PHP：具有3个参数的函数>使用较少的参数(PHP: Function with 3 arguments > Use it with less arguments)

 我刚刚创建了一个函数来添加至今|年|月|日。 但是我有一个小问题，有时候我只想添加几年，但由于函数有3个参数（年，月，日），所以我得到一个警告： 
 
 
 
 警告：缺少addDate（）的参数2，在第10行的C：\ xampp \ htdocs \ date.php中调用，并在第2行的C：\ xampp \ htdocs \ date.php中定义 
 
 
 警告：缺少addDate（）的参数3，在第10行的C：\ xampp \ htdocs \ date.php中调用，并在第2行的C：\ xampp \ htdocs \ date.php中定义 
 
 
<?php 
function addDate($years, $months, $days)
{
    $currentDate = date('Y-m-d');

    $newDate = date('Y-m-d', strtotime($currentDate. '+'. $years. ' years +'. $months. ' months +'. $days. ' days'));
    echo $newDate;
}

addDate(2);
?>
 
 我试图使用addDate（2，null，null）; 但它不起作用。 

I just created a function to add years|months|days to date. But I have a small issue, sometimes I want to add just years, but since the function has 3 arguments (years,months, days), I get a warning:
 
 
 
Warning: Missing argument 2 for addDate(), called in C:\xampp\htdocs\date.php on line 10 and defined in C:\xampp\htdocs\date.php on line 2
 
 
Warning: Missing argument 3 for addDate(), called in C:\xampp\htdocs\date.php on line 10 and defined in C:\xampp\htdocs\date.php on line 2
 
 
<?php 
function addDate($years, $months, $days)
{
    $currentDate = date('Y-m-d');

    $newDate = date('Y-m-d', strtotime($currentDate. '+'. $years. ' years +'. $months. ' months +'. $days. ' days'));
    echo $newDate;
}

addDate(2);
?>
 
I've tried to use addDate(2, null, null); but It doesn't work.

原文：https://stackoverflow.com/questions/39215846