PHP Date

[jandrews3]

I'm just missing something stupid, I know I am, but I've been working on it for more than an hour.

Why does the following code:

$month = 3;
$day = 9;
print "M: ".$month."<br>";
print "D: ".$day."<br>";
$due = $month." / ".$day;
print $due."<br>";
print date("M j", $due)."<br>";

 

yield the following output:

M: 3

D: 9

3 / 9

Dec 31

 

AUGGHHH! Shouldn't the last line be "Mar 9"???? I know. I'm stupid but my brain fried after I accidentally deleted my quiz table with nearly 90 quizzes for my students ... and NO, I was too stupid to have backed it up. So ... I'm not all here right now. But I just can't figure what's going wrong with this script. Thanks!

[Pikachu2000]

date expects a unix timestamp, not a string. It's failing and returning the beginning of the Unix epoch (zero seconds), which works out to Dec 31, 1969.

[Nyuszer]

you should use mktime() to convert your date to unix timestamp

 

$due = mktime(0,0,0,$month,$day);
print date("M j", $due);