在PHP中执行与日期时间相关的操作

对于大多数datetime操作,我通常会转换为Unixtime并对Unixtime整数执行加减等操作,但您可能需要查看Zend framework Zend_Date类。

这有很多您描述的功能。尽管Zend被宣传为一个“框架”,但它作为一个可以从中挑选元素的类库工作得非常好。我们通常将其包含在项目中,然后在需要的时候将其提取出来。

像“+1个月”或“-3天”这样的东西有时无法提供您期望的输出。

要添加日期,可以使用以下方法 日期时间::添加 ( 向DateTime对象添加日、月、年、小时、分钟和秒的数量 ),从PHP5.3.0开始提供。

要找出两个日期之间的差异,有 DateTime::diff

PHP5+的DateTime对象很有用,因为它是闰时和 夏令时意识到,但它需要一些扩展来真正实现 解决问题。我写了以下内容来解决类似的问题。

$tryme = new Extended_DateTime('2007-8-26');
$newer = new Extended_DateTime('2008-9-1');

print 'Weekdays From '.$tryme->format('Y-m-d').' To '.$newer->format('Y-m-d').': '.$tryme -> find_WeekdaysFromThisTo($newer) ."\n";
/*  Output:  Weekdays From 2007-08-26 To 2008-09-01: 265  */
print 'All Days From '.$tryme->format('Y-m-d').' To '.$newer->format('Y-m-d').': '.$tryme -> find_AllDaysFromThisTo($newer) ."\n";
/*  Output:  All Days From 2007-08-26 To 2008-09-01: 371   */
$timefrom = $tryme->find_TimeFromThisTo($newer);
print 'Between '.$tryme->format('Y-m-d').' and '.$newer->format('Y-m-d').' there are '.
      $timefrom['years'].' years, '.$timefrom['months'].' months, and '.$timefrom['days'].
      ' days.'."\n";
/*  Output: Between 2007-08-26 and 2008-09-01 there are 1 years, 0 months, and 5 days. */

class Extended_DateTime extends DateTime {

    public function find_TimeFromThisTo($newer) {
        $timefrom = array('years'=>0,'months'=>0,'days'=>0);

        // Clone because we're using modify(), which will destroy the object that was passed in by reference
        $testnewer = clone $newer;

        $timefrom['years'] = $this->find_YearsFromThisTo($testnewer);
        $mod = '-'.$timefrom['years'].' years';
        $testnewer -> modify($mod);

        $timefrom['months'] = $this->find_MonthsFromThisTo($testnewer);
        $mod = '-'.$timefrom['months'].' months';
        $testnewer -> modify($mod);

        $timefrom['days'] = $this->find_AllDaysFromThisTo($testnewer);
        return $timefrom;
    } // end function find_TimeFromThisTo


    public function find_YearsFromThisTo($newer) {
        /*
        If the passed is:
        not an object, not of class DateTime or one of its children,
        or not larger (after) $this
        return false
        */
        if (!is_object($newer) || !($newer instanceof DateTime) || $newer->format('U') < $this->format('U'))
            return FALSE;
        $count = 0;

        // Clone because we're using modify(), which will destroy the object that was passed in by reference
        $testnewer = clone $newer;

        $testnewer -> modify ('-1 year');
        while ( $this->format('U') < $testnewer->format('U')) {
            $count ++;
            $testnewer -> modify ('-1 year');
        }
        return $count;
    } // end function find_YearsFromThisTo


    public function find_MonthsFromThisTo($newer) {
        /*
        If the passed is:
        not an object, not of class DateTime or one of its children,
        or not larger (after) $this
        return false
        */
        if (!is_object($newer) || !($newer instanceof DateTime) || $newer->format('U') < $this->format('U'))
            return FALSE;

        $count = 0;
        // Clone because we're using modify(), which will destroy the object that was passed in by reference
        $testnewer = clone $newer;
        $testnewer -> modify ('-1 month');

        while ( $this->format('U') < $testnewer->format('U')) {
            $count ++;
            $testnewer -> modify ('-1 month');
        }
        return $count;
    } // end function find_MonthsFromThisTo


    public function find_AllDaysFromThisTo($newer) {
        /*
        If the passed is:
        not an object, not of class DateTime or one of its children,
        or not larger (after) $this
        return false
        */
        if (!is_object($newer) || !($newer instanceof DateTime) || $newer->format('U') < $this->format('U'))
            return FALSE;

