我需要帮助使用 PHP 将今天的日期添加 30 天
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I need help adding 30 days to today's date using PHP
提问by Charles Harris
I have a user system for my website. I need to make something that adds 30 days to todays date.
我的网站有一个用户系统。我需要做一些让今天的日期增加 30 天的东西。
UPDATE: I decided to go with an integer instead of a date.
更新：我决定使用整数而不是日期。
Here's what I did. I set things up so that when someone pays their bill it sets their "Days" value to (value) + 30
这就是我所做的。我进行了设置，以便在有人支付账单时将他们的“天数”值设置为 (value) + 30
I made a cron job that takes away 1 day from every user's "Days". Obviously, I set the cron job to run once per day.
我做了一个 cron 工作，从每个用户的“天”中带走了 1 天。显然，我将 cron 作业设置为每天运行一次。
回答by Dipesh Parmar
Use strtotimeand dateas below.
使用strtotime和date如下。
echo date('Y-m-d', strtotime("+30 days"));
OR
或者
<?php
echo date('Y-m-d h:i:s')."\n";
echo date('Y-m-d h:i:s', mktime(date('h'),date('i'),date('s'),date('m'),date('d')+30,date('Y')))."\n";
?>
回答by Prashank
First you can add date like this
首先你可以像这样添加日期
$iSecsInDay = 86400;
$iTotalDays = 30;
$user_signup = time() + ($iSecsInDays * $iTotalDays);
Then you can use the timestamp to generate date in any way you want, like
然后您可以使用时间戳以您想要的任何方式生成日期，例如
$date = date('d-m-Y', $user_signup);
Full reference can be found here. http://php.net/manual/en/function.date.php
完整参考可以在这里找到。http://php.net/manual/en/function.date.php
回答by álvaro Ferreira Pérez
If you want to add +30 days to the actual date and introduce in mysql you could use:
如果您想将 +30 天添加到实际日期并在 mysql 中引入，您可以使用：
$nextMonthDate = date("Y-m-d H:i:s", strtotime("+30 days",time()));
$nextMonthDate = date("Ymd H:i:s", strtotime("+30 days",time()));
Y-m-d H:i:s is the format to MySQL DATETIME.
Ymd H:i:s 是 MySQL DATETIME 的格式。
Example:
例子：
Actual date: echo echo date("Y-m-d H:i:s",time()); --> 2013-03-08 00:37:23
实际日期：echo echo date("Ymd H:i:s",time()); --> 2013-03-08 00:37:23
+30 days: echo date("Y-m-d H:i:s", strtotime("+30 days",time())); --> 2013-04-07 00:37:23
+30 天：echo date("Ymd H:i:s", strtotime("+30 days",time())); --> 2013-04-07 00:37:23
回答by álvaro Ferreira Pérez
$NewDate=Date('y:m:d', strtotime("+30 days"));
$NewDate=Date('y:m:d', strtotime("+30 days"));
int strtotime ( string $time [, int $now = time() ] )
The function expects to be given a string containing an English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.
该函数期望得到一个包含英文日期格式的字符串，并将尝试将该格式解析为 Unix 时间戳（自 1970 年 1 月 1 日 00:00:00 UTC 以来的秒数），相对于 now 给出的时间戳，或如果现在没有提供当前时间。
回答by Thomas Hambach
You cannot assign a function directly to a variable of a class. This is why you are getting the additional error. Consider moving it to your constructor.
您不能将函数直接分配给类的变量。这就是您收到额外错误的原因。考虑将其移动到您的构造函数。
...
private $user_signup = null;
public function __construct() {
$date = new DateTime();
$date->add(new DateInterval('P30D'));
$this->user_signup = $date->format('Y-m-d');
}
...
Also see Assigning a function's result to a variable within a PHP class? OOP Weirdnessfor more information.
另请参阅将函数的结果分配给 PHP 类中的变量？OOP 怪异以获取更多信息。
