I have a cron "time definition"
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)
And I have an unix timestamp.
我有一个unix时间戳。
Is there an obvious way to find (calculate) the next time (after that given timestamp) the job is due to be executed?
有没有一种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?
I'm using PHP, but the problem should be fairly language-agnostic.
我正在使用PHP,但问题应该与语言无关。
[Update]
The class "PHP Cron Parser" (suggested by Ray) calculates the LAST time the CRON job was supposed to be executed, not the next time.
“PHP Cron Parser”类(由Ray建议)计算CRON作业应该执行的最后时间,而不是下次。
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.
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。
8 个解决方案
#1
23
This is basically doing the reverse of checking if the current time fits the conditions. so something like:
这基本上与检查当前时间是否符合条件相反。所以像这样:
//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;
}
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:
我可能写了一点倒退。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;
}
#2
31
Here's a PHP project that is based on dlamblin's psuedo code.
这是一个基于dlamblin的伪代码的PHP项目。
It can calculate the next run date of a CRON expression, the previous run date of a CRON expression, and determine if a CRON expression matches a given time. You can skip This CRON expression parser fully implements CRON:
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:
- Increments of ranges (e.g. */12, 3-59/15)
- Intervals (e.g. 1-4, MON-FRI, JAN-MAR )
- Lists (e.g. 1,2,3 | JAN,MAR,DEC)
- Last day of a month (e.g. L)
- Last given weekday of a month (e.g. 5L)
- Nth given weekday of a month (e.g. 3#2, 1#1, MON#4)
- Closest weekday to a given day of the month (e.g. 15W, 1W, 30W)
范围增量(例如* / 12,3-59 / 15)
间隔时间(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)
最后给出一个月的工作日(例如5L)
Nth给出一个月的工作日(例如3#2,1-1,MON#4)
工作日最接近一个月的某一天(例如15W,1W,30W)
https://github.com/mtdowling/cron-expression
Usage (PHP 5.3+):
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();
#3
8
For anyone interested, here's my final PHP implementation, which pretty much equals dlamblin pseudo code:
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', $cron->timestamp) . '<hr>';
#4
6
Use this function:
使用此功能:
function parse_crontab($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 *'));
Edit Maybe this is more readable:
编辑也许这更具可读性:
<?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};");
}
Usage:
<?php
if (parse_crontab('*/5 2 * * *')) {
// should run cron
} else {
// should not run cron
}
#5
4
Check this out:
看一下这个:
It can calculate the next time a scheduled job is supposed to be run based on the given cron definitions.
#6
4
Created javascript API for calculating next run time based on @dlamblin idea. Supports seconds and years. Have not managed to test it fully yet so expect bugs but let me know if find any.
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。没有设法完全测试它所以期待错误,但让我知道如果找到任何。
Repository link: https://bitbucket.org/nevity/cronner
存储库链接:https://bitbucket.org/nevity/cronner
#7
2
Thanks for posting this code. It definitely helped me out, even 6 years later.
感谢您发布此代码。它确实帮助了我,即使是6年之后。
Trying to implement I found a small bug.
试图实现我发现了一个小bug。
date('i G j n w', $time)
returns a 0 padded integer for the minutes.
date('i G j n w',$ time)返回分钟的0填充整数。
Later in the code, it does a modulus on that 0 padded integer. PHP doesn't seem to handle this as expected.
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
As you can see, 08 % 5
returns 0, whereas 8 % 5
returns the expected 3. I couldn't find a non padded option for the date command. I tried fiddling with the {$time[$k]} % $1 === 0
line (like changing {$time[$k]}
to ({$time[$k]}+0)
, but couldn't get it to drop the 0 padding during the modulus.
如您所见,08%5返回0,而8%5返回预期的3.我找不到date命令的非填充选项。我尝试摆弄{$ time [$ k]}%$ 1 === 0行(比如将{$ time [$ k]}更改为({$ time [$ k]} + 0),但无法获得它在模数期间掉落0填充。
So, I ended up just changing the original value returned by the date function and removed the 0 by running $time[0] = $time[0] + 0;
.
所以,我最后只是改变了日期函数返回的原始值,并通过运行$ time [0] = $ time [0] + 0;删除了0。
Here is my test.
