40

我有$adate;其中包含：

Tue Jan 4 07:59:59 2011

我想在这个日期添加以下内容：

$duration=674165; // in seconds

添加秒数后，我需要将结果恢复为日期格式。

我不知道我在做什么，但我得到了奇怪的结果。

注意：两个变量都是动态的。现在它们等于给定的值，但下一个查询它们将具有不同的值。

4

9 回答 9

82

如果您使用的是 php 5.3+，您可以使用一种新方法来完成它。

<?php 
$date = new DateTime();
echo $date->getTimestamp(). "<br>";
$date->add(new DateInterval('PT674165S')); // adds 674165 secs
echo $date->getTimestamp();
?>
于 2010-12-27T12:44:27.677 回答
28

只需使用一些不错的 PHP 日期/时间函数：

$adate="Tue Jan 4 07:59:59 2011";
$duration=674165;
$dateinsec=strtotime($adate);
$newdate=$dateinsec+$duration;
echo date('D M H:i:s Y',$newdate);
于 2010-12-27T12:47:36.957 回答
11

鉴于这$adate是一个时间戳（如果是这种情况），您可以执行以下操作：

$duration = 674165;
$result_date = strtotime(sprintf('+%d seconds', $duration), $adate);
echo date('Y-m-d H:i:s', $result_date);
于 2010-12-27T12:43:00.610 回答
6 
// add 20 sec to now
$duration = 20;
echo date("Y-m-d H:i:s", strtotime("+$duration sec"));
于 2016-06-28T13:24:05.100 回答
4

做这个：

$seconds = 1;
$date_now = "2016-06-02 00:00:00";

echo date("Y-m-d H:i:s", (strtotime(date($date_now)) + $seconds));
于 2016-06-02T14:15:05.303 回答
3 
$current_time_zone = 150;
date("Y-m-d H:i:s",strtotime(date("Y-m-d H:i:s"))+$current_time_zone);
于 2017-06-15T10:04:41.170 回答
1

我为时区制作了这个示例，但是如果您更改某些部分，它可能会对您有所帮助：

$seconds_to_add = 30;
$time = new DateTime();
$time->setTimezone(new DateTimeZone('Europe/London'));
$time2 = $time->format("Y/m/d G:i:s");
$time->add(new DateInterval('PT' . $seconds_to_add . 'S'));
$timestamp = $time->format("Y/m/d G:i:s");
echo $timestamp;
echo '========';
echo $time2;

结果：

2018/06/17 3:16:23========2018/06/17 3:15:53
于 2018-06-17T02:22:40.107 回答
0

When I need to implement some calculation, I always keep in mind what is it that I need. For example, if I need to add some seconds to a datetime, I need a datetime somewhere in the future. That is, I need a class called Future. How is it identified? What data does it need to operate? It seems that there should be at least two values: a datetime relative to which I need a future date, and an interval which defines a time distance between start datetime and desired datetime.

So here is a code:

use Meringue\ISO8601DateTime\FromISO8601 as DateTimeFromISO8601String;

$f =
    new Future(
        new DateTimeFromISO8601String('2011-01-04T07:59:59+00'),
        new NSeconds(674165)
    );

Then you can output it in ISO8601 format:

$f->value(); // returns 2011-01-12T03:16:04+00:00

If your initial datetime is not in ISO8601 format, which is the case in your example, you should create a datetime from a custom format, hence a name of the class -- FromCustomFormat. Since it is a datetime which represents itself in ISO8601 format, it extends an abstract class called ISO8601DateTime. Here is a complete example:

(new Future(
    new FromCustomFormat(
        'D M j H:i:s Y',
        'Tue Jan 4 07:59:59 2011'
    ),
    new NSeconds(674165)
))
    ->value();

Here is some more about this approach.

于 2020-07-01T17:15:04.923 回答
0

我strtotime()无法解决在当前时间添加动态数据/时间值的问题

这是我的解决方案：

$expires = 3600; //my dynamic time variable (static representation here)
$date = date_create(date('Y-m-d H:i:s')); //create a date/time variable (with the specified format - create your format, see (1))
echo date_format($date, 'Y-m-d H:i:s')."<br/>"; //shows the date/time variable without add seconds/time
date_add($date, date_interval_create_from_date_string($expires.' seconds')); //add dynamic quantity of seconds to data/time variable
echo date_format($date, 'Y-m-d H:i:s'); //shows the new data/time value

字体：https ://secure.php.net/manual/en/datetime.add.php （也可以参考面向对象的样式，Elzo Valugi解决方案）

(1) https://secure.php.net/manual/en/function.date.php

于 2018-03-20T13:58:53.813 回答