Question

我有一个php日历，它在页面上完美地显示了所有日期，但我需要将青少年日期显示为2位数字（01,02,03等），但它们没有显示出来 - 我怎么能这样做，因为它对结束链接很重要，即href =“events.php？date = 20131201”而不是href =“events.php？date = 2013121”。

这是我的完整代码，如果这会有所帮助：

<?php
            $currDay = date("j");
            $today = date("d"); // Current day
            $month = date("m"); // Current month
            $displaymonth = date("F");
            $year = date("Y"); // Current year
            $days = cal_days_in_month(CAL_GREGORIAN,$month,$year); // Days in current month

            $lastmonth = date("t", mktime(0,0,0,$month-1,1,$year)); // Days in previous month

            $start = date("N", mktime(0,0,0,$month,1,$year)); // Starting day of current month
            $finish = date("N", mktime(0,0,0,$month,$days,$year)); // Finishing day of  current month
            $laststart = $start - 1; // Days of previous month in calander

            $counter = 1;
            $nextMonthCounter = 1;

            echo '

                <div class="calendarHeader"><strong>'.$displaymonth.' '.$year.'</strong></div>
                <div class="calendarDates">
                    <div class="calendarCell"><b>M</b></div>
                    <div class="calendarCell"><b>T</b></div>
                    <div class="calendarCell"><b>W</b></div>
                    <div class="calendarCell"><b>T</b></div>
                    <div class="calendarCell"><b>F</b></div>
                    <div class="calendarCell"><b>S</b></div>
                    <div class="calendarCell"><b>S</b></div>

                    <div class="clearLeft"></div>
                </div>
                <div class="calendar">
                ';

                if($start > 5){ $rows = 6; }else {$rows = 5; }

                for($i = 1; $i <= $rows; $i++){
                    for($x = 1; $x <= 7; $x++){             

                        if((int)$currDay > (int)$date) {
                           $class  ='calendarBlank';
                        }

                        if(($counter - $start) < 0){
                            $date = (($lastmonth - $laststart) + $counter);
                            $class = 'calendarBlank';
                        }else if(($counter - $start) >= $days){
                            $date = ($nextMonthCounter);
                            $nextMonthCounter++;

                            $class = 'calendarBlank';

                        }else if($counter <= $today){
                            $date = ($counter - $start + 1);
                            $class = 'calendarBlank';
                        }else{
                            $date = ($counter - $start + 1);
                            if($today == $counter - $start + 1){
                                $class = 'calendarToday';
                            }
                        }                               

                        echo '<a href="events.php?type='.$type.'&from='.$year,$month,$date.'&var='.$var.'" target="_parent"><div class="calendarCell '.$class.'">'.$date.'</div></a>';

                        $counter++;
                        $class = '';
                    }
                }

                echo '
                <div class="clearLeft"></div>
                </div>
            ';
            ?>

Answer 1

您可以使用str_pad()（假设我正确阅读您的代码，$date是代表您日期的日期部分的变量）

$date = str_pad($date , 2, "0", STR_PAD_LEFT);

您也可以使用sprintf()

$date = sprintf("%02d", $date );

Answer 2

$date = sprintf("%02d", $date);   // e.g., 05

$date = sprintf("%4d%02d%02d", $year, $month, $day);  // e.g., 20130704

Answer 3

而不是'&from='.$year,$month,$date使用

'&from='.sprintf("%04d%02d%02d", $year, $month, $date)

或

'&from='.$year.str_pad($month, 2, "0", STR_PAD_LEFT).str_pad($date, 2, "0", STR_PAD_LEFT)

或

'&from='.date("Ymd", mktime(0,0,0,$month,$date,$year))

Answer 4

对于这种类型的日期操作，我强烈建议使用DateTime类：

$today = new DateTime("2013-01-01");
echo $today->format("Ymd");
// 20130101

PHP日期显示青少年前数字的2位数

4 个答案: