设置 strtotime 添加变量到天数

Question

我正在尝试导出特定天数并将其设置为变量 $max_future。目前它是固定的天数，但我希望用户输入变量是它拥有的数字。

目前设置为：

$max_future = date("Y-m-d", strtotime($today . "+6 days"));

我想要这样的东西：

$exportDays = '9'; //(or whatever the user input was)
$max_future = date("Y-m-d", strtotime($today . "+$exportDays days"));

这可能吗？感谢您的帮助

Answer 1

尝试：

$max_future = date("Y-m-d", strtotime($today . "+" . $exportDays . " days"));

Answer 2

像这样：

<?php
$output = 9; // you can take this value from input.
$today = date("Y-m-d");
echo date('Y-m-d', strtotime($today. ' +'.$output. ' days'));
?>

Answer 3

<?php
    $exportDays = $_REQUEST['exportDays']; // or however you want to get it from user input
    $max_date = date('Y-m-d', strtotime(date('Y-m-d') . ' +' . $exportDays . ' days'));

设置 strtotime 添加变量到天数

Setting strtotime adding a variable to amount of days

php

strtotime