Question

我创建了一个表managers：

create table managers(
    ManagerId int identity(1,1) not for replication not null,
    M_name varchar(20),
    Salary varchar(20),
    joining_year varchar(20),
    city varchar(20),
    IdCode int
)

然后在此表中插入一些数据：

ManagerId   | M_name  |  Salary  | joining_year |  city   |  IdCode
----------------------------------------------------------------------
1          | riva    | 50000    | 1998         |  pune   | 4
2          | tanmay  | 48500    | 1990         |gurgaon  | 2
3          | david   | 49000    | 2001         | goa     | 2
4          | null    | null     | null         | null    | null
5          | null    | null     | null         | null    | null
6          | dannial | 52185    | 2010         | kanpur  | 6

并有第二张表managerEmp

create table managerEmp(
    employeId int identity(1,1) not for replication not null,
    family_member varchar(20),
    wife_name varchar(20),
    age int
)

我在该表中有一些数据：

employeId | family_member |wife_name | age
--------------------------------------------
1        | 6             |mrs.kapoor| 31
2        | 5             |mrs.mishra| 25
3        | null          |nll       | null
4        | 2             |mrs.khan  | 21
5        | 4             |mrs.bajaj | 22

现在，我想从该表中选择不常见的数据。结果将是：

M_name | Salary | city | wife_name | age
-----------------------------------------
null   | null   | null | mrs.khan  | 21
null   | null   | null | mrs.bajaj | 22
dannial| 52185  |kanpur| null      |null

Answer 1

根据您的输出进行查询：

from functools import reduce # Okay, okay, 2 lines!
reduce (lambda r, x: r + [{'weekday_name': days_of_the_week[x]}] if x in days_of_the_week 
                     else r + [{'weekday_name': r[-1]['weekday_name'], 'starting_time': x}] if len(r[-1]) == 3 
                     else (r[-1].update({'starting_time': x}), r)[1] if len(r[-1]) == 1 
                     else (r[-1].update({'ending_time': x}), r)[1], 
        filter(lambda x: x in days_of_the_week or x[0].isdigit(), input_array), [])

链接到SQL小提琴http://sqlfiddle.com/#!6/e1776/1

如何从两个表中检索不匹配的数据

1 个答案: