MySQL 从日期开始的当前和上个月的第一天和最后一天（无时间戳）

Question

我希望以下查询能让您知道我在寻找什么-

SELECT SUM(t1.hours) AS totalhours FROM
(
    SELECT (time_to_sec(timediff(time_out, time_in)) / 3600) AS hours FROM bb_work_log 
    WHERE user_id = 6 AND (working_date BETWEEN '2014-04-01' AND '2014-04-31')
) AS t1

在我的查询中，您可以看到我在这里手动给出的working_date 。 但是，我不想手动完成。 我想动态选择当月的第一天和最后一天。

Answer 1

您可以使用LAST_DAY(NOW() - INTERVAL 1 MONTH) + INTERVAL 1 DAY ，这将从现在减去一个月，并通过在上个月的LAST_DAY添加 1 天将为您提供当月的第一天

SELECT SUM(t1.hours) AS totalhours FROM
(
    SELECT (time_to_sec(timediff(time_out, time_in)) / 3600) AS hours FROM bb_work_log 
    WHERE user_id = 6 
    AND (working_date BETWEEN LAST_DAY(NOW() - INTERVAL 1 MONTH) 
    AND LAST_DAY(NOW()))
) AS t1

LAST_DAY(NOW() - INTERVAL 1 MONTH) 这会给你上个月的最后一天

每月的第一天/最后一天 Fiddle Demo

Answer 2

上个月的第一天

select last_day(curdate() - interval 2 month) + interval 1 day

上个月的最后一天

select last_day(curdate() - interval 1 month)

当月的第一天

select last_day(curdate() - interval 1 month) + interval 1 day

当月的最后一天

select last_day(curdate())

Answer 3

你可以通过这些方式实现它----

/* Current month*/
SELECT DATE_SUB(LAST_DAY(NOW()),INTERVAL DAY(LAST_DAY(NOW()))-1 DAY),CONCAT(LAST_DAY(NOW()),' 23:59:59');

SELECT LAST_DAY(CURDATE()) - INTERVAL DAY(LAST_DAY(CURDATE()))-1 DAY ,CONCAT(LAST_DAY(NOW()),' 23:59:59');

/* previous month*/
SELECT  DATE_FORMAT(CURDATE() - INTERVAL 1 MONTH,'%Y-%m-01 00:00:00'),DATE_FORMAT(LAST_DAY(CURDATE()-INTERVAL 1 MONTH),'%Y-%m-%d 23:59:59');

Answer 4

-- first day of previous month
set @start_date = date_format(NOW() - INTERVAL 1 MONTH, '%Y-%m-01');
-- last day of previous month
set @end_date = date_format(NOW() , '%Y-%m-01') - INTERVAL 1 day;
select @start_date ,@end_date ;
--  first day of current month
set @start_date = date_format(NOW(), '%Y-%m-01');
-- last dat of current month
set @end_date = date_format(NOW() + INTERVAL 1 MONTH, '%Y-%m-01') - INTERVAL 1 day;
select @start_date ,@end_date ;