Question

我有以下表格评估 -

SLNO  EID  Period_From   Period_To
 1    101  2017-06-01    2017-11-14
 2    102  2017-07-01    2017-09-30
 3    103  2017-05-01    2017-07-31

如果Period_To和currentdate等于75天，那么row应该返回1 else 0我在下面使用查询 -

 SELECT SLNO,EID, Period_From,Period_To,(CASE WHEN (PERIOD_TO = (select DATE_ADD(PERIOD_TO,INTERVAL 75 DAY))) THEN 1 ELSE 0 END) AS ASSESSMENT_ENABLE from assessment;

我的结果为

    SLNO  EID  Period_From   Period_To    ASSESSMENT_ENABLE
     1    101  2017-06-01    2017-11-14    0
     2    102  2017-07-01    2017-09-30    0
     3    103  2017-05-01    2017-07-31    0

我的结果错了。请帮我。

Answer 1

我认为你不需要子查询。试试这个对我有用

SELECT SLNO,EID, Period_From,Period_To,
(CASE WHEN (PERIOD_TO = DATE_FORMAT(CURDATE(), '%Y-%m-%d') - INTERVAL 75 DAY) THEN 1 ELSE 0 END) 
AS ASSESSMENT_ENABLE

Answer 2

使用CURRENT_DATE和DATEDIFF函数的简短解决方案：

SELECT 
    SLNO, EID, Period_From, Period_To,
    IF(DATEDIFF(CURRENT_DATE(), date_to) = 75, 1, 0) AS ASSESSMENT_ENABLE
FROM
    ASSESSMENT

https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html

Answer 3

你可以试试这个 -

select
    SLNO,
    EID,
    Period_From,
    Period_To,
    case when (DATEDIFF(now(),PERIOD_TO)=75) then 1 else 0 end as ASSESSMENT_ENABLE
from
    assessment;

mysql - 返回日期等于某个值的行

3 个答案: