php 将日期添加到当前日期
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add day to current date
提问by Brad
add a day to date, so I can store tomorrow's date in a variable.
添加日期，以便我可以将明天的日期存储在变量中。
$tomorrow = date("Y-m-d")+86400;
I forgot.
我忘了。
回答by lonesomeday
I'd encourage you to explore the PHP 5.3 DateTimeclass. It makes dates and times far easier to work with:
我鼓励您探索 PHP 5.3DateTime类。它使日期和时间更容易使用：
$tomorrow = new DateTime('tomorrow');
// e.g. echo 2010-10-13
echo $tomorrow->format('d-m-Y');
Furthermore, you can use the + 1 daysyntax with any date:
此外，您可以将+ 1 day语法用于任何日期：
$xmasDay = new DateTime('2010-12-24 + 1 day');
echo $xmasDay->format('Y-m-d'); // 2010-12-25
回答by casablanca
datereturns a string, whereas you want to be adding 86400 seconds to the timestamp. I think you're looking for this:
date返回一个字符串，而您希望将 86400 秒添加到时间戳。我想你正在寻找这个：
$tomorrow = date("Y-m-d", time() + 86400);
回答by Matchu
date()returns a string, so adding an integer to it is no good.
date()返回一个字符串，因此向它添加一个整数是不好的。
First build your tomorrow timestamp, using strtotimeto be not only clean but more accurate (see Pekka's comment):
首先建立你明天的时间戳，使用strtotime不仅干净而且更准确（见 Pekka 的评论）：
$tomorrow_timestamp = strtotime("+ 1 day");
Then, use it as the second argument for your datecall:
然后，将其用作date调用的第二个参数：
$tomorrow_date = date("Y-m-d", $tomorrow_timestamp);
Or, if you're in a super-compact mood, that can all be pushed down into
或者，如果你的心情非常紧凑，那么这一切都可以推到
$tomorrow = date("Y-m-d", strtotime("+ 1 day"));
回答by David Snabel-Caunt
Nice and obvious:
漂亮而明显：
$tomorrow = strtotime('tomorrow');
回答by Koushik Das
You can use the addmethod datetimeclass.
Eg, you want to add one day to current date and time.
您可以使用add方法datetime类。例如，您想在当前日期和时间上增加一天。
$today = new DateTime();
$today->add(new DateInterval('P1D'));
Further reference php datetime add
进一步参考php datetime add
Hope this helps.
希望这可以帮助。
回答by Mat Barnett
I find mktime()most useful for this sort of thing. E.g.:
我发现mktime()对这类事情最有用。例如：
$tomorrow=date("Y-m-d", mktime(0, 0, 0, date("m"), date("d")+1, date("Y")));
