php $日期 + 1 年？
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$date + 1 year?
提问by Matt
I'm trying to get a date that is one year from the date I specify.
我试图获得一个自我指定的日期起一年的日期。
My code looks like this:
我的代码如下所示：
$futureDate=date('Y-m-d', strtotime('+one year', $startDate));
It's returning the wrong date. Any ideas why?
它返回错误的日期。任何想法为什么？
回答by Misho
$futureDate=date('Y-m-d', strtotime('+1 year'));
$futureDate is one year from now!
$futureDate 是一年后！
$futureDate=date('Y-m-d', strtotime('+1 year', strtotime($startDate)) );
$futureDate is one year from $startDate!
$futureDate 是 $startDate 的一年！
回答by Nidhin Baby
To add one year to todays date use the following:
要将一年添加到今天的日期，请使用以下命令：
$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));
For the other examples you must initialize $StartingDate with a timestamp value for example:
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
Try this
尝试这个
$newEndingDate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($StaringDate)) . " + 365 day"));
or
或者
$newEndingDate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($StaringDate)) . " + 1 year"));
回答by K Prime
Try: $futureDate=date('Y-m-d',strtotime('+1 year',$startDate));
尝试： $futureDate=date('Y-m-d',strtotime('+1 year',$startDate));
回答by AnarchyOutlaw
// 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;
回答by Gardenee
just had the same problem, however this was the simplest solution:
刚刚遇到了同样的问题，但这是最简单的解决方案：
<?php (date('Y')+1).date('-m-d'); ?>
回答by Andrew Atkinson
I prefer the OO approach:
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'))
Use DateTimeImmutableotherwise you will modify the original date too!
more on DateTimeImmutable: http://php.net/manual/en/class.datetimeimmutable.php
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
If you just want from todays date then you can always do:
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');
回答by Developer
//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'));
hope this simpler bit of code helps someone in future :)
希望这段更简单的代码可以帮助将来的某个人:)
回答by Frank Farmer
strtotime()is returning bool(false), because it can't parse the string '+one year'(it doesn't understand "one"). falseis then being implicitly cast to the integertimestamp 0. It's a good idea to verify strtotime()'s output isn't bool(false)before you go shoving it in other functions.
strtotime()正在返回bool(false)，因为它无法解析字符串'+one year'（它不理解“一”）。 false然后被隐式转换为integer时间戳0。在你将它推入其他函数之前验证它的strtotime()输出不是一个好主意bool(false)。
Return Values
Returns a timestamp on success, FALSE otherwise. Previous to PHP 5.1.0, this function would return -1 on failure.
返回值
成功时返回时间戳，否则返回 FALSE。在 PHP 5.1.0 之前，此函数将在失败时返回 -1。
回答by SeanJA
If you are using PHP 5.3, it is because you need to set the default time zone:
如果您使用的是 PHP 5.3，那是因为您需要设置默认时区：
date_default_timezone_set()
回答by Treby
Try This
尝试这个
$nextyear = date("M d,Y",mktime(0, 0, 0, date("m",strtotime($startDate)), date("d",strtotime($startDate)), date("Y",strtotime($startDate))+1));
