[PHP] date problem
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,
since the statement should be false. 01/03/2012 is not less than 09/16/2012.
Marc
Re: [PHP] date problem
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 = MUST begin after . WSOFFBEGIN . \n;
}
I cannot figure out why the $error is being assigned inside the if
statement, since the statement should be false. 01/03/2012 is not less than
09/16/2012.
Marc
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 TRUNCATE PARTITION in SQL Server
https://connect.microsoft.com/SQLServer/feedback/details/417926/truncate-partition-of-partitioned-table
2013/1/3 Marc Fromm marc.fr...@wwu.edu
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, since the statement should be false. 01/03/2012 is not less than
09/16/2012.
Marc
Re: [PHP] date problem
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 is being assigned inside the if
statement, since the statement should be false. 01/03/2012 is not
less than 09/16/2012.
You shouldn't be comparing the date strings, but the UNIX timestamp values:
define('WSOFFBEGIN','09/16/2012');
$jes = 01/03/2012;
if ( strtotime($jes) strtotime(WSOFFBEGIN) )
{
$error = MUST begin after . WSOFFBEGIN . \n;
}
Ken
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RE: [PHP] date problem
Thanks for the reply.
Every example on comparing dates in PHP that I found uses the strtotime
function which I am using. What other type can I use?
When is this example below supposed to work?
// your first date coming from a mysql database (date fields)
$dateA = '2008-03-01 13:34';
// your second date coming from a mysql database (date fields)
$dateB = '2007-04-14 15:23';
if(strtotimehttp://www.php.net/strtotime($dateA)
strtotimehttp://www.php.net/strtotime($dateB)){
// 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 than
HTH
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Microsoft!
They need to add TRUNCATE PARTITION in SQL Server
https://connect.microsoft.com/SQLServer/feedback/details/417926/truncate-partition-of-partitioned-table
2013/1/3 Marc Fromm marc.fr...@wwu.edumailto:marc.fr...@wwu.edu
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,
since the statement should be false. 01/03/2012 is not less than 09/16/2012.
Marc
Re: [PHP] date problem
Marc,
When you take a date and do a strtotime you are converting it to an int
which you can compare to each other much easier. So for your above example
you would be best doing.
define('WSOFFBEGIN','09/16/2012');
$jes = 01/03/2012;
if ( strtotime($jes) strtotime(WSOFFBEGIN) )
{
$error = MUST begin after . WSOFFBEGIN . \n;
}
On Thu, Jan 3, 2013 at 4:12 PM, Marc Fromm marc.fr...@wwu.edu wrote:
Thanks for the reply.
Every example on comparing dates in PHP that I found uses the strtotime
function which I am using. What other type can I use?
When is this example below supposed to work?
// your first date coming from a mysql database (date fields)
$dateA = '2008-03-01 13:34';
// your second date coming from a mysql database (date fields)
$dateB = '2007-04-14 15:23';
if(strtotimehttp://www.php.net/strtotime($dateA) strtotime
http://www.php.net/strtotime($dateB)){
// 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 than
HTH
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Microsoft!
They need to add TRUNCATE PARTITION in SQL Server
https://connect.microsoft.com/SQLServer/feedback/details/417926/truncate-partition-of-partitioned-table
2013/1/3 Marc Fromm marc.fr...@wwu.edumailto:marc.fr...@wwu.edu
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, since the statement should be false. 01/03/2012 is not less than
09/16/2012.
Marc
RE: [PHP] date problem
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 a strtotime you are converting it to an int which
you can compare to each other much easier. So for your above example you would
be best doing.
define('WSOFFBEGIN','09/16/2012');
$jes = 01/03/2012;
if ( strtotime($jes) strtotime(WSOFFBEGIN) )
{
$error = MUST begin after . WSOFFBEGIN . \n;
}
On Thu, Jan 3, 2013 at 4:12 PM, Marc Fromm
marc.fr...@wwu.edumailto:marc.fr...@wwu.edu wrote:
Thanks for the reply.
Every example on comparing dates in PHP that I found uses the strtotime
function which I am using. What other type can I use?
When is this example below supposed to work?
// your first date coming from a mysql database (date fields)
$dateA = '2008-03-01 13:34';
// your second date coming from a mysql database (date fields)
$dateB = '2007-04-14 15:23';
if(strtotimehttp://www.php.net/strtotime($dateA)
strtotimehttp://www.php.net/strtotime($dateB)){
// bla bla
}
Thanks
From: Serge Fonville
[mailto:serge.fonvi...@gmail.commailto:serge.fonvi...@gmail.com]
Sent: Thursday, January 03, 2013 2:05 PM
To: Marc Fromm
Cc: php-general@lists.php.netmailto: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 TRUNCATE PARTITION in SQL Server
https://connect.microsoft.com/SQLServer/feedback/details/417926/truncate-partition-of-partitioned-table
2013/1/3 Marc Fromm
marc.fr...@wwu.edumailto:marc.fr...@wwu.edumailto:marc.fr...@wwu.edumailto:marc.fr...@wwu.edu
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,
since the statement should be false. 01/03/2012 is not less than 09/16/2012.
Marc
Re: [PHP] date problem
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
Marc,
When you take a date and do a strtotime you are converting it to an int which
you can compare to each other much easier. So for your above example you would
be best doing.
define('WSOFFBEGIN','09/16/2012');
$jes = 01/03/2012;
if ( strtotime($jes) strtotime(WSOFFBEGIN) )
{
$error = MUST begin after . WSOFFBEGIN . \n;
}
On Thu, Jan 3, 2013 at 4:12 PM, Marc Fromm
marc.fr...@wwu.edumailto:marc.fr...@wwu.edu wrote:
Thanks for the reply.
Every example on comparing dates in PHP that I found uses the strtotime
function which I am using. What other type can I use?
When is this example below supposed to work?
// your first date coming from a mysql database (date fields)
$dateA = '2008-03-01 13:34';
// your second date coming from a mysql database (date fields)
$dateB = '2007-04-14 15:23';
if(strtotimehttp://www.php.net/strtotime($dateA)
strtotimehttp://www.php.net/strtotime($dateB)){
// bla bla
}
Thanks
From: Serge Fonville
[mailto:serge.fonvi...@gmail.commailto:serge.fonvi...@gmail.com]
Sent: Thursday, January 03, 2013 2:05 PM
To: Marc Fromm
Cc: php-general@lists.php.netmailto: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 TRUNCATE PARTITION in SQL Server
https://connect.microsoft.com/SQLServer/feedback/details/417926/truncate-partition-of-partitioned-table
2013/1/3 Marc Fromm
marc.fr...@wwu.edumailto:marc.fr...@wwu.edumailto:marc.fr...@wwu.edumailto:marc.fr...@wwu.edu
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,
since the statement should be false. 01/03/2012 is not less than 09/16/2012.
Marc
And hopefully put quotes around 01/03/2012.
--
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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 General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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 H:i:s,strtotime('fourth Tuesday of March
2011')).\n;
print fifth: .date(d-m-Y H:i:s,strtotime('fifth Tuesday of March
2011')).\n;
2011/4/2 Dan Dan dani.mani...@gmail.com:
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
Tuesday')).\n;
print second: .date(d-m-Y H:i:s,strtotime('March 2011 second
Tuesday')).\n;
print third: .date(d-m-Y H:i:s,strtotime('March 2011 third
Tuesday')).\n;
print fourth: .date(d-m-Y H:i:s,strtotime('March 2011 fourth
Tuesday')).\n;
first Tuesday :01-03-2011 00:00:00
first: 08-03-2011 00:00:00
second: 15-03-2011 00:00:00
third: 22-03-2011 00:00:00
fourth: 29-03-2011 00:00:00
Thanks
-dani
On Fri, Apr 1, 2011 at 9:37 AM, Daniel Brown danbr...@php.net wrote:
On Fri, Apr 1, 2011 at 12:35, Dan Dan dani.mani...@gmail.com wrote:
Hi Folks,
I am trying to get the day of month for a particular day of week (e.g.
Tuesday) for the first, second, third, fourth week in a month. The code i
have seems issues in March, but works e.g. in April:
print date(d-m-Y H:i:s,strtotime('1 March 2011 Tuesday'));
01-03-2011 00:00:00
print date(d-m-Y H:i:s,strtotime('1 March 2011 first Tuesday'));
08-03-2011 00:00:00
While in April, I have
print date(d-m-Y H:i:s,strtotime('1 April 2011 Tuesday'));
05-04-2011 00:00:00
print date(d-m-Y H:i:s,strtotime('1 April 2011 first Tuesday'));
05-04-2011 00:00:00
Could someone help whats wrong with the technique i am trying to find
that
day of month. Is there any better way ?
Because you're combining the date with the day of the week. It so
happens that 1 March was a Tuesday, but today - 1 April - is a Friday.
Pick one or the other, not both.
--
/Daniel P. Brown
Network Infrastructure Manager
http://www.php.net/
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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: 31-12-1969 16:00:00
third: 22-03-2011 00:00:00
fourth: 31-12-1969 16:00:00
fifth: 31-12-1969 16:00:00
However it works if reorder Year and Month like:
echo first: .date(d-m-Y H:i:s,strtotime('2011 March first
wednesday')).\n;
echo second: .date(d-m-Y H:i:s,strtotime('2011 March second
wednesday')).\n;
echo third: .date(d-m-Y H:i:s,strtotime('2011 March third
wednesday')).\n;
echo fourth: .date(d-m-Y H:i:s,strtotime('2011 March fourth
wednesday')).\n;
echo fifth: .date(d-m-Y H:i:s,strtotime('2011 March fifth
wednesday')).\n;
first: 02-03-2011 00:00:00
second: 09-03-2011 00:00:00
third: 16-03-2011 00:00:00
fourth: 23-03-2011 00:00:00
fifth: 30-03-2011 00:00:00
Thanks
-dani
On Sat, Apr 2, 2011 at 1:17 AM, Louis Huppenbauer
louis.huppenba...@gmail.com wrote:
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 H:i:s,strtotime('fourth Tuesday of March
2011')).\n;
print fifth: .date(d-m-Y H:i:s,strtotime('fifth Tuesday of March
2011')).\n;
2011/4/2 Dan Dan dani.mani...@gmail.com:
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
Tuesday')).\n;
print second: .date(d-m-Y H:i:s,strtotime('March 2011 second
Tuesday')).\n;
print third: .date(d-m-Y H:i:s,strtotime('March 2011 third
Tuesday')).\n;
print fourth: .date(d-m-Y H:i:s,strtotime('March 2011 fourth
Tuesday')).\n;
first Tuesday :01-03-2011 00:00:00
first: 08-03-2011 00:00:00
second: 15-03-2011 00:00:00
third: 22-03-2011 00:00:00
fourth: 29-03-2011 00:00:00
Thanks
-dani
On Fri, Apr 1, 2011 at 9:37 AM, Daniel Brown danbr...@php.net wrote:
On Fri, Apr 1, 2011 at 12:35, Dan Dan dani.mani...@gmail.com wrote:
Hi Folks,
I am trying to get the day of month for a particular day of week (e.g.
Tuesday) for the first, second, third, fourth week in a month. The
code i
have seems issues in March, but works e.g. in April:
print date(d-m-Y H:i:s,strtotime('1 March 2011 Tuesday'));
01-03-2011 00:00:00
print date(d-m-Y H:i:s,strtotime('1 March 2011 first Tuesday'));
08-03-2011 00:00:00
While in April, I have
print date(d-m-Y H:i:s,strtotime('1 April 2011 Tuesday'));
05-04-2011 00:00:00
print date(d-m-Y H:i:s,strtotime('1 April 2011 first Tuesday'));
05-04-2011 00:00:00
Could someone help whats wrong with the technique i am trying to find
that
day of month. Is there any better way ?
Because you're combining the date with the day of the week. It so
happens that 1 March was a Tuesday, but today - 1 April - is a Friday.
Pick one or the other, not both.
--
/Daniel P. Brown
Network Infrastructure Manager
http://www.php.net/
Re: [PHP] date problem
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
Tuesday')).\n;
print second: .date(d-m-Y H:i:s,strtotime('March 2011 second
Tuesday')).\n;
print third: .date(d-m-Y H:i:s,strtotime('March 2011 third
Tuesday')).\n;
print fourth: .date(d-m-Y H:i:s,strtotime('March 2011 fourth
Tuesday')).\n;
first Tuesday :01-03-2011 00:00:00
first: 08-03-2011 00:00:00
second: 15-03-2011 00:00:00
third: 22-03-2011 00:00:00
fourth: 29-03-2011 00:00:00
Thanks
-dani
On Fri, Apr 1, 2011 at 9:37 AM, Daniel Brown danbr...@php.net wrote:
On Fri, Apr 1, 2011 at 12:35, Dan Dan dani.mani...@gmail.com wrote:
Hi Folks,
I am trying to get the day of month for a particular day of week (e.g.
Tuesday) for the first, second, third, fourth week in a month. The code i
have seems issues in March, but works e.g. in April:
print date(d-m-Y H:i:s,strtotime('1 March 2011 Tuesday'));
01-03-2011 00:00:00
print date(d-m-Y H:i:s,strtotime('1 March 2011 first Tuesday'));
08-03-2011 00:00:00
While in April, I have
print date(d-m-Y H:i:s,strtotime('1 April 2011 Tuesday'));
05-04-2011 00:00:00
print date(d-m-Y H:i:s,strtotime('1 April 2011 first Tuesday'));
05-04-2011 00:00:00
Could someone help whats wrong with the technique i am trying to find
that
day of month. Is there any better way ?
Because you're combining the date with the day of the week. It so
happens that 1 March was a Tuesday, but today - 1 April - is a Friday.
Pick one or the other, not both.
--
/Daniel P. Brown
Network Infrastructure Manager
http://www.php.net/
[PHP] Date Problem with \n
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 is supposed to be. Is there any way to get an n in there? -- Kevin Murphy Webmaster: Information and Marketing Services Western Nevada College www.wnc.edu 775-445-3326 P.S. Please note that my e-mail and website address have changed from wncc.edu to wnc.edu.
