Question

Is there any php function available where I can add days to a date to make up another date? For example, I have a date in the following format: 27-December-2011

If I add 7 to the above, it should give: 03-January-2012.

很多人

Answer 1

引言

$add_days = 7;
$date = date( Y-m-d ,strtotime($date) + (24*3600*$add_days));

Answer 2

看看这种简单的幻灯

$date = date("Y-m-d");// current date

$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 day");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 week");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +2 week");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 month");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days");

Answer 3

您可使用add 日期方法。任何途径,这种解决办法都用于“S.C.”版本,即5.3。

Answer 4

date( Y-m-d , strtotime( +6 days , strtotime($original_date)));

Answer 5

实际上,这比一切都容易。

$some_var = date("Y-m-d",strtotime("+7 day"))

You can use a variable instead of the string, of course. It will be great if the people answering the questions, won t complicate things. Less code, means less time to waste on the server ;).

Answer 6

$date = new DateTime( 27-December-2011 );
$date->add(new DateInterval( P7D ));
echo $date->format( d-F-Y ) . "
";

改变描述的形式是你想要的。 (见date()。

Answer 7

$registered = $udata->user_registered;
$registered = date( "d m Y", strtotime( $registered ));
$challanexpiry = explode(   , $registered);
$day   = $challanexpiry[0];
$month = $challanexpiry[1];
$year  = $challanexpiry[2];
$day = $day+10;
$bankchallanexpiry = $day . " " . $month . " " . $year;

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