计算下一次将执行cron作业的时间 [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.

有没有明显的方法来找到（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PHP，但是问题应该是相当的语言无关的。

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

[更新]

类 PHP Cron Parser （由Ray建议）计算LAST

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

为了方便起见：在我的例子中，cron时间参数只是绝对的，单个的数字或 *。没有时间范围，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

当前时间符合条件。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。此外，它可以是一个更短的，如果在每个主要如果而不是做大于检查你只是将当前时间等级增加一，设置较小的时间等级为0，然后继续;然而，你会循环更多。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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