每月第二天到最后一天的Cron表达式 [英] Cron expression for the second to last day of week of the month

问题描述

我想触发每月的倒数第二天。

I want to trigger the second to last day of week of the month.

通常，这是该月的最后一个星期五（例如：30/06 / 2017）

Typically, here is the last friday of the month (ex : 30/06/2017)

0 0 0 ? * FRIL 

我想倒数第二个（例如：2017年6月23日）

And I want the second to last (ex : 23/06/2017)

0 0 0 ? * FRIL-1 

但是此语法返回与以前相同的结果（使用Quartz Scheduler和< a href = http://www.cronmaker.com/ rel = nofollow noreferrer> cronmaker ）

But this syntax return the same result as before (with the Quartz scheduler and with cronmaker)

一周的第二天到最后一天的月份可以出现在该月的第3周或第4周。
因此：

The second to last day of week of the month can appear either the 3rd or the 4th week of the month. So it's :

either : 0 0 0 ? * FRI#3 
or     : 0 0 0 ? * FRI#4 

您有什么建议吗？

推荐答案

我认为您无法使用cron语法来表达这一点。

我可以想到一些解决方法，但是：

I don't think you can express that using cron syntax.
I can think of some work-arounds, though:

• 您可以安排每个星期五的工作，并有一些工作中的逻辑，以便在继续工作之前检查它是否倒数第二。

• You could schedule your job every Friday and have some in-job logic to check whether it is actually the second-to-last before going on.

另一种选择是为最后一个星期五创建一个虚拟的cron触发器，检索下一个触发时间，减去该日期之后的7天并创建使用该确切日期的实际触发器-但您必须每周执行一次（通过预先安排几个触发器，或者在每次运行后重新安排自己的工作）。

Another option would be to create a dummy cron trigger for the last Friday, retrieve the "next fire time", substract 7 days from that date and create the actual trigger using that exact date -- but you'd have to do that for every week (either by pre-scheduling several triggers, or by having your job re-schedule itself after each run).

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