php - 添加 3 分钟到日期变量

发布于 2024-12-10 18:14:57 字数 971 浏览 0 评论 0原文

我想向已有的日期/时间变量添加 3 分钟，但我不知道如何执行此操作。我从这样的字符串创建了变量：（顺便说一句，它采用 RFC 2822 日期格式）

$date = 2011-10-18T19:56:00+0200

我使用以下命令将该字符串转换为日期：

$time = date_format(DateTime::createFromFormat("Y-m-d\TH:i:sO", $date), "G:i")

现在，我想向该变量添加 3 分钟，但我不知道怎么办。我之前在脚本中使用过以下命令，但这适用于当前日期/时间，所以我不确定如何将其用于我的时间变量：

$currenttime = date('G:i', strtotime('+2 hours'));

那么，如何向 $time 变量添加三分钟？

我之前尝试过：

$date = '2011-10-18T19:56:00+0200';
$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
echo date('G:i', strtotime('+3 minutes', $time->getTimestamp()));

但是这给出了添加了 3 分钟的当前时间，它不使用 $date 变量...

我尝试过：

$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
$time = $time->add(new DateInterval('P2H'));

但是当我

echo date_format($time, 'G:i');

什么都不做时就会回显...

这里有什么帮助吗？

原文

I want to add 3 minutes to a date/time variable I have, but I'm not sure how to do this. I made the variable from a string like this: (which is in the RFC 2822 date format btw)

$date = 2011-10-18T19:56:00+0200

I converted that string into date using this command:

$time = date_format(DateTime::createFromFormat("Y-m-d\TH:i:sO", $date), "G:i")

Now, I'd like to add 3 minutes to that variable, but I I'm not sure how. I've used the following command in my script before, but that applies to the current date/time, so I'm not sure how to use that for my time variable:

$currenttime = date('G:i', strtotime('+2 hours'));

So, how can I add three minutes to the $time variable?

I tried this before:

$date = '2011-10-18T19:56:00+0200';
$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
echo date('G:i', strtotime('+3 minutes', $time->getTimestamp()));

but that gives the current time with 3 minutes added, it doesnt use the $date variable...

And I tried:

$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
$time = $time->add(new DateInterval('P2H'));

But then when I do

echo date_format($time, 'G:i');

nothing is echoed...

Any help here?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

ゝ杯具 2024-12-17 18:14:57

您只需使用 strtotime 两次：

$date = strtotime('2011-10-18T19:56:00+0200');
echo date('G:i', strtotime('+3 minutes', $date));

You could just use strtotime twice:

$date = strtotime('2011-10-18T19:56:00+0200');
echo date('G:i', strtotime('+3 minutes', $date));

回复收藏 0 原文

苄①跕圉湢 2024-12-17 18:14:57

而不是

$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
$time = $time->add(new DateInterval('P2H'));

尝试（添加 3 分钟）

$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
$time->add(new DateInterval('PT3M'));

首先，由于您使用的是 PHP 的 DateTime 类，因此您不需要将 add 方法的输出分配给变量 - 它会修改您的 DateTime传递到构造函数中。其次，如果您使用同一个类修改时间，则必须确保时间定义之前有一个 T 。对于您的示例，DateInterval('P2H') 无效 - 它应该是 DateInterval('PT2H')。

Instead of

$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
$time = $time->add(new DateInterval('P2H'));

try (for adding 3 minutes)

$time = DateTime::createFromFormat("Y-m-d\TH:i:sO", $date);
$time->add(new DateInterval('PT3M'));

First, since you're using PHP's DateTime class, you don't need to assign the output of the add method to a variable - it will modify the DateTime you passed into the constructor. Second, if you're making modifications to time using the same class, you have to make sure there's a T before your time definition. For your example, DateInterval('P2H') is invalid - it should be DateInterval('PT2H').

回复收藏 0 原文

~没有更多了~