PHP时间算术 [英] PHP Time Arithmetic

问题描述

我的数据库中有两次：

  $ time1 =02:00:03; 
 $ time2 =04:00:04; 
  

，我想添加他们 $ time1 + $ time2 最好的方法是什么？

我没有任何问题，因为我可以做这样的事情：

  $ end = new DateTime（$ this-> hora_final）; 
 $ start = new DateTime（$ this-> hora_inicio）; 
 $ diff = $ end-> diff（$ start）; 
 
 $ diff->格式（'％H：％I：％S'）; 
  

它的工作完美...但我似乎找不到一种方法来添加它们。 ..任何想法？

注意

解决方案

DateTime :: add（） 。添加两个日期真的没有什么意义，因此您不能将 $ time2 视为 DateTime 添加它到 time1 。通常你会发现像 01.04。 + 2天，但不是 01.04。 + 02.00 。该方法接受类型为 DateInterval 的对象。要创建它，我建议使用类似

 列表（$ hour，$ min，$ second）= explode（'：' ，$ time2）; 
 $ interval = new DateInterval（PT {$ hour} H {$ min} M {$ second} S）; 
  

现在您应该可以将间隔添加到日期

  $ x = new DateTime（$ time1）; 
 $ y = $ x-> add（$ interval）; 
  

（未经测试）

I have two times in my database:

$time1 = "02:00:03";
$time2 = "04:00:04";

and I want to make add them $time1 + $time2 what is the best way to do that?

I have no problem with the diference cause I can do something like this:

$end = new DateTime($this->hora_final);
$start = new DateTime($this->hora_inicio);
$diff = $end->diff($start);

$diff->format('%H:%I:%S');

and it works perfectly... but i can't seem to find a way to add them... any ideas?

Regards

解决方案

DateTime::add(). It really doesn't make much sense to add two dates, thus you cannot treat $time2 as DateTime, to add it to time1. Usually you would say something like 01.04. + 2 days, but not 01.04. + 02.00. The method accepts an object of type DateInterval. To create it I suggest to use something like

list($hour, $min, $second) = explode(':', $time2);
$interval = new DateInterval("PT{$hour}H{$min}M{$second}S");

Now you should be able to add the interval to the date

$x = new DateTime($time1);
$y = $x->add($interval);

(untested)

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