Cron作业运行每月的第一个或最后一个选定的星期几 [英] Cron job to run every first or last chosen day of week of every month

问题描述

我有一个表单，用户选择何时运行脚本。我有字段{time，week of day，first / last} 。现在当用户选择例如：10:45 AM每个第一个星期一 - 这意味着在每个月的第一个星期一运行脚本，我该如何转换为cron作业。

I have a form where user chooses when to run a script. I've fields {time,day of week, first/last}. Now when the user chooses for example: 10:45 AM every first Monday - Which means run script on every first Monday of every month, how can I convert this to cron job.

谢谢。

推荐答案

不幸的是，crontab条目中的日期和星期字段是ORed一起，而不是与。所以你不能真正做一个第一个星期一的月与crontab领域单独;你需要额外的逻辑。你可以这样做：

Unfortunately, the day-of-month and day-of-week fields in a crontab entry are ORed together rather than ANDed. So you can't really do a "first Monday of the month" with crontab fields alone; you need additional logic. You could do something like this:

min hr 1-7 * * [ `date +%w` -eq 1 ] && first-monday.sh

查找工作日的最后一个实例有点棘手，因为它取决于很多天是在这个月：

Finding the last instance of a weekday is a bit trickier since it depends on how many days are in the month:

min hr 25-31 1,3,5,7,8,10,12 * [ `date +%w` -eq 5 ] && last-friday.sh
min hr 24-30 4,6,9,11        * [ `date +%w` -eq 5 ] && last-friday.sh
min hr 22-28 2               * [ `date +%w` -eq 5 ] && last-friday.sh

在闰年，2月29日星期五，运行脚本在22nd，而不是因为cron作业不能指定一年，这是尽可能接近你可以得到。

On leap years in which February 29th falls on Friday, that last entry will actually run the script on the 22nd instead, but since cron jobs can't specify a year, that's about as close as you can get.

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