Re: [PHP] Date Problem with \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 month) where the n is supposed to be. Is there any way to get an n in there? Use single quotes, then your \n will not be converted to a NL char -- Jim Lucas Some men are born to greatness, some achieve greatness, and some have greatness thrust upon them. Twelfth Night, Act II, Scene V by William Shakespeare -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date Problem with \n
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 j, Y); I get the number 8 (current month) where the n is supposed to be. Is there any way to get an n in there? As noted, apostrophes will work in this case. If you need to embed a variable, however, you may want to re-read this from the manual a couple times: You can prevent a recognized character in the format string from being expanded by escaping it with a preceding backslash. If the character with a backslash is already a special sequence, you may need to also escape the backslash. So, for example: date(g:i:s a \o\\n l F j, Y); Here's what happens: PHP's string parser eats the \\ and turns it into a single back-slash: \ *THEN* you have \n being passed to the C date() function, which then eats the \n and says: This is then not the n for Numeric representation of a month, without leading zeros but just a regular old n as text. -- Some people have a gift link here. Know what I want? I want you to buy a CD from some indie artist. http://cdbaby.com/browse/from/lynch Yeah, I get a buck. So? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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, making it 29 days long, in years where the quotient has no remainder when divided by 4, excluding years where the quotient has no remainder when divided by 100, but including years where the quotient has no remainder when divided by 400. So 1996, 2000, and 2400 are leap years but 1800, 1899, 1900 and 2100 are not. Touche! One more reason not to use the OP's method but rather the one posted by Richard Lynch. kgt -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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?
?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;
?
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;
$now = date('Y-m-d');
list($by, $bm, $bd) = explode('-', $DOB);
list($ny, $nm, $nd) = explode('-', $now);
$age = $ny - $by;
if ($age){
if ($bm $nm){
// do nothing, they were born before this month
// so subtracting the years is correct
}
elseif ($nm $bm){
$age--; //They were born in a later month, so not quite a year yet
}
else{
//They were born this month. Count the days.
if ($bd $nd){
//Do nothing. They were born before this date.
}
elseif ($nd $bd){
$age--; //They were born later this month, so not quite a year yet
}
else{
//It's their birthday! Go ahead and count it as a year,
//unless you want to futz with the hour of their birth...
}
}
}
if (!$age){
//Do whatever you want with children less than 1 year old...
//return it as X months or just call it infant or pretend they're 1
//or whatever.
//Maybe even just use $age (which is 0) and call it done.
}
?
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[PHP] date() problem
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; ? Thanks, Ryan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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); $dif_d = ($dif_s/60/60/24); $age = floor(($dif_d/365.24)); echo $age; ? Thanks, Ryan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php If I change $day2 to =time(); instead of strtotime() I get 17. Could be some weirdness going on with using date()?. Edward Vermillion [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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 = strtotime(date(Y-m-d)); $dif_s = ($day2-$day1); $dif_d = ($dif_s/60/60/24); $age = floor(($dif_d/365.24)); echo $age; ? Thanks, Ryan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php If I change $day2 to =time(); instead of strtotime() I get 17. Could be some weirdness going on with using date()?. Actually it's in the math... With the numbers you have here $dif_d would need to be 6209.24/25 to get up to 17, so the fault lies in the 365.24/25 assumption for a day. Don't ask me how to get around it... it makes my head hurt. :D Edward Vermillion [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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);
$dif_d = ($dif_s/60/60/24);
$age = floor(($dif_d/365.24));
echo $age;
?
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;
$now = date('Y-m-d');
list($by, $bm, $bd) = explode('-', $DOB);
list($ny, $nm, $nd) = explode('-', $now);
$age = $ny - $by;
if ($age){
if ($bm $nm){
// do nothing, they were born before this month
// so subtracting the years is correct
}
elseif ($nm $bm){
$age--; //They were born in a later month, so not quite a year yet
}
else{
//They were born this month. Count the days.
if ($bd $nd){
//Do nothing. They were born before this date.
}
elseif ($nd $bd){
$age--; //They were born later this month, so not quite a year yet
}
else{
//It's their birthday! Go ahead and count it as a year,
//unless you want to futz with the hour of their birth...
}
}
}
if (!$age){
//Do whatever you want with children less than 1 year old...
//return it as X months or just call it infant or pretend they're 1
//or whatever.
//Maybe even just use $age (which is 0) and call it done.
}
?
