计算何时执行 cron 作业然后下次 [英] Calculate when a cron job will be executed then next time

问题描述

我有一个 cron时间定义"

I have a cron "time definition"

1 * * * * (every hour at xx:01)
2 5 * * * (every day at 05:02)
0 4 3 * * (every third day of the month at 04:00)
* 2 * * 5 (every minute between 02:00 and 02:59 on fridays)

我有一个 Unix 时间戳.

And I have an unix timestamp.

是否有明显的方法可以找到(计算)下一次(在给定的时间戳之后)作业将被执行?

Is there an obvious way to find (calculate) the next time (after that given timestamp) the job is due to be executed?

我使用的是 PHP，但问题应该与语言无关.

I'm using PHP, but the problem should be fairly language-agnostic.

[更新]

类PHP Cron Parser"(由 Ray 建议)计算上次应该执行 CRON 作业，而不是下一次.

The class "PHP Cron Parser" (suggested by Ray) calculates the LAST time the CRON job was supposed to be executed, not the next time.

为了更容易:在我的例子中，cron 时间参数只是绝对的、单个数字或*".没有时间范围，也没有*/5"间隔.

To make it easier: In my case the cron time parameters are only absolute, single numbers or "*". There are no time-ranges and no "*/5" intervals.

推荐答案

这基本上与检查当前时间是否符合条件相反.所以像:

This is basically doing the reverse of checking if the current time fits the conditions. so something like:

//Totaly made up language
next = getTimeNow();
next.addMinutes(1) //so that next is never now
done = false;
while (!done) {
  if (cron.minute != '*' && next.minute != cron.minute) {
    if (next.minute > cron.minute) {
      next.addHours(1);
    }
    next.minute = cron.minute;
  }
  if (cron.hour != '*' && next.hour != cron.hour) {
    if (next.hour > cron.hour) {
      next.hour = cron.hour;
      next.addDays(1);
      next.minute = 0;
      continue;
    }
    next.hour = cron.hour;
    next.minute = 0;
    continue;
  }
  if (cron.weekday != '*' && next.weekday != cron.weekday) {
    deltaDays = cron.weekday - next.weekday //assume weekday is 0=sun, 1 ... 6=sat
    if (deltaDays < 0) { deltaDays+=7; }
    next.addDays(deltaDays);
    next.hour = 0;
    next.minute = 0;
    continue;
  }
  if (cron.day != '*' && next.day != cron.day) {
    if (next.day > cron.day || !next.month.hasDay(cron.day)) {
      next.addMonths(1);
      next.day = 1; //assume days 1..31
      next.hour = 0;
      next.minute = 0;
      continue;
    }
    next.day = cron.day
    next.hour = 0;
    next.minute = 0;
    continue;
  }
  if (cron.month != '*' && next.month != cron.month) {
    if (next.month > cron.month) {
      next.addMonths(12-next.month+cron.month)
      next.day = 1; //assume days 1..31
      next.hour = 0;
      next.minute = 0;
      continue;
    }
    next.month = cron.month;
    next.day = 1;
    next.hour = 0;
    next.minute = 0;
    continue;
  }
  done = true;
}

我可能写得有点倒退.此外，如果在每个 main 中，如果不是执行大于检查，而是将当前时间等级增加 1 并将较小的时间等级设置为 0，然后继续，则它可能会短得多；但是，您将循环更多.像这样:

I might have written that a bit backwards. Also it can be a lot shorter if in every main if instead of doing the greater than check you merely increment the current time grade by one and set the lesser time grades to 0 then continue; however then you'll be looping a lot more. Like so:

//Shorter more loopy version
next = getTimeNow().addMinutes(1);
while (true) {
  if (cron.month != '*' && next.month != cron.month) {
    next.addMonths(1);
    next.day = 1;
    next.hour = 0;
    next.minute = 0;
    continue;
  }
  if (cron.day != '*' && next.day != cron.day) {
    next.addDays(1);
    next.hour = 0;
    next.minute = 0;
    continue;
  }
  if (cron.weekday != '*' && next.weekday != cron.weekday) {
    next.addDays(1);
    next.hour = 0;
    next.minute = 0;
    continue;
  }
  if (cron.hour != '*' && next.hour != cron.hour) {
    next.addHours(1);
    next.minute = 0;
    continue;
  }
  if (cron.minute != '*' && next.minute != cron.minute) {
    next.addMinutes(1);
    continue;
  }
  break;
}

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