根据周数和年份获取特定日期

Question

我从外部来源获取年份和 'week-of-the-year'，我无法更改。我想获得那个具体周的星期一和星期日的日期。现在我有下面的代码，但它看起来很复杂。

<?php

// simulate the server data
$year = date('Y');
$week = 19;

$weekNow = (int) date('W');
$weekDiff = $week - $weekNow; // can be positive or negative
$daysDiff = $weekDiff * 30;

$dayInDesiredWeek = (int) date('N', strtotime($daysDiff . ' days'));  
if($dayInDesiredWeek !== 1) {
    $daysDiff += $daysDiff > 0 ? -($dayInDesiredWeek - 1) : $dayInDesiredWeek - 1;
}
$monday = date('Y-m-d', strtotime($daysDiff . ' days'));
$sunday = date('Y-m-d', strtotime($monday . '7 days'));

我可以使用任何其他（更简单的）方法来获得相同的结果吗？提前谢谢你。

Answer 1

所以，这是我迄今为止看到的最简单的解决方案，当时我正在阅读 PHPDocs of DateTime Object。

所以步骤是：

创建new DateTime()对象
运行它是带有参数的 setISODate 方法：($year, $weak, $offsetFromMonday)
然后运行 format 方法并得到所需的日期

所以，一起：

$monday = ($dt = new DateTime)->setISODate($year, $weak, 1)->format('Y-m-d');
$sunday = $dt->setISODate($year, $weak, 7)->format('Y-m-d');

Answer 2

 $dt = date("Y-m-d", strtotime("2017W01"));

使用Carbon

 $dt = new Carbon($dt);

 var_dump($dt->startOfWeek());
 var_dump($dt->endOfWeek());

根据周数和年份获取特定日期

Get specific day's date based on the week number and the year

php

date

php-7.1