从某个日期开始每天运行的 Cron 表达式 [英] Cron expression to run every day starting from a date

问题描述

我需要一个从 2016 年 1 月 25 日起每天中午 12 点触发的 cron 表达式.这就是我想出的:

I need a cron expression which will fire every day at 12 pm starting from jan 25 2016. This is what I came up with:

0 0 12 25/1 * ? *

但在 1 月 31 日之后，下次发射时间是 2 月 25 日.

but after jan 31, the next firing time is feb 25.

是否有用于执行此操作的 cron 表达式?如果不是，我可以使用什么?

Is there a cron expression expression for doing this? If not what can I use?

推荐答案

假设你想在 1 月 25 日之后永远运行这个进程(即 2032 年，那时服务器可能已经被替换)，我会用三个表达式:

Assuming that after January 25th you want to run this process forever after (i.e. 2032, when probably the server will be already substituted), I would do it with three expressions:

0 0 12 25-31 1 * 2016  command # Will run the last days of Jan 2016 after the 25th
0 0 12 * 2-12 * 2016   command # Will run the rest of the months of 2016
0 0 12 * * * 2017-2032 command # will run for every day of years 2017 and after.

我希望这会有所帮助.

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