SQL SELECT JOIN

Question

员工日志表

Employeeno        | day_in      | day_out
0123              | 2017-06-20  | 2017-06-21
0123              | 2017-06-21  | 2017-06-22
0122              | 2017-06-20  | 2017-06-21
0122              | 2017-06-22  | 2017-06-23
0121              | 2017-06-23  | 2017-06-24

我想查询day_in（日期时间）小于2017-06-21的员工以及大于2017-06-21的员工编号0122的日期。

Answer 1

只需使用子查询即可获得结果。

 SELECT Employeeno FROM [Employee_logs] where day_in<' 2017-06-21' and Employeeno in(
    SELECT Employeeno FROM  [Employee_logs]  where day_in>' 2017-06-21')

Answer 2

创建表脚本

  Create table Attendance(Employeeno int ,day_in datetime,day_out datetime)

插入记录脚本

  Insert into Attendance
    Select 0123,'2017-06-20 ','2017-06-21' union
    Select 0123,'2017-06-21 ','2017-06-22' union
    Select 0122,'2017-06-20 ','2017-06-21' union
    Select 0122,'2017-06-22 ','2017-06-23' union
    Select 0121,'2017-06-23 ','2017-06-24'  

方法1

SELECT Employeeno FROM Attendance where day_in<' 2017-06-21' INTERSECT
SELECT Employeeno FROM Attendance where day_in>' 2017-06-21'

对于我的上述方法，我使用了INTERSECT，它提供了常见的记录。 完全使用可以在这里看到[https://blog.sqlauthority.com/2008/10/17/sql-server-get-common-records-from-two-tables-without-using-join/]

方法2

WITH x AS 
(
    -- Gives the result for Employees with Days_in less 2017-06-21
    SELECT * FROM Attendance where day_in<'2017-06-21'
), 
y AS 
(
      -- Gives the result for Employees with Days_in greater than 2017-06-21
     SELECT * FROM Attendance where day_in>'2017-06-21'
)
SELECT * FROM y join x on y.Employeeno=x.Employeeno 
-- Give the result based on the required query for all employees.

Answer 3

如果您使用派生表找到最早和最新的day-in值，则可以轻松找到符合条件的值：

-- Create test data
declare @t table(Employeeno nvarchar(5), day_in date, day_out date);
insert into @t values
 ('0123','2017-06-20','2017-06-21')
,('0123','2017-06-21','2017-06-22')
,('0122','2017-06-20','2017-06-21')
,('0122','2017-06-22','2017-06-23')
,('0121','2017-06-23','2017-06-24');

-- Query
with d as
(
    select Employeeno
            ,min(day_in) as EarliestDayIn
            ,max(day_in) as LatestDayIn
    from @t
    group by Employeeno
)
select *
from d
where EarliestDayIn < '20170621'
    and LatestDayIn > '20170621';

输出：

+------------+---------------+-------------+
| Employeeno | EarliestDayIn | LatestDayIn |
+------------+---------------+-------------+
|       0122 | 2017-06-20    | 2017-06-22  |
+------------+---------------+-------------+