Question

创建下面的制作时间以显示每两周一次的日期

<?php
  $date1 = "07/05/2013";
  $date2 = date('M j, Y', strtotime($date1 . " + 14 day"));
  $date3 = date('M j, Y', strtotime($date2 . " + 14 day"));
  $date4 = date('M j, Y', strtotime($date3 . " + 14 day"));
  $date5 = date('M j, Y', strtotime($date4 . " + 14 day"));
  $date6 = date('M j, Y', strtotime($date5 . " + 14 day"));
  $date7 = date('M j, Y', strtotime($date6 . " + 14 day"));
  $date8 = date('M j, Y', strtotime($date7 . " + 14 day"));
  $date9 = date('M j, Y', strtotime($date8 . " + 14 day"));
  $date10 = date('M j, Y', strtotime($date9 . " + 14 day"));
  $date11 = date('M j, Y', strtotime($date10 . " + 14 day"));
  $date12 = date('M j, Y', strtotime($date11 . " + 14 day"));
  $date13 = date('M j, Y', strtotime($date12 . " + 14 day"));
  $date14 = date('M j, Y', strtotime($date13 . " + 14 day"));
  $date15 = date('M j, Y', strtotime($date14 . " + 14 day"));
  $date16 = date('M j, Y', strtotime($date15 . " + 14 day"));
  $date17 = date('M j, Y', strtotime($date16 . " + 14 day"));
  $date18 = date('M j, Y', strtotime($date17 . " + 14 day"));
?>

如何按月将它组合在一起？假设我想知道8月或12月有多少日期。如果我想知道到今年年底有多少日期？非常感谢助手。

Answer 1

试试这个：

<?php
// initiate months
$month_arr = Array();
for ($i=1; $i<=12; $i++){
    // no. of dates
    $month_arr[$i] = 0;
}
$date_arr = Array();
$date_start = "07/05/2013";
$date_arr[] = date('M j, Y', strtotime($date_start));

for ($i=1; $i<=17; $i++){
    $date_temp = date('M j, Y', strtotime($date_arr[$i-1] . " + 14 day"));
    $month = date('n', strtotime($date_temp));

    $month_arr[$month] += 1;
    $date_arr[] = $date_temp;

  }

 foreach ($month_arr as $k => $v){
    echo "<BR>Month: " . $k . ", No. of dates: " . $v;
 }

// all dates
echo "<BR>All dates<BR>";
var_dump ($date_arr);
?>

您可以扩展此逻辑，将实际日期按月分组，而不是仅获取一个月中的日期计数。这是解决方案：

<?php
// initiate months
$month_arr = Array(
            'January' => Array('num_dates'=>0, 'dates'=>Array()) , 
            'February' => Array('num_dates'=>0, 'dates'=>Array()), 
            'March' => Array('num_dates'=>0, 'dates'=>Array()), 
            'April' => Array('num_dates'=>0, 'dates'=>Array()), 
            'May' => Array('num_dates'=>0, 'dates'=>Array()), 
            'June' => Array('num_dates'=>0, 'dates'=>Array()), 
            'July' => Array('num_dates'=>0, 'dates'=>Array()), 
            'August' => Array('num_dates'=>0, 'dates'=>Array()), 
            'September' => Array('num_dates'=>0, 'dates'=>Array()), 
            'October' => Array('num_dates'=>0, 'dates'=>Array()), 
            'November' => Array('num_dates'=>0, 'dates'=>Array()), 
            'December' => Array('num_dates'=>0, 'dates'=>Array())
        );
$date_arr = Array();

$date_start = "07/05/2013";
$date_arr[] = date('M j, Y', strtotime($date_start));

for ($i=1; $i<=17; $i++){
    $date_temp = date('M j, Y', strtotime($date_arr[$i-1] . " + 14 day"));
    $month = date('F', strtotime($date_temp));

    $month_arr[$month]['dates'][] = $date_temp;
    $month_arr[$month]['num_dates'] += 1;
    $date_arr[] = $date_temp;
}

foreach ($month_arr as $k => $v){
    if (!empty($v)){
        if ($v['num_dates'] != 0){
            echo "<BR><BR>Month: " . $k;
            echo "<BR>No. of dates: " . $v['num_dates'];
            foreach ($v['dates'] as $k1=>$v1){
                echo "<BR>" . $v1;
            }
        }
     }
}
?>

