ,DATEDIFF(Year,Date_birth,Getdate()) Age
 ,Avg(Cast(DATEDIFF(Year,Date_birth,Getdate()) as decimal(5,2))) Over (Partition by first_name)  Agv_ByName
 ,Avg(Cast(DATEDIFF(Year,Date_birth,Getdate()) as decimal(5,2))) Over (Partition by 1) AllUsers
 ,Case when first_name = 'Paul' THEN 
            Avg(Cast(DATEDIFF(Year,Date_birth,Getdate()) as decimal(5,2))) Over (Partition by first_name) 
 Else NULL end Agv_Pauls_age
 FROM 
 @M
 )
 SELECT 
  membre_number 
 ,first_name    
 ,Last_name 
 ,Date_birth    
 ,Age   
 ,Agv_ByName    
 ,AllUsers  
 ,MAX(IsNUll(Agv_Pauls_age,0)) Over (Partition By 1 ) Agv_Pauls_age
 from cte

结果

membre_number first_name Last_name  Date_birth Age         Agv_ByName                              AllUsers                                Agv_Pauls_age
------------- ---------- ---------- ---------- ----------- --------------------------------------- --------------------------------------- ---------------------------------------
6             Alana      XX         1975-05-21 42          42.000000                               28.166666                               30.750000
2             dadada     Surname 1  2013-01-01 4           4.000000                                28.166666                               30.750000
3             Paul       Surname 2  2009-03-01 8           30.750000                               28.166666                               30.750000
4             Paul       Surname 3  1989-04-01 28          30.750000                               28.166666                               30.750000
5             Paul       Surname 4  1930-05-21 87          30.750000                               28.166666                               30.750000
1             Paul       Surname 1  2017-01-01 0           30.750000                               28.166666                               30.750000

(6 rows affected)

团体的平均日期

1 个答案: