Question

我有一个cron“时间定义”

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。

是否有一种明显的方法可以在下一次（在给定的时间戳之后）找到（计算）作业将要执行的时间？

我正在使用PHP，但问题应该与语言无关。

[更新]

类“PHP Cron Parser”（Ray建议）计算CRON作业应该执行的最后时间，而不是下次。

为了更容易：在我的情况下，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。

Answer 1

这是一个基于dlamblin的伪代码的PHP项目。

它可以计算CRON表达式的下一个运行日期，CRON表达式的上一个运行日期，并确定CRON表达式是否与给定时间匹配。您可以跳过此CRON表达式解析器完全实现CRON：

范围增量（例如* / 12,3-59 / 15）
间隔期（例如1-4，MON-FRI，JAN-MAR）
列表（例如1,2,3 | JAN，MAR，DEC）
一个月的最后一天（例如L）
最后一个工作日（例如5L）
Nth给出一个月的工作日（例如3＃2,1-1，MON＃4）
最接近工作日至该月的某一天（例如15W，1W，30W）

https://github.com/mtdowling/cron-expression

用法（PHP 5.3 +）：

<?php

// Works with predefined scheduling definitions
$cron = Cron\CronExpression::factory('@daily');
$cron->isDue();
$cron->getNextRunDate();
$cron->getPreviousRunDate();

// Works with complex expressions
$cron = Cron\CronExpression::factory('15 2,6-12 */15 1 2-5');
$cron->getNextRunDate();

Answer 2

这基本上反过来检查当前时间是否符合条件。所以像：

//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;
}

我可能已经写了一点倒退。如果在每个主要部分中，如果不是大于检查而只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，则它可以更短。然而，你会循环更多。像这样：

//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;
}

Answer 3

对于任何感兴趣的人，这是我的最终PHP实现，它几乎等于dlamblin伪代码：

class myMiniDate {
    var $myTimestamp;
    static private $dateComponent = array(
                                    'second' => 's',
                                    'minute' => 'i',
                                    'hour' => 'G',
                                    'day' => 'j',
                                    'month' => 'n',
                                    'year' => 'Y',
                                    'dow' => 'w',
                                    'timestamp' => 'U'
                                  );
    static private $weekday = array(
                                1 => 'monday',
                                2 => 'tuesday',
                                3 => 'wednesday',
                                4 => 'thursday',
                                5 => 'friday',
                                6 => 'saturday',
                                0 => 'sunday'
                              );

    function __construct($ts = NULL) { $this->myTimestamp = is_null($ts)?time():$ts; }

    function __set($var, $value) {
        list($c['second'], $c['minute'], $c['hour'], $c['day'], $c['month'], $c['year'], $c['dow']) = explode(' ', date('s i G j n Y w', $this->myTimestamp));
        switch ($var) {
            case 'dow':
                $this->myTimestamp = strtotime(self::$weekday[$value], $this->myTimestamp);
                break;

            case 'timestamp':
                $this->myTimestamp = $value;
                break;

            default:
                $c[$var] = $value;
                $this->myTimestamp = mktime($c['hour'], $c['minute'], $c['second'], $c['month'], $c['day'], $c['year']);
        }
    }


    function __get($var) {
        return date(self::$dateComponent[$var], $this->myTimestamp);
    }

    function modify($how) { return $this->myTimestamp = strtotime($how, $this->myTimestamp); }
}


$cron = new myMiniDate(time() + 60);
$cron->second = 0;
$done = 0;

echo date('Y-m-d H:i:s') . '<hr>' . date('Y-m-d H:i:s', $cron->timestamp) . '<hr>';

$Job = array(
            'Minute' => 5,
            'Hour' => 3,
            'Day' => 13,
            'Month' => null,
            'DOW' => 5,
       );

while ($done < 100) {
    if (!is_null($Job['Minute']) && ($cron->minute != $Job['Minute'])) {
        if ($cron->minute > $Job['Minute']) {
            $cron->modify('+1 hour');
        }
        $cron->minute = $Job['Minute'];
    }
    if (!is_null($Job['Hour']) && ($cron->hour != $Job['Hour'])) {
        if ($cron->hour > $Job['Hour']) {
            $cron->modify('+1 day');
        }
        $cron->hour = $Job['Hour'];
        $cron->minute = 0;
    }
    if (!is_null($Job['DOW']) && ($cron->dow != $Job['DOW'])) {
        $cron->dow = $Job['DOW'];
        $cron->hour = 0;
        $cron->minute = 0;
    }
    if (!is_null($Job['Day']) && ($cron->day != $Job['Day'])) {
        if ($cron->day > $Job['Day']) {
            $cron->modify('+1 month');
        }
        $cron->day = $Job['Day'];
        $cron->hour = 0;
        $cron->minute = 0;
    }
    if (!is_null($Job['Month']) && ($cron->month != $Job['Month'])) {
        if ($cron->month > $Job['Month']) {
            $cron->modify('+1 year');
        }
        $cron->month = $Job['Month'];
        $cron->day = 1;
        $cron->hour = 0;
        $cron->minute = 0;
    }

    $done = (is_null($Job['Minute']) || $Job['Minute'] == $cron->minute) &&
            (is_null($Job['Hour']) || $Job['Hour'] == $cron->hour) &&
            (is_null($Job['Day']) || $Job['Day'] == $cron->day) &&
            (is_null($Job['Month']) || $Job['Month'] == $cron->month) &&
            (is_null($Job['DOW']) || $Job['DOW'] == $cron->dow)?100:($done+1);
}

echo date('Y-m-d H:i:s', $cron->timestamp) . '<hr>';

Answer 4

使用此功能：

function parse_crontab($time, $crontab)
         {$time=explode(' ', date('i G j n w', strtotime($time)));
          $crontab=explode(' ', $crontab);
          foreach ($crontab as $k=>&$v)
                  {$v=explode(',', $v);
                   foreach ($v as &$v1)
                           {$v1=preg_replace(array('/^\*$/', '/^\d+$/', '/^(\d+)\-(\d+)$/', '/^\*\/(\d+)$/'),
                                             array('true', '"'.$time[$k].'"==="\0"', '(\1<='.$time[$k].' and '.$time[$k].'<=\2)', $time[$k].'%\1===0'),
                                             $v1
                                            );
                           }
                   $v='('.implode(' or ', $v).')';
                  }
          $crontab=implode(' and ', $crontab);
          return eval('return '.$crontab.';');
         }
var_export(parse_crontab('2011-05-04 02:08:03', '*/2,3-5,9 2 3-5 */2 *'));
var_export(parse_crontab('2011-05-04 02:08:03', '*/8 */2 */4 */5 *'));

修改也许这更具可读性：

<?php

    function parse_crontab($frequency='* * * * *', $time=false) {
        $time = is_string($time) ? strtotime($time) : time();
        $time = explode(' ', date('i G j n w', $time));
        $crontab = explode(' ', $frequency);
        foreach ($crontab as $k => &$v) {
            $v = explode(',', $v);
            $regexps = array(
                '/^\*$/', # every 
                '/^\d+$/', # digit 
                '/^(\d+)\-(\d+)$/', # range
                '/^\*\/(\d+)$/' # every digit
            );
            $content = array(
                "true", # every
                "{$time[$k]} === 0", # digit
                "($1 <= {$time[$k]} && {$time[$k]} <= $2)", # range
                "{$time[$k]} % $1 === 0" # every digit
            );
            foreach ($v as &$v1)
                $v1 = preg_replace($regexps, $content, $v1);
            $v = '('.implode(' || ', $v).')';
        }
        $crontab = implode(' && ', $crontab);
        return eval("return {$crontab};");
    }

用法：

<?php
if (parse_crontab('*/5 2 * * *')) {
    // should run cron
} else {
    // should not run cron
}

Answer 5

检查this out：

它可以根据给定的cron定义计算下次计划作业的运行时间。

Answer 6

根据@dlamblin的想法创建了用于计算下一次运行时间的JavaScript API。支持秒和年。还没有设法完全测试它所以期待错误，但如果发现任何问题，请告诉我。

存储库链接：https://bitbucket.org/nevity/cronner

Answer 7

感谢您发布此代码。它确实帮助了我，即使是6年之后。

尝试实施我发现了一个小错误。

date('i G j n w', $time)返回分钟的0填充整数。

稍后在代码中，它对该0填充整数执行模数。 PHP似乎没有像预期的那样处理它。

$ php
<?php
print 8 % 5 . "\n";
print 08 % 5 . "\n";
?>
3
0

如您所见，08 % 5返回0，而8 % 5返回预期的3.我找不到date命令的非填充选项。我尝试摆弄{$time[$k]} % $1 === 0行（比如将{$time[$k]}更改为({$time[$k]}+0)，但无法在模数期间删除0填充。

所以，我最后只是更改了日期函数返回的原始值，并通过运行$time[0] = $time[0] + 0;删除了0。

这是我的测试。

<?php

function parse_crontab($frequency='* * * * *', $time=false) {
    $time = is_string($time) ? strtotime($time) : time();
    $time = explode(' ', date('i G j n w', $time));
    $time[0] = $time[0] + 0;
    $crontab = explode(' ', $frequency);
    foreach ($crontab as $k => &$v) {
        $v = explode(',', $v);
        $regexps = array(
            '/^\*$/', # every 
            '/^\d+$/', # digit 
            '/^(\d+)\-(\d+)$/', # range
            '/^\*\/(\d+)$/' # every digit
        );
        $content = array(
            "true", # every
            "{$time[$k]} === $0", # digit
            "($1 <= {$time[$k]} && {$time[$k]} <= $2)", # range
            "{$time[$k]} % $1 === 0" # every digit
        );
        foreach ($v as &$v1)
            $v1 = preg_replace($regexps, $content, $v1);
            $v = '('.implode(' || ', $v).')';
    }
    $crontab = implode(' && ', $crontab);
    return eval("return {$crontab};");
}

for($i=0; $i<24; $i++) {
    for($j=0; $j<60; $j++) {
        $date=sprintf("%d:%02d",$i,$j);
        if (parse_crontab('*/5 * * * *',$date)) {
             print "$date yes\n";
        } else {
             print "$date no\n";
        }
    }
}

?>

Answer 8

我的回答不是唯一的。只是用java编写的@BlaM答案的复制品，因为PHP的日期和时间与Java略有不同。

该程序假定CRON表达式很简单。它只能包含数字或*。

Minute = 0-60
Hour = 0-23
Day = 1-31
MONTH = 1-12 where 1 = January.
WEEKDAY = 1-7 where 1 = Sunday.

代码：

package main;

import java.util.Calendar;
import java.util.Date;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class CronPredict
{
    public static void main(String[] args)
    {
        String cronExpression = "5 3 27 3 3 ls -la > a.txt";
        CronPredict cronPredict = new CronPredict();
        String[] parsed = cronPredict.parseCronExpression(cronExpression);
        System.out.println(cronPredict.getNextExecution(parsed).getTime().toString());
    }

    //This method takes a cron string and separates entities like minutes, hours, etc.
    public String[] parseCronExpression(String cronExpression)
    {
        String[] parsedExpression = null;
        String cronPattern = "^([0-9]|[1-5][0-9]|\\*)\\s([0-9]|1[0-9]|2[0-3]|\\*)\\s"
                        + "([1-9]|[1-2][0-9]|3[0-1]|\\*)\\s([1-9]|1[0-2]|\\*)\\s"
                        + "([1-7]|\\*)\\s(.*)$";
        Pattern cronRegex = Pattern.compile(cronPattern);

