$pubdate is probably null or something.
Regards,
–Josh
Joshua Kehn | @joshkehn
http://joshuakehn.com
On Dec 7, 2011, at 1:48 PM, Jack wrote:
Hello All,
I have a problem where Dates are coming out as 12.31.1969 19:00:00 which of
course we didn't
On 12/7/2011 10:48 AM, Jack wrote:
Hello All,
I have a problem where Dates are coming out as 12.31.1969 19:00:00 which of
course we didn't have PC's in 1969
I'm not able to see where the date is getting screwed up, any ideas??
//
How about a little debugging here (and possibly elsewhere):
if (isset($pubdate) ($pubdate 0)) {
$pubdate=strtotime($pubdate);
} else {
die(Barf. Can't run a string to time conversion on 0 or -1.);
}
Thanks Kevin,
This bombs and gives me
How about a little debugging here (and possibly elsewhere):
if (isset($pubdate) ($pubdate 0)) {
$pubdate=strtotime($pubdate);
} else {
die(Barf. Can't run a string to time conversion on 0 or
-1.);
}
Thanks Kevin,
This bombs and gives
To: PHP
Subject: RE: [PHP] Problem with date
How about a little debugging here (and possibly elsewhere):
if (isset($pubdate) ($pubdate 0)) {
$pubdate=strtotime($pubdate);
} else {
die(Barf. Can't run a string to time conversion on 0 or
-1
-Original Message-
From: Jack [mailto:jacklistm...@gmail.com]
Sent: Wednesday, December 07, 2011 1:49 PM
To: PHP
Subject: [PHP] Problem with date
Hello All,
I have a problem where Dates are coming out as 12.31.1969 19:00:00
which of
course we didn't have PC's in 1969
On 12/10/2010 11:52, Rado Oršula wrote:
I do not know good English.
In the attached source code.
Here is erroneous statement:
date: 2010-10-31 00:00:00
date+*0*h: 2010-10-31 *00*:00:00
date+*1*h: 2010-10-31 *01*:00:00
*date+2h: 2010-10-31 02:00:00
date+3h: 2010-10-31 02:00:00 *
date+*4*h:
On 12 October 2010 10:52, Rado Oršula rado.ors...@gmail.com wrote:
I do not know good English.
In the attached source code.
Here is erroneous statement:
date: 2010-10-31 00:00:00
date+0h: 2010-10-31 00:00:00
date+1h: 2010-10-31 01:00:00
date+2h: 2010-10-31 02:00:00
date+3h: 2010-10-31
Adding the number of seconds in a day could fall in the same day due
to daylight saving time, a more reliable way of adding one day (or a
given number of days) is this:
$tomorrow = mktime(0, 0, 0, date(m) , date(d)+1, date(Y));
Then use date() with $tomorrow to format it.
(Taken from example
strtotime would be a neater way of solving the problem...
ie (untested):
$n = 0;
while ($n = 9)
{
$date = d.$n;
$$date = date(Y-m-d, strtotime(+ $n days)
$n++;
}
If you are just enquiring about the maths though, I'm not sure!
-Original Message-
From: Korgan
I have a problem with the date function
?
$theday = date(Y-m-d, time());
echo date((w), $theday);
?
I'm pretty sure the optional second argument for date is a unix timestamp
that you would generate by using either time or mktime or strtotime. You
are passing it something in the form of
i found out that using strtotime did the trick !!
?
strtotime($theday);
$theday = date(Y-m-d, time());
echo date((w), $theday);
?
I have a problem with the date function
?
$theday = date(Y-m-d, time());
echo date((w), $theday);
?
I'm pretty sure the optional second argument for
On Tuesday 03 June 2003 17:40, Jack wrote:
q: how i can do a date subtraction in php? (eg. 2003-06-02 - 3 =
2003-05-30)
strtotime()
--
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications
-Original Message-
From: Gareth Mulholland [mailto:[EMAIL PROTECTED]]
I'm having problems with mktime and Midnight on 30th March 2003.
The code I'm using is:
echo mktime(0,0,1,'03','29','2003');
echo mktime(0,0,1,'03','30','2003');
echo
All parameters whould be integers. Unquote them and use intval to strip the
leading zeros.
On 1/29/03 11:51, Gareth Mulholland [EMAIL PROTECTED] wrote:
I'm having problems with mktime and Midnight on 30th March 2003.
The code I'm using is:
echo mktime(0,0,1,'03','29','2003');
echo
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