这是我的考试。
<?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";
}
}
}
?>
#8
2
My answer is not unique. Just a replica of @BlaM answer written in java because PHP's date and time is a bit different from Java.
我的回答并不是唯一的。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。
This program assumes that the CRON expression is simple. It can only contain digits or *.
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.
Code:
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;
}
}
#1
23
This is basically doing the reverse of checking if the current time fits the conditions. so something like:
这基本上与检查当前时间是否符合条件相反。所以像这样:
//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;
}
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:
我可能写了一点倒退。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;
}
#2
31
Here's a PHP project that is based on dlamblin's psuedo code.
这是一个基于dlamblin的伪代码的PHP项目。
It can calculate the next run date of a CRON expression, the previous run date of a CRON expression, and determine if a CRON expression matches a given time. You can skip This CRON expression parser fully implements CRON:
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:
- Increments of ranges (e.g. */12, 3-59/15)
- Intervals (e.g. 1-4, MON-FRI, JAN-MAR )
- Lists (e.g. 1,2,3 | JAN,MAR,DEC)
- Last day of a month (e.g. L)
- Last given weekday of a month (e.g. 5L)
- Nth given weekday of a month (e.g. 3#2, 1#1, MON#4)
- Closest weekday to a given day of the month (e.g. 15W, 1W, 30W)
范围增量(例如* / 12,3-59 / 15)
间隔时间(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)
最后给出一个月的工作日(例如5L)
Nth给出一个月的工作日(例如3#2,1-1,MON#4)
工作日最接近一个月的某一天(例如15W,1W,30W)
https://github.com/mtdowling/cron-expression
Usage (PHP 5.3+):
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();
#3
8
For anyone interested, here's my final PHP implementation, which pretty much equals dlamblin pseudo code:
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', $cron->timestamp) . '<hr>';
#4
6
Use this function:
使用此功能:
function parse_crontab($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 *'));
Edit Maybe this is more readable:
编辑也许这更具可读性:
<?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};");
}
Usage:
<?php
if (parse_crontab('*/5 2 * * *')) {
// should run cron
} else {
// should not run cron
}
#5
4
Check this out:
看一下这个:
It can calculate the next time a scheduled job is supposed to be run based on the given cron definitions.
#6
4
Created javascript API for calculating next run time based on @dlamblin idea. Supports seconds and years. Have not managed to test it fully yet so expect bugs but let me know if find any.
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。没有设法完全测试它所以期待错误,但让我知道如果找到任何。
Repository link: https://bitbucket.org/nevity/cronner
存储库链接:https://bitbucket.org/nevity/cronner
#7
2
Thanks for posting this code. It definitely helped me out, even 6 years later.
感谢您发布此代码。它确实帮助了我,即使是6年之后。
Trying to implement I found a small bug.
试图实现我发现了一个小bug。
date('i G j n w', $time)
returns a 0 padded integer for the minutes.
date('i G j n w',$ time)返回分钟的0填充整数。
Later in the code, it does a modulus on that 0 padded integer. PHP doesn't seem to handle this as expected.
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
As you can see, 08 % 5
returns 0, whereas 8 % 5
returns the expected 3. I couldn't find a non padded option for the date command. I tried fiddling with the {$time[$k]} % $1 === 0
line (like changing {$time[$k]}
to ({$time[$k]}+0)
, but couldn't get it to drop the 0 padding during the modulus.
如您所见,08%5返回0,而8%5返回预期的3.我找不到date命令的非填充选项。我尝试摆弄{$ time [$ k]}%$ 1 === 0行(比如将{$ time [$ k]}更改为({$ time [$ k]} + 0),但无法获得它在模数期间掉落0填充。
So, I ended up just changing the original value returned by the date function and removed the 0 by running $time[0] = $time[0] + 0;
.
所以,我最后只是改变了日期函数返回的原始值,并通过运行$ time [0] = $ time [0] + 0;删除了0。
Here is my test.
这是我的考试。
<?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";
}
}
}
?>
#8
2
My answer is not unique. Just a replica of @BlaM answer written in java because PHP's date and time is a bit different from Java.
我的回答并不是唯一的。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。
This program assumes that the CRON expression is simple. It can only contain digits or *.
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.
Code:
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;
}
}