1/3/2012 is in fact less then 9/16/2012.
On Thu, Jan 3, 2013 at 3:57 PM, Marc Fromm marc.fr...@wwu.edu wrote:
I am comparing to dates.
define('WSOFFBEGIN','09/16/2012');
$jes = 01/03/2012;
if ( date(m/d/Y, strtotime($jes)) date(m/d/Y, strtotime(WSOFFBEGIN))
)
{
$error
Hi.
date returns a string
You should compare a different type for bigger/smaller than
HTH
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Microsoft!
They need to add TRUNCATE PARTITION in SQL Server
At 04:57 PM 1/3/2013, Marc Fromm wrote:
I am comparing to dates.
define('WSOFFBEGIN','09/16/2012');
$jes = 01/03/2012;
if ( date(m/d/Y, strtotime($jes)) date(m/d/Y, strtotime(WSOFFBEGIN)) )
{
$error = MUST begin after . WSOFFBEGIN . \n;
}
I cannot figure out why the $error
:05 PM
To: Marc Fromm
Cc: php-general@lists.php.net
Subject: Re: [PHP] date problem
Hi.
date returns a string
You should compare a different type for bigger/smaller than
HTH
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Microsoft!
They need to add
)){
// bla bla
}
Thanks
From: Serge Fonville [mailto:serge.fonvi...@gmail.com]
Sent: Thursday, January 03, 2013 2:05 PM
To: Marc Fromm
Cc: php-general@lists.php.net
Subject: Re: [PHP] date problem
Hi.
date returns a string
You should compare a different type for bigger/smaller
Thanks Jonathan. I removed the date() syntax function and it works.
From: Jonathan Sundquist [mailto:jsundqu...@gmail.com]
Sent: Thursday, January 03, 2013 2:16 PM
To: Marc Fromm
Cc: Serge Fonville; php-general@lists.php.net
Subject: Re: [PHP] date problem
Marc,
When you take a date and do
On 1/3/2013 5:22 PM, Marc Fromm wrote:
Thanks Jonathan. I removed the date() syntax function and it works.
From: Jonathan Sundquist [mailto:jsundqu...@gmail.com]
Sent: Thursday, January 03, 2013 2:16 PM
To: Marc Fromm
Cc: Serge Fonville; php-general@lists.php.net
Subject: Re: [PHP] date problem
On 01/03/2013 01:57 PM, Marc Fromm wrote:
$jes = 01/03/2012;
# php -r echo 01/03/2012;
0.00016567263088138
You might want to put quotes around that value so it is actually a
string and does not get evaluated.
--
Jim Lucas
http://www.cmsws.com/
http://www.cmsws.com/examples/
--
PHP
Just try of March. Worked for me.
print first: .date(d-m-Y H:i:s,strtotime('first Tuesday of March
2011')).\n;
print second: .date(d-m-Y H:i:s,strtotime('second Tuesday of March
2011')).\n;
print third: .date(d-m-Y H:i:s,strtotime('third Tuesday of March
2011')).\n;
print fourth: .date(d-m-Y
It seems different php versions have different outputs for this code:
Fedora Core 14 (x86):
first: 01-03-2011 00:00:00
second: 08-03-2011 00:00:00
third: 22-03-2011 00:00:00
fourth: 22-03-2011 00:00:00
fifth: 29-03-2011 00:00:00
Fedora Core11 (x86_64):
first: 31-12-1969 16:00:00
second:
I removed the day (1 before the March), but its still giving the same
result, i.e. different days of month with and without the 'first'. Any
further help ?
print first Tuesday :.date(d-m-Y H:i:s,strtotime('March 2011
Tuesday')).\n;
print first: .date(d-m-Y H:i:s,strtotime('March 2011 first
Kevin Murphy wrote:
Small issue with formatting a date. If I type in this:
echo date(g:i:s a \o\n l F j, Y);
the n character in the word on doesn't appear, but instead what I
get is a new line in the source code. If I type it as:
echo date(g:i:s a \on l F j, Y);
I get the number 8 (current
On Mon, August 13, 2007 12:50 pm, Kevin Murphy wrote:
Small issue with formatting a date. If I type in this:
echo date(g:i:s a \o\n l F j, Y);
the n character in the word on doesn't appear, but instead what I
get is a new line in the source code. If I type it as:
echo date(g:i:s a \on l F
Philip Hallstrom wrote:
of leap years between the two dates. Leap years occur every 4 years,
and 17 / 4 = 4.25, so there were 4 leap years between 7/6/88 and
7/6/05 and
Just to nitpick... :-)
http://en.wikipedia.org/wiki/Leap_year
The Gregorian calendar adds an extra day to February,
Hey Richard,
Thanks, you've pulled my butt outa the fire again :-)
Cheers,
Ryan
On 7/6/2005 10:59:36 PM, Richard Lynch ([EMAIL PROTECTED]) wrote:
On Wed, July 6, 2005 12:07 pm, Ryan A said:
I'm confused, this should give me the age as 17 instead of 16...but it
does
not...any ideas why?
On Jul 6, 2005, at 2:07 PM, Ryan A wrote:
Hi,
I'm confused, this should give me the age as 17 instead of 16...but it
does
not...any ideas why?
?php print date(Y:m:d);
$age=1988-07-06;
$day1=strtotime($age);
$day2 = strtotime(date(Y-m-d));
$dif_s = ($day2-$day1);
$dif_d =
On Jul 6, 2005, at 2:35 PM, Edward Vermillion wrote:
On Jul 6, 2005, at 2:07 PM, Ryan A wrote:
Hi,
I'm confused, this should give me the age as 17 instead of 16...but
it does
not...any ideas why?
?php print date(Y:m:d);
$age=1988-07-06;
$day1=strtotime($age);
$day2 =
On Wed, July 6, 2005 12:07 pm, Ryan A said:
I'm confused, this should give me the age as 17 instead of 16...but it
does
not...any ideas why?
?php print date(Y:m:d);
$age=1988-07-06;
$day1=strtotime($age);
$day2 = strtotime(date(Y-m-d));
$dif_s = ($day2-$day1);
$dif_d =
Ryan A wrote:
Hi,
I'm confused, this should give me the age as 17 instead of 16...but it does
not...any ideas why?
?php print date(Y:m:d);
$age=1988-07-06;
$day1=strtotime($age);
$day2 = strtotime(date(Y-m-d));
$dif_s = ($day2-$day1);
$dif_d = ($dif_s/60/60/24);
$age =
On Jul 6, 2005, at 3:59 PM, Richard Lynch wrote:
365.24 is an appoximation.
Sooner or later, it's gonna bit you in the butt.
If you want somebody's age accurately, you're probably going to have
to do
it the hard way.
Something like this might work:
?php
$DOB = 1988-07-06;
$now =
of leap years between the two dates. Leap years occur every 4 years, and 17
/ 4 = 4.25, so there were 4 leap years between 7/6/88 and 7/6/05 and
Just to nitpick... :-)
http://en.wikipedia.org/wiki/Leap_year
The Gregorian calendar adds an extra day to February, making it 29 days
long, in
On Jul 6, 2005, at 4:44 PM, Philip Hallstrom wrote:
of leap years between the two dates. Leap years occur every 4 years,
and 17 / 4 = 4.25, so there were 4 leap years between 7/6/88 and
7/6/05 and
Just to nitpick... :-)
http://en.wikipedia.org/wiki/Leap_year
The Gregorian calendar adds
On Jul 6, 2005, at 5:17 PM, Edward Vermillion wrote:
On Jul 6, 2005, at 4:44 PM, Philip Hallstrom wrote:
of leap years between the two dates. Leap years occur every 4
years, and 17 / 4 = 4.25, so there were 4 leap years between 7/6/88
and 7/6/05 and
Just to nitpick... :-)
On Jul 6, 2005, at 5:31 PM, Edward Vermillion wrote:
On Jul 6, 2005, at 5:17 PM, Edward Vermillion wrote:
On Jul 6, 2005, at 4:44 PM, Philip Hallstrom wrote:
of leap years between the two dates. Leap years occur every 4
years, and 17 / 4 = 4.25, so there were 4 leap years between 7/6/88
Try this instead...
echo date('F',strtotime(+1 month));
read through the user comments at...
http://us2.php.net/manual/en/function.strtotime.php
there are some things pertaining to your situtation
-Chris
On Wed, 22 Dec 2004 15:38:38 -0800, PHP [EMAIL PROTECTED] wrote:
echo
Hi,
I did use the +1 month, but I was just curious as to what was up.
Try this instead...
echo date('F',strtotime(+1 month));
read through the user comments at...
http://us2.php.net/manual/en/function.strtotime.php
there are some things pertaining to your situtation
-Chris
On Wed, 22 Dec 2004
Ok,
I was using the download version of the manual and it didn't have any user
comments on this, but the online one does.
Hi,
I did use the +1 month, but I was just curious as to what was up.
Try this instead...
echo date('F',strtotime(+1 month));
read through the user comments at...
try +one month
- Original Message -
From: PHP
To: php
Sent: Wednesday, December 22, 2004 6:38 PM
Subject: [PHP] Date problem?
echo date('F',strtotime(next month));
This is printing February right now. Does this sound right or is this a but
in strtotime()? is next month
* Thus wrote Shaun ([EMAIL PROTECTED]):
Hi,
Why does the following code print '00', surely it should print '08', I'm
baffled!
date(H, mktime(8, 0, 0, 0, 0, 0));
?php
echo date(r, mktime(8,0,0,0,0,0)), \n;
//Wed, 31 Dec 1969 23:59:59 +
echo date(r, mktime(0,0,0,0,0,0)), \n;
//Wed, 31
I am storing dates in an Access database in a field with a Date/Time Type
the date is being generated using date(n/d/Y h:i a). It appears to be
stored in Access correctly but when I output it to the page using PHP it
seems to be changing. It is being stored in the database as 6/19/2003
1:44:00
Hi,
That looks like a unix timestamp. Try this:
Echo date(Y/m/d h:i:s, 1056044640);
www.php.net/date
Check that link to see the lettering codes and the syntax, I
might be off on what I typed up there, but you'll get the concept.
-Dan Joseph
-Original
Hi there, can anyone tell me how to fix my code so that on the last day of
the month, my code doesn't repeat the months...
I'm not sure what you mean by this, but I can be a moron on this list
sometimes and the clear answer usually comes to me about 2 seconds after I
hit the send button. Do you
change
for ($i=1; $i=12; $i++)
to
for ($i=1; $i12; $i++)
--- Vinesh Hansjee [EMAIL PROTECTED] wrote:
Hi there, can anyone tell me how to fix my code so
that on the last day of
the month, my code doesn't repeat the months...
What the code suppose to do is, makes a drop down
box from which
- Original Message -
From: Vinesh Hansjee [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, March 31, 2003 6:55 AM
Subject: [PHP] Date Problem - Last Day Of Month
Hi there, can anyone tell me how to fix my code so that on the last day of
the month, my code doesn't repeat the
From: Alexander Tsonev [EMAIL PROTECTED]
Hello,
I would ask you a question about date type
if I have a variable from date type ($newdate) such as 2003-02-17
how can I separate $newdate into 3 different variables? I want to create
such variables:
$day=17
$month=2
$year=2003
I searched a
At 05:23 PM 12/31/02 +0800, Denis L. Menezes wrote:
Hello friends.
Is there a routine in PHP I can use to find if today's date fits between
the commencing date and the ending date?
SELECT * FROM Table WHERE NOW() = StartDate AND NOW() = EndDate
Rick
--
PHP General Mailing List
Our school holds many seminars of varying durations and dates. I want
to
make a page which says What's on today which will show all the
seminars
that are on today. However the entry in the database will show two
fields
Commencing Date and Ending date.
Is there a routine in PHP I can use to
At 18:44 02/12/2002 -0500, John W. Holmes wrote:
Can you help me for this ?
Yeah, I already did:
Perhaps you could throw a clue my way, in that case.
I have an actor database (mySQL again) which has actor dates of birth in
'-MM-DD' format. I want to be able to query against that
On Tuesday 03 December 2002 18:29, James Coates wrote:
Perhaps you could throw a clue my way, in that case.
I have an actor database (mySQL again) which has actor dates of birth in
'-MM-DD' format. I want to be able to query against that ignoring the
year:
* give me actors who have
At 19:32 03/12/2002 +0800, Jason Wong wrote:
Assuming the column holding the the birthday is called 'birthday':
[snip]
Many thanks!
* give me actors whose birthdays fall between two dates (ie: Aries)
This one is easy, use the BETWEEN clause in your SELECT statement. Consult
manual for
Thanks for this,
I understand how to update in date in database, but I need when I get date
from database to increase or decrease before to save in database.
Can you help me for this ?
John W. Holmes [EMAIL PROTECTED] wrote in message
Thanks for this,
I understand how to update in date in database, but I need when I get
date
from database to increase or decrease before to save in database.
Can you help me for this ?
Yeah, I already did:
You can select out the date you have now, use strtotime() to make it
into a unix
I have one problem:
Date field in MySql database with value as 2002-31-12.
I want to increment or decrement this date and to put it again in
table.
Can someone help me to increment or decrement date with some days?
UPDATE yourtable SET yourcolumn = yourcolumn + INTERVAL 1 DAY WHERE ...
Thanks,
But I need before to save date in database to do some checks with the
inc/dec date.
Cal you help me ?
Thanks,
Rosen
John W. Holmes [EMAIL PROTECTED] wrote in message
002301c29960$21d6a360$7c02a8c0@coconut">news:002301c29960$21d6a360$7c02a8c0@coconut...
I have one problem:
Date field
What exactly do you want to do? I'm not a mind reader...
---John Holmes...
-Original Message-
From: Rosen [mailto:[EMAIL PROTECTED]]
Sent: Sunday, December 01, 2002 12:54 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Date problem
Thanks,
But I need before to save date in database
01c29965$7862e950$7c02a8c0@coconut...
What exactly do you want to do? I'm not a mind reader...
---John Holmes...
-Original Message-
From: Rosen [mailto:[EMAIL PROTECTED]]
Sent: Sunday, December 01, 2002 12:54 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Date problem
Thanks,
on 02/12/02 9:59 AM, Rosen ([EMAIL PROTECTED]) wrote:
I want to get date from database, to increment ot decrement it with some
days, to show the date and after thath
if user confirm it to save it to database.
And in what format is the date currently stored? -MM-DD? MySQL
timestamp? Unix
It's in -MM-YY
Justin French [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
on 02/12/02 9:59 AM, Rosen ([EMAIL PROTECTED]) wrote:
I want to get date from database, to increment ot decrement it with some
days, to show the date and after thath
if user
I want to get date from database, to increment ot decrement it with
some
days, to show the date and after thath
if user confirm it to save it to database.
There are a ton of ways you can do it. You can select the date and it's
inc/dec value in the same statement:
SELECT datecol, datecol +
say it again?
--
Maxim Maletsky
[EMAIL PROTECTED]
On Sun, 10 Nov 2002 12:37:48 +0800 Michael P. Carel [EMAIL PROTECTED] wrote:
to all;
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
On Friday 16 August 2002 17:24, Kae Verens wrote:
when I place date(h:i a) in a page on my server, and view it in a
browser, I am returned: 02:51 pm. According to the server itself,
though (logged in through ssh), using date, it is 3:51. gmdate(h:i
a) returns the same 2:51 time. They are both
So what did you get? Did it work? (No. I doubt it...) Don't just post a
question like Here's what I'm doing...fix it
Anyway, your really wasting your time by not using a date or timestamp
column. If you don't understand why or realize how easy it is to change
it, you've got a lot of learning to
SELECT COUNT(*) AS c
FROM users_table
WHERE UNIX_TIMESTAMP( user_regdate ) '1022882400'
The only way you can do it with a char column is to select the entire
database, load it into a PHP array, using strtotime() to (hopefully)
convert May 29, 2002, etc, into a unix timestamp, and then
SELECT COUNT(*) AS c
FROM users_table
WHERE UNIX_TIMESTAMP( user_regdate ) '1022882400'
The only way you can do it with a char column is to select the
entire
database, load it into a PHP array, using strtotime() to (hopefully)
convert May 29, 2002, etc, into a unix timestamp, and
On Thu, 6 Jun 2002, andy wrote:
I would like to count the users out of a mysql db who registered after a
certain date.
The column I have in the db is a char and I do not want to change this
anymore.
This is how a typical entry looks like: May 29, 2002
This is how I tryed it:
// while
First, the timestamp that is retrieved from mysql is NOT the same as is
required for PHP.
Second, do you REALLY wan the Day of the Week (Mon, Tue, Wed) or the date of
the month (1,2,3,4,5,6,...)?
Third, let Mysql do the conversion for you. For instance if you timestamp
field is named mydate:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
* and then Richard Emery declared
Third, let Mysql do the conversion for you. For instance if you timestamp
field is named mydate:
SELECT DATE_FORMAT(mydate,%e %b %Y) AS thedate FROM mytable
Nice one, cheers Richard.
I didn't know about
On Sunday 28 April 2002 21:10, Nick Wilson wrote:
* and then Richard Emery declared
Third, let Mysql do the conversion for you. For instance if you
timestamp field is named mydate:
SELECT DATE_FORMAT(mydate,%e %b %Y) AS thedate FROM mytable
Nice one, cheers Richard.
I didn't
On 28 Apr 2002 at 14:48, Nick Wilson wrote:
I have a field in MySQL db like this: date TIMESTAMP,
and it looks pretty regular like this: 20020428011911
If you've got the data in your database then do the date/time
conversions as part of your sql query - it's more efficient.
Something like
date() is your answer, use it in the piece of code generating the query.
$DateTime = date(Y-m-d H:i:s); // 2002-02-18 16:10:43
Niklas
-Original Message-
From: eoghan [mailto:[EMAIL PROTECTED]]
Sent: 18. helmikuuta 2002 16:14
To: [EMAIL PROTECTED]
Subject: [PHP] date problem
hello
can't think of exactly the math, but get the weekdayday of the 1st,
divide how many days in the month/7, and it should be easy enuff.
use a combination of date() and mktime() to get the first day of the
month methinks.
Gfunk - http://www.gfunk007.com/
I sense much beer in you.
All months have more than 4 weeks except February (but only when it is
not a leap year).
I'm probably not understanding what exactly you are trying to do.
By your definition then, wouldn't this month only have 3 weeks?
January 2001
Su Mo Tu We Th Fr Sa
1 2 3 4 5 6
7 8 9 10 11
For a newspaper, a week start on sunday and ends in a saturday.
Media planners divide ads in newspapers by monhs and than by weeks,
respectively.
Well, this _seems_ contradictory to your original "definition", but I
got it.
so let's say:
january 2001 started in a monday
Su Mo Tu We
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