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 is being assigned inside the if statement,
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
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 month) where the n
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?
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);
$age = floor(($dif_d/365.24));
echo $age;
?
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
Hi all
I have a database that holds car rental info
DB:
carrental_from (datetime field)
carrental_to (datetime field)
carrental_price (datetime field) [rates are per hour]
The values I have are like:
-00-00 00:00:00,-00-00 07:00:00,10 (all year around 00:00-07:00)
-00-00
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" a valid
parameter?
and yes, I am sure our server is set to
December(todays date).
echo date('F',strtotime("last
month"));
Does give me
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
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));
Thanks for your help
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To unsubscribe, visit: http://www.php.net/unsub.php
* 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
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
Message-
From: Logan McKinley [mailto:[EMAIL PROTECTED]
Sent: Monday, June 23, 2003 5:06 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Date problem
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
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 you can
select the month, and if its the current month
the box is already selected.
select name=month
?
for
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; $i++)
--- Vinesh Hansjee [EMAIL PROTECTED] wrote:
Hi there, can anyone tell me how to fix my code so
that on the last day of
the month, my code doesn't repeat the months...
What the code suppose to do is, makes a drop down
box from which
- Original Message -
From: Vinesh Hansjee [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, March 31, 2003 6:55 AM
Subject: [PHP] Date Problem - Last Day Of Month
Hi there, can anyone tell me how to fix my code so that on the last day of
the month, my code doesn't repeat
hi,
using date(dS); how can i can increase the days so that it shows
19th 20th 21st
I have tried
while ($i 2){
$day++;
echo' td'.$day.'/td';
$i++;
}
but i get:
19th 19ti 19tj
thanks for your help
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PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit:
Hello,
I would ask you a question about date type
if I have a variable from date type ($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
$year=2003
I searched a lot, but I didn't find how to do this.
I'll be very
From: Alexander Tsonev [EMAIL PROTECTED]
Hello,
I would ask you a question about date type
if I have a variable from date type ($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
Hello friends.
Happy New Year to you all.
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.
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
Hi,
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?
Thanks,
Rosen
--
PHP General Mailing List (http://www.php.net/)
To
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 +
to all;
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say it again?
--
Maxim Maletsky
[EMAIL PROTECTED]
On Sun, 10 Nov 2002 12:37:48 +0800 Michael P. Carel [EMAIL PROTECTED] wrote:
to all;
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Hi,
I'm using ImpAkt2 (PHP4, MySql, Apache on W2K)
I have update form (which has date-field). So now I have to enter date on
format yy.mm.dd (ex. today 02.09.27), and in the database has then correct
format -mm-dd.
Ok! I have also listing-page, which has date-field (it format is
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 one hour off from the
server time.
Any ideas of
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
Hi, im making a tab/lyric portal, and for viewing tabs i want to display the
time the lyric/tab was submitted. So I retrive it from a MySQL database (as
a timestamp) and format it using the date function. The problem is, that the
date: 19-01-2038 04:14:07 is allways returned, even though in the
don't
have any more queries based on time or date...
---John Holmes...
-Original Message-
From: andy [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 06, 2002 2:02 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: [PHP] date problem
Hi there,
I would like to count the users out
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
Hi there,
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 '10...' is unix timestamp june
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
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi all
I have a field in MySQL db like this: date TIMESTAMP,
and it looks pretty regular like this: 20020428011911
when I come to use php date(D M Y, $myTimeStamp) though I get
TUE JAN 2038!
I see something in the manual re this date but I'm
:
SELECT DATE_FORMAT(mydate,%e %b %Y) AS thedate FROM mytable
- Original Message -
From: Nick Wilson [EMAIL PROTECTED]
To: php-general [EMAIL PROTECTED]
Sent: Sunday, April 28, 2002 7:48 AM
Subject: [PHP] date problem
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi all
I have a field
-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
hello
i am running into some trouble with a very basic problem.
i want to insert the current date/time into my db. i have
the field set up as a datetime field. when i submit info,
i just get a blank date, i mean it all zeros, like
-00-00 00.00:00:00. how do i insert the current datetime
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
Hi,
I am reading a date from an input in format 'DD-MM-' ex. 10-11-2001.
Now I want to add 3 months to the date. I have tested mktime and strftime
etc and no matter what I do I get the year as 1970. (Systemdate works
fine). How would I go about adding 3 months to a date in that format?
For reference:
OS: OpenBSD 2.9
Web Server: Apache1.3.19
PHP Version: 4.0.6
My problem is that date() and all the other time functions return GMT instead
of localtime. system(date) returns the correct localtime. Those functions
used to return localtime since GMT. The problem seems to have
Hey!
A little bit of definition:
- a week start on sunday and ends on saturday;
- a year has 365/7 ~= 52 weeks;
How do I discover what months have 5 weeks and what have 4?
Rom
. Beer leads to intoxication, intoxication to
hangovers, and hangovers to... suffering.
- Original Message -
From: "Romulo Roberto Pereira" [EMAIL PROTECTED]
To: "php-general" [EMAIL PROTECTED]
Sent: Friday, January 19, 2001 11:07 AM
Subject: [PHP] date problem: some
ckdown.com/
On Thu, 18 Jan 2001, Romulo Roberto Pereira wrote:
Date: Thu, 18 Jan 2001 19:07:31 -0500
From: Romulo Roberto Pereira [EMAIL PROTECTED]
To: php-general [EMAIL PROTECTED]
Subject: [PHP] date problem: some months have 5 weeks... how I discover which
ones?
Hey!
A little bit of
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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