        $count = 0;
        // Clone because we're using modify(), which will destroy the object that was passed in by reference
        $testnewer = clone $newer;
        $testnewer -> modify ('-1 day');

        while ( $this->format('U') < $testnewer->format('U')) {
            $count ++;
            $testnewer -> modify ('-1 day');
        }
        return $count;
    } // end function find_AllDaysFromThisTo


    public function find_WeekdaysFromThisTo($newer) {
        /*
        If the passed is:
        not an object, not of class DateTime or one of its children,
        or not larger (after) $this
        return false
        */
        if (!is_object($newer) || !($newer instanceof DateTime) || $newer->format('U') < $this->format('U'))
            return FALSE;

        $count = 0;

        // Clone because we're using modify(), which will destroy the object that was passed in by reference
        $testnewer = clone $newer;
        $testnewer -> modify ('-1 day');

        while ( $this->format('U') < $testnewer->format('U')) {
            // If the calculated day is not Sunday or Saturday, count this day
            if ($testnewer->format('w') != '0' && $testnewer->format('w') != '6')
                $count ++;
            $testnewer -> modify ('-1 day');
        }
        return $count;
    } // end function find_WeekdaysFromThisTo

    public function set_Day($newday) {
        if (is_int($newday) && $newday > 0 && $newday < 32 && checkdate($this->format('m'),$newday,$this->format('Y')))
            $this->setDate($this->format('Y'),$this->format('m'),$newday);
    } // end function set_Day


    public function set_Month($newmonth) {
        if (is_int($newmonth) && $newmonth > 0 && $newmonth < 13)
            $this->setDate($this->format('Y'),$newmonth,$this->format('d'));
    } // end function set_Month


    public function set_Year($newyear) {
        if (is_int($newyear) && $newyear > 0)
            $this->setDate($newyear,$this->format('m'),$this->format('d'));
    } // end function set_Year
} // end class Extended_DateTime

PEAR::Date 看起来它可能有一些有用的功能。

PEAR::Calendar 也可能有用。

最简单的方法是使用时间戳,表示自2008年1月1日以来的秒数。使用时间戳类型,您可以执行以下操作。。。

now = time();
tomorrow = now + 24 * 60 * 60; // 24 hours * 60 minutes * 60 seconds

查看文档以了解更多信息 time() , date() 和 mktime()

您可以使用以下选项的组合: strtotime , mktime 和 date 做算术题

下面是一个使用组合来进行一些运算的示例 http://rushi.wordpress.com/2008/04/13/php-print-out-age-of-date-in-words/

if ($timestamp_diff < (60*60*24*7)) {
   echo floor($timestamp_diff/60/60/24)." Days";
} elseif ($timestamp_diff > (60*60*24*7*4)) {
   echo floor($timestamp_diff/60/60/24/7)." Weeks";
} else {
   $total_months = $months = floor($timestamp_diff/60/60/24/30);
   if($months >= 12) {
      $months = ($total_months % 12);
      $years&nbsp; = ($total_months - $months)/12;
      echo $years . " Years ";
   }
   if($months > 0)
      echo $months . " Months";
}
?>

@如石我个人不喜欢strotime。。我不知道为什么,但我今天早上发现,将这样的字符串“2008-09-11 9:5 AM”传递给strottime会返回一个错误的。。。

我认为您提供的代码不能解决示例问题“给定两个日期,两个日期之间有多少个工作日?”?在SQL或$PETLLAN中实现,我没有考虑是否有公共假日列表…

您可以获得两个日期之间的天数,如下所示:

$days = (strtotime("2008-09-10") - strtotime("2008-09-12")) / (60 * 60 * 24);

您可以将函数设置为类似的形式(我的工作计算机中没有安装php,所以我不能保证语法100%正确)

function isWorkDay($date)
{
 // check if workday and return true if so
}

function numberOfWorkDays($startdate, $enddate)
{
  $workdays = 0;
  $tmp = strtotime($startdate);
  $end = strtotime($enddate);
  while($tmp <= $end)
  {
    if ( isWorkDay( date("Y-m-d",$tmp) ) ) $workdays++;
    $tmp += 60*60*24;
  }
  return $workdays;
}

如果您不喜欢strotime,并且您总是使用相同格式的日期,您可以使用explode函数,如

list($year, $month, day) = explode("-", $date);

我强烈建议使用PHP5.2 DateTime objects ,而不是在进行日期计算时使用UNIX时间戳。当您使用返回UNIX时间戳的PHP日期函数时,可以使用的范围非常有限(例如,1970年以前没有)。

如果你看一看 http://php.net/date ,您将找到一些使用 mktime() 执行操作。

mktime() 详情如下:

$tomorrow = date("Y-m-d", mktime(0, 0, 0, date("m"), date("d") + 1, date("Y")));

mktime() 使生活变得更加轻松,并使您不必执行嵌套的if语句和其他此类麻烦的编码。

要获取工作日/假期,请参阅postgresql CTE ftw http://osssmb.wordpress.com/2009/12/02/business-days-working-days-sql-for-postgres-2/