如何从 PHP 中的日期时间对象中减去 24 小时
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how do I subtract 24 hour from date time object in PHP
提问by adit
I have the following code:
我有以下代码：
$now = date("Y-m-d H:m:s");
$date = date("Y-m-d H:m:s", strtotime('-24 hours', $now));
However, now it gives me this error:
但是，现在它给了我这个错误：
A non well formed numeric value encountered in...
why is this?
为什么是这样？
回答by vascowhite
$date = (new \DateTime())->modify('-24 hours');
or
或者
$date = (new \DateTime())->modify('-1 day');
(The latter takes into account this commentas it is a valid point.)
（后者考虑了这个评论，因为它是一个有效的观点。）
Should work fine for you here. See http://PHP.net/datetime
在这里应该适合你。见http://PHP.net/datetime
$date will be an instance of DateTime, a real DateTime object.
$date 将是DateTime 的一个实例，一个真正的 DateTime 对象。
回答by Joel Harkes
strtotime()expects a unix timestamp (which is number seconds since Jan 01 1970)
strtotime()需要一个 unix 时间戳（即number seconds since Jan 01 1970）
$date = date("Y-m-d H:i:s", strtotime('-24 hours', time())); ////time() is default so you do not need to specify.
i would suggest using the datetime library though, since it's a more object oriented approach.
不过，我建议使用 datetime 库，因为它是一种更面向对象的方法。
$date = new DateTime(); //date & time of right now. (Like time())
$date->sub(new DateInterval('P1D')); //subtract period of 1 day
The advantage of this is that you can reuse the DateInterval:
这样做的好处是您可以重用DateInterval：
$date = new DateTime(); //date & time of right now. (Like time())
$oneDayPeriod = new DateInterval('P1D'); //period of 1 day
$date->sub($oneDayPeriod);
$date->sub($oneDayPeriod); //2 days are subtracted.
$date2 = new DateTime();
$date2->sub($oneDayPeriod); //can use the same period, multiple times.
Carbon (update 2020)
碳（2020 年更新）
Most popular library for processing DateTimes in PHP is Carbon.
在 PHP 中处理日期时间的最流行的库是Carbon。
Here you would simply do:
在这里，您只需执行以下操作：
$yesterday = Carbon::now()->subDay();
回答by Sumit Bijvani
you can do this in many ways...
你可以通过多种方式做到这一点......
echo date('Y-m-d H:i:s',strtotime('-24 hours')); // "i" for minutes with leading zeros
OR
或者
echo date('Y-m-d H:i:s',strtotime('last day')); // 24 hours (1 day)
Output
输出
2013-07-17 10:07:29
回答by Sundar
This may be helpful for you:
这可能对您有帮助：
//calculate like this
$date = date("Y-m-d H:m:s", (time()-(60*60*24)));
//check the date
echo $date;
回答by Sreedhu Madhu
Simplest way to sub or add time,
最简单的方法来减少或增加时间，
<?php
**#Subtract 24 hours**
$dtSub = new DateTime('- 24 hours');
var_dump($dtSub->format('Y-m-d H:m:s'));
**#Add 24 hours**
$dtAdd = new DateTime('24 hours');
var_dump($dtAdd->format('Y-m-d H:m:s'));die;
?>
回答by Dipen Soni
$now = date("Y-m-d H:i:s");
$date = date("Y-m-d H:i:s", strtotime('-24 hours', strtotime($now)));
Add "strtotime" before $now, and Y-m-d H:m:s replace with Y-m-d H:i:s
在 $now 前添加 "strtotime"，并将 Ymd H:m:s 替换为 Ymd H:i:s
回答by steven
this should work, too
这也应该有效
$date = date("Y-m-d H:m:s", strtotime('-24 hours'));
回答by Amal Murali
You can simply use time()to get the current timestamp.
您可以简单地使用time()来获取当前时间戳。
$date = date("Y-m-d H:m:s", strtotime('-24 hours', time()));
回答by Mohamed Salah
all you have to do is to alter your code to be
你所要做的就是改变你的代码
$now = strtotime(date("Y-m-d H:m:s"));
$date = date("Y-m-d H:m:s", strtotime('-24 hours', $now));
