如何在php中获取一系列日期？

您上面的代码将使用一个额外的字符-您在这一行遗漏了一个

$

：

<?php

$start_date = '2012-01-01';
$end_date = '2012-12-31';
$total_days = round(abs(strtotime($end_date) - strtotime($start_date)) / 86400, 0) + 1;

if ($end_date >= $start_date)
{
  for ($day = 0; $day < $total_days; $day++)
  {
    echo "<br />" . date("Y-m-d", strtotime("{$start_date} + {$day} days"));
                                 //     You missed the $ here ^
  }
}

echo“
”。日期（“Y-m-d”，STROTIME（{$start_date}+{$day}天”）；
//^这个不见了

我将使用该类（以及and）来实现以下目的：

echo "<br />" . date("Y-m-d", strtotime("{$start_date} + {$day} days"));
//                                                        ^ This was missing

---

@艾莉森·克里斯蒂安森：那你应该接受它作为答案。是的，我知道，它在最初的五分钟里不允许我这样做，我接着做了其他事情。谢谢你提醒我。

<?php

$start_date = '2012-01-01';
$end_date = '2012-12-31';
$total_days = round(abs(strtotime($end_date) - strtotime($start_date)) / 86400, 0) + 1;

if ($end_date >= $start_date)
{
  for ($day = 0; $day < $total_days; $day++)
  {
    echo "<br />" . date("Y-m-d", strtotime("{$start_date} + {$day} days"));
                                 //     You missed the $ here ^
  }
}

echo "<br />" . date("Y-m-d", strtotime("{$start_date} + {$day} days"));
//                                                        ^ This was missing

$start_date = '2012-01-01';
$end_date = '2012-12-31';

$start    = new DateTime($start_date);
$end      = new DateTime($end_date);
$interval = new DateInterval('P1D'); // 1 day interval
$period   = new DatePeriod($start, $interval, $end);

foreach ($period as $day) {
    // Do stuff with each $day...
    echo $day->format('Y-m-d'), "\n";
}