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Re: [PHP] date() problem
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); $age = floor(($dif_d/365.24)); echo $age; ? Thanks, Ryan Subtracting the timestamps gives you 536457600 seconds, which is correct, but ( $dif_s / 60 / 60 / 24 ) gives you the actual number of days between these two dates. We say on *average* there are 365.25 days in a year, so 17 years (the correct age in years between these two dates) has about 17 * 365.25 = 6209.25 days. In actuality, there are 6209 days between these two dates. This seems close, but note that 6209.25 / 365.25 = 17 but 6209 / 365.25 = 16.9993155 And of course floor on 16.9993155 is still just 16. I don't think you can accurately calculate dates in this way without calculating the actual number 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 therefore 365 * 17 + 4 = 6209 days, the actual number of days between these two dates. But note that ( 6209 - 4 ) / 365 = 17, which is correct. If you subtract the number of extra days due to leap years, then there are precisely 365 days in each year. It boils down to the fact that though there are on *average* 365.25 days in a calendar year, there are never *actually* 365.25 days in a calendar year. There are only either 365 or 366. kgt -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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;
$now = date('Y-m-d');
list($by, $bm, $bd) = explode('-', $DOB);
list($ny, $nm, $nd) = explode('-', $now);
$age = $ny - $by;
if ($age){
if ($bm $nm){
// do nothing, they were born before this month
// so subtracting the years is correct
}
elseif ($nm $bm){
$age--; //They were born in a later month, so not quite a year
yet
}
else{
//They were born this month. Count the days.
if ($bd $nd){
//Do nothing. They were born before this date.
}
elseif ($nd $bd){
$age--; //They were born later this month, so not quite a year
yet
}
else{
//It's their birthday! Go ahead and count it as a year,
//unless you want to futz with the hour of their birth...
}
}
}
if (!$age){
//Do whatever you want with children less than 1 year old...
//return it as X months or just call it infant or pretend
they're 1
//or whatever.
//Maybe even just use $age (which is 0) and call it done.
}
?
Actually, I was thinking it could be done inside a simple formula since
his original formula works on the leap years, ie. years divisible
by four. I guess the trick would be figuring out the actual seconds in
a year and doing most of the calculations on the unix timestamp.
Something is tickling the back of my brain on this but I can't see it
just yet. Eh...
Edward Vermillion
[EMAIL PROTECTED]
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Re: [PHP] date() problem
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 years where the quotient has no remainder when divided by 4, excluding years where the quotient has no remainder when divided by 100, but including years where the quotient has no remainder when divided by 400. So 1996, 2000, and 2400 are leap years but 1800, 1899, 1900 and 2100 are not. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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 an extra day to February, making it 29 days long, in years where the quotient has no remainder when divided by 4, excluding years where the quotient has no remainder when divided by 100, but including years where the quotient has no remainder when divided by 400. So 1996, 2000, and 2400 are leap years but 1800, 1899, 1900 and 2100 are not. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php I always wondered what kind of drugs those guys were on when they came up with the leap-year system... :P One interesting side note to the op's problem, since I'm not going to be able to do anything else till I figure this out now..., if one of the dates is a leap year the 365.25 works fine, which make me wonder how the strtotime() function is working... or maybe that's what it's supposed to do... ;) Edward Vermillion [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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... :-) http://en.wikipedia.org/wiki/Leap_year The Gregorian calendar adds an extra day to February, making it 29 days long, in years where the quotient has no remainder when divided by 4, excluding years where the quotient has no remainder when divided by 100, but including years where the quotient has no remainder when divided by 400. So 1996, 2000, and 2400 are leap years but 1800, 1899, 1900 and 2100 are not. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php I always wondered what kind of drugs those guys were on when they came up with the leap-year system... :P One interesting side note to the op's problem, since I'm not going to be able to do anything else till I figure this out now..., if one of the dates is a leap year the 365.25 works fine, which make me wonder how the strtotime() function is working... or maybe that's what it's supposed to do... ;) But then, even if I do figure this out, it's still going to tell me I'm only 35... which is a bit off... As usual Richard's way is probably the best for any real world age deduction. ;) Edward Vermillion [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date() problem
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 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 years where the quotient has no remainder when divided by 4, excluding years where the quotient has no remainder when divided by 100, but including years where the quotient has no remainder when divided by 400. So 1996, 2000, and 2400 are leap years but 1800, 1899, 1900 and 2100 are not. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php I always wondered what kind of drugs those guys were on when they came up with the leap-year system... :P One interesting side note to the op's problem, since I'm not going to be able to do anything else till I figure this out now..., if one of the dates is a leap year the 365.25 works fine, which make me wonder how the strtotime() function is working... or maybe that's what it's supposed to do... ;) But then, even if I do figure this out, it's still going to tell me I'm only 35... which is a bit off... As usual Richard's way is probably the best for any real world age deduction. ;) Well, I'll let you test through it but it seems to work with the two or three dates I ran through... ?php $bd=1988-07-07; // birth date $td=2005-07-06; // today $day1 = strtotime($bd); $day2 = strtotime($td); $sy = 31536000; // short year 365 days $spd = 86400; // seconds per day $realAge = floor(($day2-$day1-($spd*floor((($day2 - $day1)/$sy)/4)))/$sy); print $realAge; exit; ? I left $sy and $spd in so you could see where those values were coming from but you could just as well stick the numbers in for the vars and only have $day1 and $day2 hanging around. of course it only works on most systems with birth dates before the epoch... :P Edward Vermillion [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] date problem
Hi all I have a database that holds car rental info DB: carrental_from (datetime field) carrental_to (datetime field) carrental_price (datetime field) [rates are per hour] The values I have are like: -00-00 00:00:00,-00-00 07:00:00,10 (all year around 00:00-07:00) -00-00 07:00:00,-00-00 00:00:00,20 (all year around 07:00-00:00) 2005-12-22 07:00:00,2006-01-02 00:00:00,15 (christmas period 00:00-07:00) 2005-12-22 00:00:00,2006-01-02 07:00:00,25 (christmas period 07:00-00:00) The user selects dates ($from - $to) to rent a car and he gets the price accordingly. I can do a (($to-$from)/60/60) and get the total number of hours but depending on the date and time you get a different result. Can anyone help with the SQL? Thanks in advance Mario -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[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" 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 November as it should
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Re: [PHP] Date problem?
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 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 November as it should
No virus found in this outgoing message.
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Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
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Re: [PHP] Date problem?
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 15:38:38 -0800, PHP [EMAIL PROTECTED] wrote:
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 November as it should
No virus found in this outgoing message.
Checked by AVG Anti-Virus.
Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
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Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
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Re: [PHP] Date problem?
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...
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 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 November as it should
No virus found in this outgoing message.
Checked by AVG Anti-Virus.
Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
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No virus found in this incoming message.
Checked by AVG Anti-Virus.
Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
--
No virus found in this outgoing message.
Checked by AVG Anti-Virus.
Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
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Version: 7.0.298 / Virus Database: 265.6.4 - Release Date: 12/22/2004
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Re: [PHP] Date problem?
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 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 November as it should
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[PHP] date problem
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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
* 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 Dec 1969 23:59:59 + echo date(r, mktime(8,0,0,1,1,0)), \n; //Sat, 1 Jan 2000 08:00:00 + ? kinda wierd, I say blame it on the y1969 bug. make sure your day and month don't go backwards, that seems to be what the problem is. Curt -- I used to think I was indecisive, but now I'm not so sure. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[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 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 PM but being displayed on the page as 1056044640 Thanks in advance, ~Logan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [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 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 PM but being displayed on the page as 1056044640 Thanks in advance, That's a timestamp - it's the number of seconds since the Unix Epoch (1/1/1970 GMT), and you can read up about conversion here: php.net/date - michal migurski- contact info and pgp key: sf/cahttp://mike.teczno.com/contact.html -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 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 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 PM but being displayed on the page as 1056044640 Thanks in advance, ~Logan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[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 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 ($i=1; $i=12; $i++) {
if ($i == date(m)) {
print option selected value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
} else {
print option value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
}
}
?
/select
Thanks
Vinesh
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Re: [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 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 mean that you don't want any months in the
SELECT control when it's the last of the month? Or you don't want the
SELECTED attribute in the OPTION tags? As far as I can tell, you're not
repeating any months, but you're only listing them once.
Try this (and IMHO a *slight* *slight* improvement in clarity, but again,
that's my HO):
select name=month
?
for ($counter=1; $counter=12; $counter++) {
print option value=\. $counter . \;
if (($counter == date(m)) || (date(d) date(t))) {
print selected;
}
print . date(F,mktime(0,0,0,$counter,date(d),date(Y))) .
/option;
}
}
/select
I adjusted it so that you only print out the SELECTED attribute if it's the
current month, and I called $i as $counter, because, to me anyway, it makes
it clearer what is the $counter. If $i works for you, though, there's
nothing wrong with that, and it's not that it's unclear anyway, what with it
in the FOR loop only a few lines earlier. (I'll shut up now about $counter.)
Anyway, I added a comparison above that said if the current date is less
than the last day of the month, then print SELECTED. That's assuming I have
your question right in that you don't want to select the current month if
it's the last day of that month. Otherwise, if I'm wrong, you can take that
comparison and put it just about anywhere.
Help? Doesn't? Am I a complete goof?
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Re: [PHP] Date Problem - Last Day Of Month
change
for ($i=1; $i=12; $i++)
to
for ($i=1; $i12; $i++)
--- Vinesh Hansjee [EMAIL PROTECTED] wrote:
Hi there, can anyone tell me how to fix my code so
that on the last day of
the month, my code doesn't repeat the months...
What the code suppose to do is, makes a drop down
box from which you can
select the month, and if its the current month
the box is already selected.
select name=month
?
for ($i=1; $i=12; $i++) {
if ($i == date(m)) {
print option selected value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) .
/option;
} else {
print option value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) .
/option;
}
}
?
/select
Thanks
Vinesh
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Re: [PHP] Date Problem - Last Day Of Month
- 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 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 ($i=1; $i=12; $i++) {
if ($i == date(m)) {
print option selected value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
} else {
print option value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
}
}
?
/select
Thanks
Vinesh
Vinesh, there's a fundemental flaw in your logic. The date(d) input will
fail to produce a reliable list if it is executed on any day after the 28th
of the month. Since today is the 31st obviously this is going to cause some
problems becuase any month that ends on the 30th or earlier will return plus
one the month. So February will return March, April willl return May, and
so on. If you wait until tommorrow (the 1st) your function will magically
work. :)
One way to fix this is to build your own array of month names and reference
that instead of using the date() function.
$months = array(1=January, 2=February, 3=March, 4=April,
5=May, 6= ... and so on...);
for($i=1; $i=12; $i++)
{
$selected = ;
if ($i == date(m));{$selected = SELECTED;}
echo option $selected value=\${months[$i]}\\n;
}
HTH,
Kevin
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[PHP] date problem
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';
$i++;
}
but i get:
19th 19ti 19tj
thanks for your help
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[PHP] date problem
Hello, I would ask you a question about date type if I have a variable from date type ($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 $year=2003 I searched a lot, but I didn't find how to do this. I'll be very happy if someone helps! Thanks in advance, Alexander Tsonev -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
From: Alexander Tsonev [EMAIL PROTECTED] Hello, I would ask you a question about date type if I have a variable from date type ($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 $year=2003 I searched a lot, but I didn't find how to do this. I'll be very happy if someone helps! list($year, $month, $day) = explode(-, $newdate); -- Lowell Allen -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Date problem
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. Is there a routine in PHP I can use to find if today's date fits between the commencing date and the ending date? Thanks Denis
Re: [PHP] Date problem
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 (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 find if today's date fits between the commencing date and the ending date? SELECT * FROM table WHERE CURRENT_DATE BETWEEN commence_date AND ending_date ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 ignoring the year: * give me actors who have birthdays in the next five days * give me actors whose birthdays fall between two dates (ie: Aries) I've got pretty much everything else working but this one's *really* playing with my head. :-( I know I should've employed a db engineer! :) James. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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 birthdays in the next five days Assuming the column holding the the birthday is called 'birthday': a) you have to work out the date in 5 days time: DATE_ADD(CURRENT_DATE, INTERVAL 5 DAY) b) You have to transform the birthdays so that they are for the current year (so instead of 1981-04-01 it becomes 2002-04-01) CONCAT(YEAR(CURRENT_DATE), '-', MONTH(birthday), '-', DAYOFMONTH(birthday)) c) Combine both into your query: SELECT *, CONCAT(YEAR(CURRENT_DATE), '-', MONTH(birthday), '-', DAYOFMONTH(birthday)) as current_birthday FROM table WHERE current_birthday = CURRENT_DATE AND current_birthday = DATE_ADD(CURRENT_DATE, INTERVAL 5 DAY) ** Untested ** use with extreme caution. * 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 details. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Intuition, however illogical, is recognized as a command prerogative. -- Kirk, Obsession, stardate 3620.7 */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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 details. Ignoring the year, obviously, to get my Aries. :-) James. -- -- [EMAIL PROTECTED] - http://www.affection.net/~jamesc/ - [+33] 6 79 02 08 99 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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 001501c2999b$d24a71f0$7c02a8c0@coconut">news:001501c2999b$d24a71f0$7c02a8c0@coconut... 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 + INTERVAL 1 DAY FROM yourtable WHERE ... Display whatever you need, if the user agrees to the new day, then issue an update query: UPDATE yourtable SET datecol = datecol + INTERVAL 1 DAY WHERE ... To make those queries dynamic, you can replace the '1' with a variable and assign it's value in PHP to either -1, 1, 2, 3, etc... $inc = -1; UPDATE yourtable SET datecol = datecol + INTERVAL $inc DAY WHERE ... Or... You can select out the date you have now, use strtotime() to make it into a unix timestamp (which PHP works with), and date() to format it however you want. If the user approves the new date, you can reformat the unix timestamp back to a -MM-DD format with date() or use FROM_UNIXTIME() in your query to insert/update the new date into the database... ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 timestamp (which PHP works with), and date() to format it however you want. If the user approves the new date, you can reformat the unix timestamp back to a -MM-DD format with date() or use FROM_UNIXTIME() in your query to insert/update the new date into the database... ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Date problem
