[Resolved] Add year to current date

[mhendry]

I am trying to write a script that will insert two dates into a MySQL DB. First date being the current day, and the second being a year from the current day. The first date will be used as a "Member Since" and the second will be "Membership Expiration" In addition I plan to use a timestamp in MySQL to act as a "Account Last Modified"
No matter what I do, I cannot get the $nextyear variable to work correctly For example:

$today = mktime();
			$nextyear = mktime(0, 0, 0, date("M"), date("j"), date("Y")+1);


		echo date("M j, Y", $today);
		echo date("M j, Y", $nextyear);

Results in:
Apr 10, 2005Dec 10, 2005

Why is the day correct, but the month and year wrong? Is there something I am missing?

[bpat1434]

You could always do this:

$today = mktime();
$year = date("Y");
$next_year = $year+1;
$nextyear = mktime(0,0,0,date("M"),date("j"),$next_year);
echo date("M j, Y", $today);
echo date("M j, Y", $nextyear);

You could also just see if incrementing it works:

$nextyear = mktime(0,0,0,date("M"),date("j"),++date("Y");

You should read the user notes on the PHP website: PHP mkTime manual

[kkobashi]

I believe the problem lies here:

$month = date("M");
returns the string "Apr"

This is not what you want to send into the mktime() function...

What you want is:

$month = date("n");
which returns an integer that is suitable for mktime()

Run this script in your browser to prove that it works:

<?php

$month = date("n");
$day = date("j");
$yr = date("Y");

echo "Today:<br />";
echo "&nbsp;&nbsp;Month: " . $month . "<br />";
echo "&nbsp;&nbsp;Day: " . $day . "<br />";
echo "&nbsp;&nbsp;Year: " . $yr . "<br />";
echo "<p />";

$todayUnixTimestamp = mktime(0, 0, 0, $month, $day, $yr);
echo "Today in Unix Epoch<br />";
echo $todayUnixTimestamp;
echo "<p />";
$nextYearUnixTimestamp = mktime(0,0,0, $month, $day, $yr+1);
echo "Next year in Unix Epoch<br />";
echo $nextYearUnixTimestamp;

echo "<p />";
echo "Difference in seconds is " . ($nextYearUnixTimestamp - $todayUnixTimestamp);
echo "<br />";
echo "Number of seconds in a year is " . (60 60 24) 365;
echo "<br />";
echo "Number of seconds in a leap year is " . (60 60 24) 366;
echo "<p />";
echo "Lets see what the UNIX timestamps result in human readable form using strftime()<p />";

echo "Today: " . strftime("%B %d %Y", $todayUnixTimestamp) . "<br />";
echo "Next year: " . strftime("%B %d %Y", $nextYearUnixTimestamp) . "<br />";
?>

Output:

Today:
Month: 4
Day: 11
Year: 2005

Today in Unix Epoch
1113202800

Next year in Unix Epoch
1144738800

Difference in seconds is 31536000
Number of seconds in a year is 31536000
Number of seconds in a leap year is 31622400

Lets see what the UNIX timestamps result in human readable form using strftime()

Today: April 11 2005
Next year: April 11 2006

[Shrike]

Such a lot of code when all it takes is a little reading of the manual

$today = date('Ymd');
$nextyear = date('Ymd', strtotime($today . " +1 year")); // Leap-year safe date addition
echo "Today is $today.<br>\nNext year is $nextyear.";

[mhendry]

Thanks for all the help.
Actually what I was looking for is right in the manual , but when I copied and pasted into my text editor it gave me errors, so I sought out a new solution, turns out there is something weird going on with my text editor. Anyhow, one simple line of code will give me what I want:

$nextyear = mktime(0, 0, 0, date('n'), date('j'), date('Y')+1);

Substituting the 'n' for the 'M' in the manual's example seems to fit the bill perfectly.

[kkobashi]

Originally posted by mhendry
Thanks for all the help.
Actually what I was looking for is right in the manual , but when I copied and pasted into my text editor it gave me errors, so I sought out a new solution, turns out there is something weird going on with my text editor. Anyhow, one simple line of code will give me what I want:

$nextyear = mktime(0, 0, 0, date('n'), date('j'), date('Y')+1);

Substituting the 'n' for the 'M' in the manual's example seems to fit the bill perfectly. [/B]

Which is what I basically said in my post above:

What you want is:

$month = date("n");
which returns an integer that is suitable for mktime()

Glad that its fixed... now do us a favor and mark this thread as resolved.