Question

我有一个函数可以返回数组中两个日期之间的所有日期，但我需要在该数组中排除星期日。

public function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) { 
    $dates = array();
    $current = strtotime($first);
    $last = strtotime($last);
    while( $current <= $last ) { 
        $dates[] = date($format, $current);
        $current = strtotime($step, $current);
    }
    return $dates;
}

排除星期日后，我有一张桌子，我将存储一些日期，我也需要从阵列中排除这些日期。

喜欢，如果我输入日期范围为01-05-2012（DD-MM-YYYY）到10-05-2012， 06-05-2012将是周日＆amp;日期01-05-2012＆amp; 08-05-2012将在我上面提到的表格中，最终的出局应该是，

如何在PHP中执行此操作？我尝试了一些，但无法找到正确的方法。

Answer 1

星期日部分：

public function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) { 
    $dates = array();
    $current = strtotime($first);
    $last = strtotime($last);
    while( $current <= $last ) { 
        if (date("D", $current) != "Sun")
            $dates[] = date($format, $current);
        $current = strtotime($step, $current);
    }
    return $dates;
}

假期部分：

首先，您需要将日期加载到某种数组中，然后在每个日期循环遍历数组并检查它们是否匹配。

Answer 2

我找到了问题的答案，感谢帮助我的人。

public function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) { 
    $dates = array();
    $current = strtotime($first);
    $last = strtotime($last);
    while( $current <= $last ) {
        $sql = "SELECT * FROM ost_holidays where holiday_date='".date('Y-m-d', $current)."' LIMIT 1";
        $sql = db_query($sql);
        $sql = db_fetch_array($sql);
        if($sql['holiday_date'] != date('Y-m-d',$current))
            if (date('w', $current) != 0)
            $dates[] = date($format, $current);
            $current = strtotime($step, $current);
    }
    return $dates;
}

以上代码用于删除假期＆amp;在给定范围内的星期日。

Answer 3

我在Jquery

中做了同样的上述方法

//Convert dates into desired formatt
 function convertDates(str) {
     var date = new Date(str),
         mnth = ("0" + (date.getMonth() + 1)).slice(-2),
         day = ("0" + date.getDate()).slice(-2);
     return [date.getFullYear(), mnth, day].join("-");
 }

 // Returns an array of dates between the two dates
 var getDates = function(startDate, endDate, holidays) {
     var dates = [],
         currentDate = startDate,
         addDays = function(days) {
             var date = new Date(this.valueOf());
             date.setDate(date.getDate() + days);
             return date;
         };
     while (currentDate <= endDate) {
         dates.push(currentDate);
         currentDate = addDays.call(currentDate, 1);
     }
     return dates;
 };
 //Indise Some Function
 var datesTemp = [];
 var dates = getDates(new Date(prodDet.details.date1), new Date(prodDet.details.date2));
 dates.forEach(function(date) {
     if (date.getDay() != 0) {
         datesTemp.push(convertDates(date));
     }
 });
 datesTemp.forEach(function(date) {
     for (var j = 0; j < prodDet.holidays.length; j++) {
         if ((prodDet.holidays[j] != date)) {
             ideal.idates.push(date);
         }
     }
 });
 console.log(ideal.idates);
 //Function Ends Here

日期范围数组不包括星期日＆amp;用PHP假期

3 个答案: