PHP：平日给定一个月的数字 [英] PHP: getting weekdays numbers of a given month

问题描述

给定一个月和一个工作日，我需要建立一个可以检索星期一，星期二，星期三，星期四和星期五的日期号的功能。这个功能应该是在2012年9月的所有星期一，3月10日，17日和17日。 24

请注意，对我来说，平日1号是星期一，星期二，3是星期三，4星期四和5星期五。

到目前为止，我已经做了：获得今天的日期的一周的第一天（我发布以下功能）。但我不知道如何从这里简单地跟随，我已经有很多小时了，我怀疑有更好的方法来做到这一点。你可以告诉我怎么样？

  function getFirstDayOfWeek（$ date）{
 $ getdate = getdate（$ date） ; 
 
 //我们星期一多少天？ 
 switch（$ getdate ['wday']）{
 case 0：//我们在星期天
 $ days = 6; 
 break; 
 
默认值：//任何其他日子
 $ days = $ getdate ['wday']  -  1; 
 break; 
} 
 
 $ seconds = $ days * 24 * 60 * 60; 
 $ monday = date（$ getdate [0]） -  $ seconds; 
 
 return $ monday; 
} 
  

感谢一吨

解决方案

不是很聪明，但适用于你：

  // sept。 2012 
 $ month = 9; 
 
 //通过月份循环
（$ i = 1; $ i <= 31; $ i ++）{
 
 //给定月份时间戳
 $ timestamp = mktime（0，0，0，$ month，$ i，2012）; 
 
 //确保我们没有去下个月
 if（date（n，$ timestamp）== $ month）{
 
 //循环中的当前日期
 $ day = date（N，$ timestamp）; 
 
 //如果这是介于1到5之间，平日，1 =星期一，5 =星期五
 if（$ day == 1 OR $ day< = 5）{
 
 //现在写下来
 $ days [$ day] [] = date（j，$ timestamp）; 
} 
} 
} 
 
 //查看是否可以工作:) 
 print_r（$ days）; 
  

Given a Month and a weekday, I need to build a function that can retrieve the day number of all Mondays, Tuesdays, Wednesdays, Thursdays and Fridays.

Let's say I give the function this month, September 2012 and weekday number 1. The function should retrieve all the Mondays in September 2012 which are: 3, 10, 17 and 24

Please note that to me weekday number 1 is Monday, number 2 Tuesday, 3 is Wednesday, 4 Thursday and 5 Friday.

So far I've have done: getting the first day of the week given today's date (I post the function below). But I don't know how to follow from here in a simple way, I've been many hours on it and I suspect there's a better way to do it. Can you please show me how?

  function getFirstDayOfWeek($date) {
    $getdate = getdate($date);

    // How many days ahead monday are we?
    switch ( $getdate['wday'] ) {
        case 0: // we are on sunday
            $days = 6;
            break;

        default: // any other day
            $days = $getdate['wday']-1;
            break;
    }

    $seconds = $days*24*60*60;
    $monday = date($getdate[0])-$seconds;

    return $monday;
}

Thanks a ton

解决方案

Not very smart, but would works for you:

// sept. 2012
$month = 9;

// loop through month days
for ($i = 1; $i <= 31; $i++) {

    // given month timestamp
    $timestamp = mktime(0, 0, 0, $month, $i, 2012);

    // to be sure we have not gone to the next month
    if (date("n", $timestamp) == $month) {

        // current day in the loop
        $day = date("N", $timestamp);

        // if this is between 1 to 5, weekdays, 1 = Monday, 5 = Friday
        if ($day == 1 OR $day <= 5) {

            // write it down now
            $days[$day][] = date("j", $timestamp);
        }
    }
}

// to see if it works :)
print_r($days);

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