PHP: Merge three variables (year, month, day) into one date

Question

I have three variables:

1.) $year (e.g. "2018"),

2.) $month (e.g. "9" or "12"),

3.) $day (e.g. "5" or "21").

I want to merge them into one single variable ($date) in the format of YYYY-mm-dd, i.e. "2018-09-14".

Note that the month "9" and the day "5" now have "0" in front of them.

What would be the best way to do so?

Thank you!

The best way to do this would be - to concatenate strings into one string. — Alex Shesterov, Sep 14 '18 at 10:04

score 1 · Accepted Answer · answered Sep 14 '18 at 10:03

You can just:

$year = 1980;
$month = 9;
$day = 5;

$date = strftime("%F", strtotime($year."-".$month."-".$day));

But why not just concatenate the strings?

$date = $year . "-" . str_pad($month, 2, "0", STR_PAD_LEFT) . "-" . str_pad($day, 2, "0", STR_PAD_LEFT);

score 0 · Answer 2 · answered Sep 14 '18 at 10:06

0

echo date("d-m-Y",mktime(0, 0, 0, $day, $month, $year));

answered Sep 14 '18 at 10:06

Hòa Lê Việt

1

Notice that the OP asked for a different format from what you are providing in your answer. An explanation on why this is better than simple string concatenation would improve your answer. – CPHPython Sep 14 '18 at 10:38

2 Answers2