Question

我试图从所选日期时间中扣除5小时30分钟，但它无法正常工作

我的PHP代码

 $ambulanceDate = date('Y-m-d H:i:s', ($_POST['ambuStartDate'] . " " . $_POST['ambuStartTime']));
                $date1 = str_replace('-', '/', $ambulanceDate);
                $ambulanceUpdatedDate = date('m-d-Y',strtotime($date1 . "-5 hours 30 minutes"));
                $ambulanceUpdatedDate = new DateTime($ambulanceUpdatedDate);

Answer 1

为什么你会做$date1 = str_replace('-', '/', $ambulanceDate);？你不能做date('Y/m/d H:i:s' ...

这应该这样做：

//Change 'Y-m-d H:i:s' with the format of your ambuStartDate, and ambuStartTime.
$ambulanceDate = DateTime::createFromFormat('Y-m-d H:i:s', ($_POST['ambuStartDate'] . " " . $_POST['ambuStartTime']));
$ambulanceUpdatedDate = $ambulanceDate->sub(new DateInterval('P5DT30M'));
//Format it how you like:
$formattedUpdatedDate = $ambulanceUpdatedDate->format('m-d-Y h:iA');

请参阅https://eval.in/772578

Answer 2

modify reference使用DateTime功能：

<?php
$ambulanceDate = date('Y-m-d H:i:s', ($_POST['ambuStartDate'] . " " . $_POST['ambuStartTime']));

$date_time = new DateTime($ambulanceDate);
$date_time->modify('-5 hours');
$date_time->modify('-30 minutes');

$ambulanceUpdatedTime = $date_time->format('m-d-Y H:i:s');

echo $ambulanceDate . PHP_EOL;
echo '-5h 30mins:' . PHP_EOL;
echo $ambulanceUpdatedTime;

输出：

2017-04-11 05:22:07
-5h 30mins:
04-10-2017 23:52:07

在这里演示：https://eval.in/772567

由于问题确实想将本地时间转换为UTC时间，因此

<?php
date_default_timezone_set('Asia/Kolkata');
//$ambulanceDate = date('Y-m-d H:i:s', ($_POST['ambuStartDate'] . " " . $_POST['ambuStartTime']));
$ambulanceDate = date('Y-m-d H:i:s', time());

$indian_time = new DateTime($ambulanceDate);
$utc_time = new DateTime($ambulanceDate);

$ambulanceUpdateTime = $utc_time
  ->setTimezone(new DateTimeZone('UTC'))
  ->format('m-d-Y H:i:s');

echo 'Indian time: ' . $indian_time->format('m-d-Y H:i:s') . PHP_EOL;
echo 'UTC time: ' . $ambulanceUpdateTime;

演示：https://eval.in/772648

从所选日期时间减去5小时30分钟

2 个答案: