PHP: Function with 3 arguments > Use it with less arguments

Question

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.

Set default values in your function definition: function addDate($years, $months = 0, $days = 0) — Mark Baker
– Mark Baker, Commented Aug 29, 2016 at 22:02
addDate(2, 0, 0); should work without editing the function. But the best solution is to use default values as others suggest — Dan
– Dan, Commented Aug 29, 2016 at 22:05
If you want to add only days for example, you'll need addDate(0, 0, 2) regardless. — AbraCadaver
– AbraCadaver, Commented Aug 29, 2016 at 22:15

MrTux · Accepted Answer · 2016-08-29 22:03:28Z

3

You can declare default values for parameter:

function addDate($years, $months = 0, $days = 0)

This way you don't need to specify those or call your function like 'addDate(2,0,0)'

See http://php.net/manual/functions.arguments.php

answered Aug 29, 2016 at 22:03

MrTux

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Comments

Patrick Moore · Accepted Answer · 2016-08-29 22:08:05Z

You can define a default value for the parameter, like,

function addDate($years = 0, $months = 0, $days = 0)
{

Might be better to check whether each is >0 before building the string:

$newDateString = '';
if ( $years > 0 ) $newDateString .= " +$years years";
if ( $months > 0 ) $newDateString .= " +$months months";
if ( $days > 0 ) $newDateString .= " +$days days";
$newDate = date('Y-m-d', strtotime( date('Y-m-d') . $newDateString ) );

Lastly, you will (probably) want to be passing the value back using return rather than echo()ing - allows for more versatility later down the line:

    return $newDate;

And calling it like so:

echo addDate( 2 );

Function altogether:

function addDate($years = 0, $months = 0, $days = 0)
{
    $newDateString = '';
    if ( $years > 0 ) $newDateString .= " +$years years";
    if ( $months > 0 ) $newDateString .= " +$months months";
    if ( $days > 0 ) $newDateString .= " +$days days";
    return date('Y-m-d', strtotime($currentDate . $newDateString ) );
}

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