        Matcher matcher = cronRegex.matcher(cronExpression);
        if(matcher.matches())
        {
            String minute = matcher.group(1);
            String hour = matcher.group(2);
            String day = matcher.group(3);
            String month = matcher.group(4);
            String weekday = matcher.group(5);
            String command = matcher.group(6);

            parsedExpression = new String[6];
            parsedExpression[0] = minute;
            parsedExpression[1] = hour;
            parsedExpression[2] = day;
            //since java's month start's from 0 as opposed to PHP which starts from 1.
            parsedExpression[3] = month.equals("*") ? month : (Integer.parseInt(month) - 1) + "";
            parsedExpression[4] = weekday;
            parsedExpression[5] = command;
        }

        return parsedExpression;
    }

    public Calendar getNextExecution(String[] job)
    {
        Calendar cron = Calendar.getInstance();
        cron.add(Calendar.MINUTE, 1);
        cron.set(Calendar.MILLISECOND, 0);
        cron.set(Calendar.SECOND, 0);

        int done = 0;
        //Loop because some dates are not valid.
        //e.g. March 29 which is a Friday may never come for atleast next 1000 years.
        //We do not want to keep looping. Also it protects against invalid dates such as feb 30.
        while(done < 100)
        {
            if(!job[0].equals("*") && cron.get(Calendar.MINUTE) != Integer.parseInt(job[0]))
            {
                if(cron.get(Calendar.MINUTE) > Integer.parseInt(job[0]))
                {
                    cron.add(Calendar.HOUR_OF_DAY, 1);
                }
                cron.set(Calendar.MINUTE, Integer.parseInt(job[0]));
            }

            if(!job[1].equals("*") && cron.get(Calendar.HOUR_OF_DAY) != Integer.parseInt(job[1]))
            {
                if(cron.get(Calendar.HOUR_OF_DAY) > Integer.parseInt(job[1]))
                {
                    cron.add(Calendar.DAY_OF_MONTH, 1);
                }
                cron.set(Calendar.HOUR_OF_DAY, Integer.parseInt(job[1]));
                cron.set(Calendar.MINUTE, 0);
            }

            if(!job[4].equals("*") && cron.get(Calendar.DAY_OF_WEEK) != Integer.parseInt(job[4]))
            {
                Date previousDate = cron.getTime();
                cron.set(Calendar.DAY_OF_WEEK, Integer.parseInt(job[4]));
                Date newDate = cron.getTime();

                if(newDate.before(previousDate))
                {
                    cron.add(Calendar.WEEK_OF_MONTH, 1);
                }

                cron.set(Calendar.HOUR_OF_DAY, 0);
                cron.set(Calendar.MINUTE, 0);
            }

            if(!job[2].equals("*") && cron.get(Calendar.DAY_OF_MONTH) != Integer.parseInt(job[2]))
            {
                if(cron.get(Calendar.DAY_OF_MONTH) > Integer.parseInt(job[2]))
                {
                    cron.add(Calendar.MONTH, 1);
                }
                cron.set(Calendar.DAY_OF_MONTH, Integer.parseInt(job[2]));
                cron.set(Calendar.HOUR_OF_DAY, 0);
                cron.set(Calendar.MINUTE, 0);
            }

            if(!job[3].equals("*") && cron.get(Calendar.MONTH) != Integer.parseInt(job[3]))
            {
                if(cron.get(Calendar.MONTH) > Integer.parseInt(job[3]))
                {
                    cron.add(Calendar.YEAR, 1);
                }
                cron.set(Calendar.MONTH, Integer.parseInt(job[3]));
                cron.set(Calendar.DAY_OF_MONTH, 1);
                cron.set(Calendar.HOUR_OF_DAY, 0);
                cron.set(Calendar.MINUTE, 0);
            }

            done =  (job[0].equals("*") || cron.get(Calendar.MINUTE) == Integer.parseInt(job[0])) &&
                    (job[1].equals("*") || cron.get(Calendar.HOUR_OF_DAY) == Integer.parseInt(job[1])) &&
                    (job[2].equals("*") || cron.get(Calendar.DAY_OF_MONTH) == Integer.parseInt(job[2])) &&
                    (job[3].equals("*") || cron.get(Calendar.MONTH) == Integer.parseInt(job[3])) &&
                    (job[4].equals("*") || cron.get(Calendar.DAY_OF_WEEK) == Integer.parseInt(job[4])) ? 100 : (done + 1);
        }

        return cron;
    }
}

计算下次执行cron作业的时间

8 个答案: