在PHP中为$Date添加天数

I have a date returned as part of a mySQL query in the form 2010-09-17

我有一个日期作为mySQL查询的一部分以2010-09-17的形式返回

I would like to set the variables $Date2 to $Date5 as follows:

我想将变量$Date2设置为$Date5，如下所示:

$Date2 = $Date + 1

$Date2 = $Date + 1

$Date3 = $Date + 2

$Date3 = $Date + 2

etc..

等。

so that it returns 2010-09-18, 2010-09-19 etc...

所以它返回2010-09-18,2010-09-19等等。

I have tried

我有试过

date('Y-m-d', strtotime($Date. ' + 1 day'))

but this gives me the date BEFORE $Date.

但是这给了我$ date之前的日期。

What is the correct way to get my Dates in the format form 'Y-m-d' so that they may be used in another query?

以“Y-m-d”格式获取日期的正确方法是什么，以便在另一个查询中使用它们?

9 个解决方案

#1

294

All you have to do is use days instead of day like this:

你所要做的就是用几天代替这一天:

<?php
$Date = "2010-09-17";
echo date('Y-m-d', strtotime($Date. ' + 1 days'));
echo date('Y-m-d', strtotime($Date. ' + 2 days'));
?>

And it outputs correctly:

它输出正确:

2010-09-18
2010-09-19

#2

If you're using PHP 5.3, you can use a DateTime object and its add method:

如果使用的是PHP 5.3，可以使用DateTime对象及其添加方法:

$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->add(new DateInterval('P1D')); // P1D means a period of 1 day
$Date2 = $date->format('Y-m-d');

Take a look at the DateInterval constructor manual page to see how to construct other periods to add to your date (2 days would be 'P2D', 3 would be 'P3D', and so on).

查看DateInterval构造函数手册页，了解如何构造其他时间来添加日期(2天将是“P2D”，3将是“P3D”，等等)。

Without PHP 5.3, you should be able to use strtotime the way you did it (I've tested it and it works in both 5.1.6 and 5.2.10):

如果没有PHP 5.3，您应该能够像以前那样使用strtotime(我已经对它进行了测试，它在5.1.6和5.2.10中都可以工作):

$Date1 = '2010-09-17';
$Date2 = date('Y-m-d', strtotime($Date1 . " + 1 day"));
// var_dump($Date2) returns "2010-09-18"

#3

From PHP 5.2 on you can use modify with a DateTime object:

从PHP 5.2开始，您可以使用DateTime对象进行修改:

http://php.net/manual/en/datetime.modify.php

$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->modify('+1 day');
$Date2 = $date->format('Y-m-d');

Be careful when adding months... (and to a lesser extent, years)

增加月份时要小心……(在更小的程度上，是几年)

#4

Here is a small snippet to demonstrate the date modifications:

这里有一个小片段来演示修改日期:

$date = date("Y-m-d");
//increment 2 days
$mod_date = strtotime($date."+ 2 days");
echo date("Y-m-d",$mod_date) . "\n";

//decrement 2 days
$mod_date = strtotime($date."- 2 days");
echo date("Y-m-d",$mod_date) . "\n";

//increment 1 month
$mod_date = strtotime($date."+ 1 months");
echo date("Y-m-d",$mod_date) . "\n";

//increment 1 year
$mod_date = strtotime($date."+ 1 years");
echo date("Y-m-d",$mod_date) . "\n";

#5

You can also use the following format

您还可以使用以下格式

strtotime("-3 days", time());
strtotime("+1 day", strtotime($date));

You can stack changes this way:

你可以这样堆叠变更:

strtotime("+1 day", strtotime("+1 year", strtotime($date)));

Note the difference between this approach and the one in other answers: instead of concatenating the values +1 day and <timestamp>, you can just pass in the timestamp as the second parameter of strtotime.

注意这种方法与其他答案中的方法的区别:您可以将时间戳作为strtotime的第二个参数传入，而不是将值+1 day和连接在一起。

#6

Here is the simplest solution to your query

这是查询的最简单的解决方案

$date=date_create("2013-03-15");\\or your date string
date_add($date,date_interval_create_from_date_string("40 days"));\\add number of days 
echo date_format($date,"Y-m-d");\\set date format of the result

#7

Using a variable for Number of days

使用变量的天数

$myDate = "2014-01-16";
$nDays = 16;
$newDate = strtotime($myDate . '+ '.$nDays.'days');
echo newDate('d/m/Y', $soma); //format new date

#8

Here has an easy way to solve this.

这里有一个简单的方法来解决这个问题。

<?php
   $date = "2015-11-17";
   echo date('Y-m-d', strtotime($date. ' + 5 days'));
?>

Output will be:

输出将:

2015-11-22

Solution has found from here - How to Add Days to Date in PHP

解决方案已经找到——如何在PHP中添加日期

#9

All have to use bellow code:

都必须使用下面的代码:

$nday = time() + ( 24 * 60 * 60);    
echo 'Now:       '. date('Y-m-d') ."\n";    
echo 'Next Day: '. date('Y-m-d', $nday) ."\n";

#1

294