此时，$ month_arr应该拥有您需要的一切。

Answer 2

我不确定你对几个月的分组意味着什么但是这将使用数组来声明所有日期，您可以更改$ limit变量来增加或减少日期量。更改$ date以调整您的原始日期 $ left是数组中剩余的天数（它在第一天被触发后触发，可以在最后一天更改为触发）和金额是8月份的日期数量

<?php
$limit = 17;
$dates = array("07-05-2013");
for ($i=1; $i<=$limit; $i++){
    $dates[$i]=date('d-m-Y', strtotime($dates[$i-1] . "+ 14 days"));
}
foreach (array_keys($dates) as $key){ 
    $value = date('m', strtotime($dates[$key])); 
    if ($value == "08"){
        $amount = $amount + 1;  
    }
    if ($amount == 1){
        $left = $limit-$key;
    }
}
print_r ($dates);
echo "<br>";
echo $amount . "<br>" . $left;
?>

Answer 3

只需添加if语句并在所有12个月内增加...如果月份相同

 $n=1;
 $month=array();//dumpyour dates in  an array
 $date=array()

 while($k<30){
 if  (date('n', strtotime($date[$k]))==date('n', strtotime($date[$k-1])){
 echo $date[$k++];
 $m = date('n', strtotime($date[$k]));
 $month[$m]=++$n;
 }
 else{
 .....

 }
 }

Answer 4

当我继续重复SO时，PHP的DateTime classes使大多数日期时间操作变得微不足道。

您的问题并非100％明确，但是根据您的评论，您似乎需要一系列时间戳，按月划分14天。我不知道你是否认为这些月份可能会跨越年度债券，但我在答案中考虑了这一点。

下面的代码应该做你想要的： -

$start = \DateTime::createFromFormat('d/m/Y', '07/05/2013');
$interval = new \DateInterval('P14D');
$periods = new \DatePeriod($start, $interval, 26);

$dates = array();

foreach($periods as $day){
    /** @var \DateTime $day */
    $dates[$day->format('Y')][$day->format('M')][] = $day->getTimestamp();
}

var_dump($dates);

输出： -

array (size=2)
  2013 => 
    array (size=8)
      'May' => 
        array (size=2)
          0 => int 1367936286
          1 => int 1369145886
      'Jun' => 
        array (size=2)
          0 => int 1370355486
          1 => int 1371565086
      'Jul' => 
        array (size=3)
          0 => int 1372774686
          1 => int 1373984286
          2 => int 1375193886
      'Aug' => 
        array (size=2)
          0 => int 1376403486
          1 => int 1377613086
      'Sep' => 
        array (size=2)
          0 => int 1378822686
          1 => int 1380032286
      'Oct' => 
        array (size=2)
          0 => int 1381241886
          1 => int 1382451486
      'Nov' => 
        array (size=2)
          0 => int 1383664686
          1 => int 1384874286
      'Dec' => 
        array (size=3)
          0 => int 1386083886
          1 => int 1387293486
          2 => int 1388503086
  2014 => 
    array (size=5)
      'Jan' => 
        array (size=2)
          0 => int 1389712686
          1 => int 1390922286
      'Feb' => 
        array (size=2)
          0 => int 1392131886
          1 => int 1393341486
      'Mar' => 
        array (size=2)
          0 => int 1394551086
          1 => int 1395760686
      'Apr' => 
        array (size=2)
          0 => int 1396966686
          1 => int 1398176286
      'May' => 
        array (size=1)
          0 => int 1399385886

我建议您使用DateTime对象数组代替时间戳数组，然后根据需要操作每个对象。在这种情况下，代码如下所示： -

$start = \DateTime::createFromFormat('d/m/Y', '07/05/2013');
$interval = new \DateInterval('P14D');
$periods = new \DatePeriod($start, $interval, 26);

$dates = array();

foreach($periods as $day){
    /** @var \DateTime $day */
    $dates[$day->format('Y')][$day->format('M')][] = $day;
}

在php中将日期分组为几个月

4 个答案: