PHP mangles my dates

Question

$doba = explode("/", $dob);

$date = date("Y-m-d", mktime(0,0,0, $doba[0], $doba[1], $doba[2]));

The above code turns any date i pass through into 1999-11-30 and i know it was working yesterday. Date is correct when I echo $doba. Anyone have any ideas?

Cheers

You need to show us the values you are passing in as $dob for us to give a definitive answer. — davr
– davr, Commented Oct 29, 2008 at 14:58

jcoby · Accepted Answer · 2008-10-29 17:17:45Z

3

or even easier: $date = date('Y-m-d', strtotime($dob))

answered Oct 29, 2008 at 17:17

jcoby

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Comments

Ross · Accepted Answer · 2008-10-29 13:57:58Z

2

What is the format of $doba? Remember mktime's syntax goes hour, minute, second, month, day year which can be confusing.

Here's some examples:

$doba = explode('/', '1991/08/03');
echo(date('Y-m-d', mktime(0,0,0, $doba[1], $doba[2], $doba[0]);

$doba = explode('/', '03/08/1991');
echo(date('Y-m-d', mktime(0,0,0, $doba[1], $doba[0], $doba[2]);

answered Oct 29, 2008 at 13:57

Ross

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Comments

Andrew Moore · Accepted Answer · 2008-10-29 14:01:59Z

1

It is a bit overkill to use mktime in this case. Assuming $dob is in the following format:

MM/DD/YYYY

you could just to the following to acheive the same result (assuming $dob is always valid):

$doba = explode("/", $dob);
$date = vsprintf('%3$04d-%1$02d-%2$02d', $doba);

answered Oct 29, 2008 at 14:01

Andrew Moore

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Comments

Powerlord · Accepted Answer · 2008-10-29 17:27:50Z

0

If you have issues with what jcoby said above, the strptime() command gives you more control by allowing you to specify the format as well.

answered Oct 29, 2008 at 17:27

Powerlord

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