DateTime添加1天 [英] DateTime add 1 day
问题描述
我的表单有2个字段 - Time_from和Time_to
My Form has 2 fields - Time_from and Time_to
现在我需要在每天的数据库中添加一个条目(如果这些日期之间有差异)
ex。 Time_from = 2013-08-26 08:00:00和Time_to = 2013-08-28 16:00:00
所以在我的数据库中我应该得到3个条目
Now i need to add an entry in my database for each day (If theres is a difference between those days) ex. Time_from = 2013-08-26 08:00:00 and Time_to = 2013-08-28 16:00:00 So in my database i should get 3 entries
Time_from = 2013-08-26 08:00:00 and Time_to = 2013-08-26 16:00:00
Time_from = 2013-08-27 08:00:00 and Time_to = 2013-08-27 16:00:00
Time_from = 2013-08-28 08:00:00 and Time_to = 2013-08-28 16:00:00
所以为了这个目的,我已经做了一个方法,我首先找到这两个日期之间的区别,然后我使用一个for循环将在两个日期之间运行多少天,最后只需添加1天。
So for that purpose i have made a method, where i first find the difference between those 2 dates and then i'm using a for loop that will run as many days in difference those 2 dates have, and at the end im just adding 1 day.
public function createOffer($offer)
{
$time_from = new DateTime($offer->time_from);
$time_to = new DateTime($offer->time_to);
$interval = $time_from->diff($time_to);
for ($counter = 0; $counter <= $interval->d; $counter++) {
$offer->time_from = $time_from->format('Y-m-d H:i:s');
$offer_time_to = new DateTime($time_from->format('Y-m-d'));
$offer_time_to->setTime($time_to->format('H'), $time_to->format('i')
, $time_to->format('s'));
$offer->time_to = $offer_time_to->format('Y-m-d H:i:s');
if ($offer->validate()) {
$offer->save();
}
date_modify($time_from, '+1 day');
}
}
由于某些原因,代码不起作用。
For some reason, the code doesn't work.
当我删除date_modify字段时,只有第一天将保存在我的数据库中(Guess for循环只运行一次)
When i remove the date_modify field only the first day will be saved in my database (Guess the for loop is only running once then)
但是在我的代码中使用date_modify,只有最后一天保存在数据库中。
But with the date_modify in my code, only the last day is saved in the database.
推荐答案
您可以使用Datatime对象的添加功能
you can use Add function of Datatime object
这里,我给你一个例子,在你的发布日期中添加一个,如
here i am giving you one example to add one 1 in your post date like
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P1D'));
echo $date->format('Y-m-d') . "\n";
?>
OUTPUT
2000-01-02
希望它能帮助你解决你的问题。
Hope it will sure help you to solve your issue.
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