$ date + 1年？

Question

我正在尝试从我指定的日期开始一年的日期.

我的代码看起来像这样:

$futureDate=date('Y-m-d', strtotime('+one year', $startDate));

它回复了错误的日期.有什么想法吗？

Answer 1

$futureDate=date('Y-m-d', strtotime('+1 year'));

$ futureDate是一年后的!

$futureDate=date('Y-m-d', strtotime('+1 year', strtotime($startDate)) );

$ futureDate是$ startDate的一年!

Answer 2

要在今天的日期添加一年,请使用以下内容:

$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));

对于其他示例,您必须使用时间戳值初始化$ StartingDate,例如:

$StartingDate = mktime();  // todays date as a timestamp

试试这个

$newEndingDate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($StaringDate)) . " + 365 day"));

要么

$newEndingDate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($StaringDate)) . " + 1 year"));

Answer 3

尝试: $futureDate=date('Y-m-d',strtotime('+1 year',$startDate));

Answer 4

//1 year from today's date
echo date('d-m-Y', strtotime('+1 year'));

//1 year from from specific date
echo date('22-09-Y', strtotime('+1 year'));

希望这段更简单的代码可以帮助将来的某个人:)

Answer 5

// Declare a variable for this year 
$this_year = date("Y");
// Add 1 to the variable
$next_year = $this_year + 1;
$year_after = $this_year + 2;

// Check your code
    echo "This year is ";
    echo $this_year;
    echo "<br />";
    echo "Next year is ";
    echo $next_year;
    echo "<br />";
    echo "The year after that is ";
    echo $year_after;

Answer 6

只是有同样的问题,但这是最简单的解决方案:

<?php (date('Y')+1).date('-m-d'); ?>

Answer 7

我更喜欢面向对象的方法：

$date = new \DateTimeImmutable('today'); //'today' gives midnight, leave blank for current time.
$futureDate = $date->add(\DateInterval::createFromDateString('+1 Year'))

使用DateTimeImmutable，否则你将改变原来的日期呢！更多关于 DateTimeImmutable：http : //php.net/manual/en/class.datetimeimmutable.php

如果您只想从今天的日期开始，那么您可以随时执行以下操作：

new \DateTimeImmutable('-1 Month');