PHP - 将一周添加到用户定义的日期
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PHP - Add one week to a user defined date
提问by jarnold
There's more than likely going to be a duplicate for this question, but I'm struggling to find a precise answer for my problem.
这个问题很可能会重复，但我正在努力为我的问题找到一个准确的答案。
The user enters a starting date for a client's rent (on a form on a previous page), then it needs to generate the next date (one week later) that the client is required to pay. For example:
用户输入客户租金的开始日期（在上一页的表格中），然后需要生成客户需要支付的下一个日期（一周后）。例如：
$start_date = $_POST['start_date'];
$date_to_pay = ???
Lets say the user enters in 2015/03/02:
假设用户在 2015/03/02 输入：
$start_date = "2015/03/02";
I then want the date to pay to be equal to a week later (2015/03/09):
然后我希望支付的日期等于一周后（2015/03/09）：
$date_to_pay = "2015/03/09";
How would one go around doing this? Many thanks.
怎么做呢？非常感谢。
回答by priya786
You can try this
你可以试试这个
$start_date = "2015/03/02";
$date = strtotime($start_date);
$date = strtotime("+7 day", $date);
echo date('Y/m/d', $date);
回答by Mihir Bhatt
Please try the following:
请尝试以下操作：
date('d.m.Y', strtotime('+1 week', $start_date));
回答by Anthony
Object Oriented Style using DateTimeclasses:
使用DateTime类的面向对象风格：
$start_date = DateTime::createFromFormat('Y/m/d', $_POST['start_date']);
$one_week = DateInterval::createFromDateString('1 week');
$start_date->add($one_week);
$date_to_pay = $start_date->format('Y/m/d');
Or for those who like to have it all in one go:
或者对于那些喜欢一口气拥有这一切的人：
$date_to_pay = DateTime::createFromFormat('Y/m/d',$_POST['start_date'])
->add(DateInterval::createFromDateString('1 week'))
->format('Y/m/d');
回答by TECHNOMAN
$start_date = "2015/03/02";
$new_date= date("Y/m/d", strtotime("$start_date +1 week"));
回答by Narendrasingh Sisodia
You can use this:
你可以使用这个：
$startdate = $_POST['start_date'];
$date_to_pay = date('Y/m/d',strtotime('+1 week',$startdate));
