如何在PHP中从日期时间对象中减去24小时

Question

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));

但是,现在它给了我这个错误:

A non well formed numeric value encountered in...

为什么是这样？

Answer 1

$date = (new \DateTime())->modify('-24 hours');

要么

$date = (new \DateTime())->modify('-1 day');

(后者考虑到这个评论,因为它是一个有效点.)

应该在这里工作正常.见http://PHP.net/datetime

$ date将是DateTime的一个实例,一个真正的DateTime对象.

Answer 2

strtotime()期望一个unix时间戳(这是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.

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

这样做的好处是你可以重用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.

Answer 3

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

要么

echo date('Y-m-d H:i:s',strtotime('last day')); // 24 hours (1 day)

产量

2013-07-17 10:07:29

Answer 4

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;
?>