Question

我有两张桌子

1.Employee

EMP_NAME,

EMP_CODE

2.Vacations

EMP_NAME,

EMP_CODE,

VACATION_START_DATE-->date type

VACATION_END_DATE-->date type,

我的问题是如何查询从table1（Employee）获取EMP_NAME，其中今天不在table2（假期）的VACATION_START_DATE和VACATION_END_DATE之间..

Answer 1

请试试这个：假设您没有emp_name假期表......以及emp_code作为两个表之间的关系。所以你可以使用join s。

SQLFIDDLE DEMO

select e.emp_code, e.emp_name, 
v.start_date, v.end_Date
from emp e
inner join
vacation v
on e.emp_code = v.emp_code
where not (Now() between v.start_date 
and v.end_date)
;

| EMP_CODE | EMP_NAME |                      START_DATE |                        END_DATE |
-------------------------------------------------------------------------------------------
|        1 |     john | December, 10 2012 00:00:00+0000 | December, 20 2012 00:00:00+0000 |
|        2 |     kate | December, 20 2012 00:00:00+0000 | December, 30 2012 00:00:00+0000 |
|        3 |      tim | December, 24 2012 00:00:00+0000 |  January, 01 2013 00:00:00+0000 |
|        1 |     john |  January, 01 2013 00:00:00+0000 |  January, 08 2013 00:00:00+0000 |

Answer 2

不确定您是否认为“今天”是变量或系统日期，因此可能需要修改以下代码，但我认为它可能有效：

SELECT
    EMP_NAME
FROM
    EMPLOYEE
WHERE
    NOT EXISTS (SELECT * FROM VACATIONS WHERE EMP_NAME = EMPLOYEE.EMP_NAME AND
                VACATIONS.VACATION_START_DATE >= Today AND
                VACATIONS.VACATION_END_DATE <= Today)

怀疑查询条件

2 个答案: