我有一个名为subscriptions
的数据库表。表格中的每一行都有一个created_at
日期,从这一天起,每个月(之后)都必须开具订阅发票。
因此,在2017-05-29
上,所有订阅created_at
天=='29'的订阅都必须开具发票,无论月份或年份如何。所以我想到了这个:
SELECT * FROM subscriptions WHERE DAY(created_at) = DAY(DATE_SUB(CURDATE(), INTERVAL 1 MONTH))
但是在那种情况下,当前一个月有30天我遇到了麻烦,因为在30日,它返回30,但在31日它也返回30.所以所有订阅都有created_at
天'30 '将开具两次发票。二月也会提出问题。
另一个问题是另一个问题,如果4月份没有第31天,我将如何开具2017-03-31的发票。
我可以在PHP中进行多次检查并检查那个月是否已经开具发票等。但我想知道我是否可以通过MySQL修复此问题。
我根据上面的查询创建了一个sqlfiddle示例,其中一些日期将会失败。
# Create subscription table
CREATE TABLE `subscription` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`company` varchar(100) DEFAULT NULL,
`created_at` datetime DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
# Fill subscription table
INSERT INTO `subscription` (`id`, `company`, `created_at`)
VALUES
(1, 'Acme', '2017-04-15 09:56:00'),
(2, 'Equmbo', '2017-02-28 10:00:00'),
(3, 'Megajo', '2017-03-31 08:10:34'),
(4, 'Astrotude', '2017-04-30 08:10:49');
# This is my base query for monthly invoice
SELECT * FROM subscription WHERE DAY(created_at) = DAY(DATE_SUB(CURDATE(), INTERVAL 1 MONTH));
# On 28th March, also fine
SELECT * FROM subscription WHERE DAY(created_at) = DAY(DATE_SUB('2017-03-28', INTERVAL 1 MONTH));
# But on 29th March, Equmbo is invoice again
SELECT * FROM subscription WHERE DAY(created_at) = DAY(DATE_SUB('2017-03-29', INTERVAL 1 MONTH));
# On 30st April, Astrotude AND Megajo must be invoiced. Only Astrotude returns.
SELECT * FROM subscription WHERE DAY(created_at) = DAY(DATE_SUB('2017-04-30', INTERVAL 1 MONTH));
答案 0 :(得分:2)
WHERE
/* don't invoice new subs from this month */
LAST_DAY(created_at)<LAST_DAY(CURDATE())
AND
(
/* exactly match sub's day value with today's day value */
DAY(created_at)=DAY(CURDATE())
OR
/* on last day of month, match sub's day if greater than today's day */
(CURDATE()=LAST_DAY(CURDATE()) AND DAY(created_at)>DAY(CURDATE()))
)
CURDATE()和LAST_DAY()返回完整日期字符串(yyyy-mm-dd)。
DAY()返回一个整数(d或dd)。