PHP获得一个月的周数

Question

所以我有一个脚本可以返回特定月份和年份的周数.我如何从该月开始特定日期并确定它是否属于该月1,2,3,4或5周的一部分？

Answer 1

我曾经尝试过最令人沮丧的事情 - 但在这里!

<?php

    /**
     * Returns the amount of weeks into the month a date is
     * @param $date a YYYY-MM-DD formatted date
     * @param $rollover The day on which the week rolls over
     */
    function getWeeks($date, $rollover)
    {
        $cut = substr($date, 0, 8);
        $daylen = 86400;

        $timestamp = strtotime($date);
        $first = strtotime($cut . "00");
        $elapsed = ($timestamp - $first) / $daylen;

        $weeks = 1;

        for ($i = 1; $i <= $elapsed; $i++)
        {
            $dayfind = $cut . (strlen($i) < 2 ? '0' . $i : $i);
            $daytimestamp = strtotime($dayfind);

            $day = strtolower(date("l", $daytimestamp));

            if($day == strtolower($rollover))  $weeks ++;
        }

        return $weeks;
    }


    //
    echo getWeeks("2011-06-11", "sunday"); //outputs 2, for the second week of the month
?>

Answer 2

编辑:对"单行"这么多 - 需要变量以避免使用条件重新计算.当我在它时,抛出一个默认参数.

function weekOfMonth($when = null) {
    if ($when === null) $when = time();
    $week = date('W', $when); // note that ISO weeks start on Monday
    $firstWeekOfMonth = date('W', strtotime(date('Y-m-01', $when)));
    return 1 + ($week < $firstWeekOfMonth ? $week : $week - $firstWeekOfMonth);
}

请注意,weekOfMonth(strtotime('Oct 31, 2011'));将返回6; 与OP的预期相反,一些罕见的月份有6个星期.2017年1月是另一个月,有6个ISO周 - 从周一开始,周日第一个月是在去年的一周.

对于starshine531,要返回0该月的索引周,请更改return 1 +为return 0 +或return (int).

对于Justin Stayton来说,从星期日而不是星期一开始的几个星期我将使用,strftime('%U'而不是date('W'如下:

function weekOfMonth($when = null) {
    if ($when === null) $when = time();
    $week = strftime('%U', $when); // weeks start on Sunday
    $firstWeekOfMonth = strftime('%U', strtotime(date('Y-m-01', $when)));
    return 1 + ($week < $firstWeekOfMonth ? $week : $week - $firstWeekOfMonth);
}

对于这个版本,2017-04-30现在是4月的第6周,而2017-01-31现在是第5周.

Answer 3

public function getWeeks($timestamp)
{
    $maxday    = date("t",$timestamp);
    $thismonth = getdate($timestamp);
    $timeStamp = mktime(0,0,0,$thismonth['mon'],1,$thismonth['year']);    //Create time stamp of the first day from the give date.
    $startday  = date('w',$timeStamp);    //get first day of the given month
    $day = $thismonth['mday'];
    $weeks = 0;
    $week_num = 0;

    for ($i=0; $i<($maxday+$startday); $i++) {
        if(($i % 7) == 0){
            $weeks++;
        }
        if($day == ($i - $startday + 1)){
            $week_num = $weeks;
        }
      }     
    return $week_num;
}

大家好,我一整天都在努力想出这个代码,我终于想通了,所以我想我会和大家分享.

你需要做的只是在函数中加上一个时间戳,它会将周数返回给你.

谢谢

Answer 4

这种方法存在问题.如果通过日期(假设2012/01/01是星期日)和"$ rollover"日是"星期日",那么这个函数将返回2.其中它实际上是第1周.我想我已经修复了以下功能.请添加评论以使其更好.

function getWeeks($date, $rollover)
{
    $cut        = substr($date, 0, 8);
    $daylen     = 86400;
    $timestamp  = strtotime($date);
    $first      = strtotime($cut . "01");   
    $elapsed    = (($timestamp - $first) / $daylen)+1;
    $i          = 1;
    $weeks      = 0;
    for($i==1; $i<=$elapsed; $i++)
    {
        $dayfind        = $cut . (strlen($i) < 2 ? '0' . $i : $i);
        $daytimestamp   = strtotime($dayfind);
        $day            = strtolower(date("l", $daytimestamp));
        if($day == strtolower($rollover))
        {
            $weeks++;  
        }
    } 
    if($weeks==0)
    {
        $weeks++; 
    }
    return $weeks;  
}

Answer 5

这是一个基于sberry数学解决方案但是使用PHP DateTime类的解决方案.

function week_of_month($date) {
    $first_of_month = new DateObject($date->format('Y/m/1'));
    $day_of_first = $first_of_month->format('N');
    $day_of_month = $date->format('j');
    return floor(($day_of_first + $day_of_month - 1) / 7) + 1;
}