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CRON 作业在该月的最后一天运行

发布于 2024-11-10 01:06:27 字数 75 浏览 5 评论 0原文

我需要创建一个将在每个月的最后一天运行的 CRON 作业。我将使用 cPanel 创建它。

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终难遇 2024-11-17 01:06:27

也许最简单的方法是简单地执行三项单独的工作：

55 23 30 4,6,9,11        * myjob.sh
55 23 31 1,3,5,7,8,10,12 * myjob.sh
55 23 28 2               * myjob.sh

不过，即使在闰年，这也会在 2 月 28 日运行，因此，如果这是一个问题，您将需要找到另一种方法。

然而，在每个月的第一天尽快运行作业通常会更容易、更正确，例如：

0 0 1 * * myjob.sh

并修改脚本来处理上一个作业> 月数据。

这消除了您在确定哪一天是该月最后一天时可能遇到的任何麻烦，并且还确保该月的所有数据都可用（假设您正在处理数据）。在该月的最后一天到午夜前五分钟跑步可能会让您错过从那时到午夜之间发生的任何事情。

无论如何，对于大多数月末工作来说，这是通常的做法。

如果您仍然确实想要在该月的最后一天运行它，一个选择是简单地检测明天是否是第一个（作为脚本的一部分，或者在 crontab 本身中）。

因此，

55 23 28-31 * * [[ "$(date --date=tomorrow +\%d)" == "01" ]] && myjob.sh

假设您有一个相对智能的 date 程序，那么：应该是一个好的开始。

如果您的 date 程序不够先进，无法为您提供相对日期，您可以编写一个非常简单的程序来为您提供该月的明天（您不需要 date的全部功能），例如：

#include <stdio.h>
#include <time.h>

int main (void) {
    // Get today, somewhere around midday (no DST issues).

    time_t noonish = time (0);
    struct tm *localtm = localtime (&noonish);
    localtm->tm_hour = 12;

    // Add one day (86,400 seconds).

    noonish = mktime (localtm) + 86400;
    localtm = localtime (&noonish);

    // Output just day of month.

    printf ("%d\n", localtm->tm_mday);

    return 0;
}

然后使用（假设您将其称为 tomdom 表示“明天的某一天”）：

55 23 28-31 * * [[ "$(tomdom)" == "1" ]] && myjob.sh

尽管您可能想要考虑添加错误检查，因为两者如果出现问题，time() 和 mktime() 可能会返回 -1。为了简单起见，上面的代码没有考虑到这一点。

Possibly the easiest way is to simply do three separate jobs:

55 23 30 4,6,9,11        * myjob.sh
55 23 31 1,3,5,7,8,10,12 * myjob.sh
55 23 28 2               * myjob.sh

That will run on the 28th of February though, even on leap years so, if that's a problem, you'll need to find another way.

However, it's usually both substantially easier and correct to run the job as soon as possible on the first day of each month, with something like:

0 0 1 * * myjob.sh

and modify the script to process the previous month's data.

This removes any hassles you may encounter with figuring out which day is the last of the month, and also ensures that all data for that month is available, assuming you're processing data. Running at five minutes to midnight on the last day of the month may see you missing anything that happens between then and midnight.

This is the usual way to do it anyway, for most end-of-month jobs.

If you still really want to run it on the last day of the month, one option is to simply detect if tomorrow is the first (either as part of your script, or in the crontab itself).

So, something like:

55 23 28-31 * * [[ "$(date --date=tomorrow +\%d)" == "01" ]] && myjob.sh

should be a good start, assuming you have a relatively intelligent date program.

If your date program isn't quite advanced enough to give you relative dates, you can just put together a very simple program to give you tomorrow's day of the month (you don't need the full power of date), such as:

#include <stdio.h>
#include <time.h>

int main (void) {
    // Get today, somewhere around midday (no DST issues).

    time_t noonish = time (0);
    struct tm *localtm = localtime (&noonish);
    localtm->tm_hour = 12;

    // Add one day (86,400 seconds).

    noonish = mktime (localtm) + 86400;
    localtm = localtime (&noonish);

    // Output just day of month.

    printf ("%d\n", localtm->tm_mday);

    return 0;
}

and then use (assuming you've called it tomdom for "tomorrow's day of month"):

55 23 28-31 * * [[ "$(tomdom)" == "1" ]] && myjob.sh

Though you may want to consider adding error checking since both time() and mktime() can return -1 if something goes wrong. The code above, for reasons of simplicity, does not take that into account.

CRON 作业在该月的最后一天运行

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