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 unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 ... http://www.mysql.com/documentation/mysql/bychapter/manual_Reference.html #Date_and_time_functions ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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 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 ... http://www.mysql.com/documentation/mysql/bychapter/manual_Reference.html #Date_and_time_functions ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 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 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 ... http://www.mysql.com/documentation/mysql/bychapter/manual_Reference.html #Date_and_time_functions ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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. Can you help me ? Thanks, Rosen John W. Holmes [EMAIL PROTECTED] wrote in message 002601c29965$7862e950$7c02a8c0@coconut">news:002601c29965$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, 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 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 ... http://www.mysql.com/documentation/mysql/bychapter/manual_Reference.html #Date_and_time_functions ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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 Timestamp? Justin French http://Indent.com.au Web Development Graphic Design -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Date problem
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 confirm it to save it to database. And in what format is the date currently stored? -MM-DD? MySQL timestamp? Unix Timestamp? Justin French http://Indent.com.au Web Development Graphic Design -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Date problem
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 + INTERVAL 1 DAY FROM yourtable WHERE ... Display whatever you need, if the user agrees to the new day, then issue an update query: UPDATE yourtable SET datecol = datecol + INTERVAL 1 DAY WHERE ... To make those queries dynamic, you can replace the '1' with a variable and assign it's value in PHP to either -1, 1, 2, 3, etc... $inc = -1; UPDATE yourtable SET datecol = datecol + INTERVAL $inc DAY WHERE ... Or... You can select out the date you have now, use strtotime() to make it into a unix timestamp (which PHP works with), and date() to format it however you want. If the user approves the new date, you can reformat the unix timestamp back to a -MM-DD format with date() or use FROM_UNIXTIME() in your query to insert/update the new date into the database... ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] date problem
to all; -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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
[PHP] Date-problem?
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 dd.mm. (finland or germany). When I'll goto update form, date-field format is dd.mm.. But when I'll accept form, everything goes wrong I'll get strange date. In the listing-page date-format is now 20.09.2027. How I can correct this problem? Can I use Impakt UniformVal and RegExp-field or what I have do to. This is quite difficult problem to explain! gustavus -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] date problem
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 a solution to this? Anyone have a ready-to-use function which can add/subtract an hour or two to/from the date/time? -- Kae Verens http://www.contactjuggling.org/users/kverens/ http://www.contactjuggling.org/ (webmaster) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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 one hour off from the server time. Any ideas of a solution to this? Anyone have a ready-to-use function which can add/subtract an hour or two to/from the date/time? mktime() -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Since a politician never believes what he says, he is surprised when others believe him. -- Charles DeGaulle */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Date() Problem
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 `date` field the timestamp says (as an actual example) 20020723200919. Here is a shortened version of the script: $link = mysql_connect($dbhost, $dbuser, $dbpass); mysql_select_db($dbname, $link); $get_resource = mysql_query(SELECT `artist_id`,`title`,`content`,`user_id`,`date`,`type`,`views` FROM `resources` WHERE `id` = $id); $values = mysql_fetch_row($get_resource); ?php $submitdate = date(d-m-Y H:i:s, $values[4]); echo(Submitted on $submitdate); ? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] date problem
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 do. 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 sort by that timestamp. You just make things more difficult with a char column. I hope you 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 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 1, 02 SELECT COUNT(*) AS c FROM users_table WHERE UNIX_TIMESTAMP( user_regdate ) '1022882400' Thanx for any help on that, andy -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] date problem
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 sort by that timestamp. Alternatively you could use the query you are now (if its returning the correct subset of rows from the table). Replace COUNT(*) AS c with your primary key field (eg: id), and then use mysql_count_row() [rtfm for more details] rather than pulling the rows. From here it looks like your fastest option, but your not providing enough information. -- Dan Hardiker [[EMAIL PROTECTED]] ADAM Software Systems Engineer First Creative Ltd -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] date problem
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 sort by that timestamp. Alternatively you could use the query you are now (if its returning the correct subset of rows from the table). Replace COUNT(*) AS c with your primary key field (eg: id), and then use mysql_count_row() [rtfm for more details] rather than pulling the rows. From here it looks like your fastest option, but your not providing enough information. The query shouldn't be returning anything, b/c a unix_timestamp of a char type column is going to be zero. So zero will never be 1022882400. Also, there is no MySQL_count_row(), function. Do you mean MySQL_num_rows()? You should use COUNT(*) in your query instead of MySQL_num_rows(), if the count is all your after. ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] date problem
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 1, 02 SELECT COUNT(*) AS c FROM users_table WHERE UNIX_TIMESTAMP( user_regdate ) '1022882400' Thanx for any help on that, andy -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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 '10...' is unix timestamp june 1, 02 SELECT COUNT(*) AS c FROM users_table WHERE UNIX_TIMESTAMP( user_regdate ) '1022882400' You can only call UNIX_TIMESTAMP on a DATE or DATETIME field, not on just any generic CHAR/VARCHAR/TEXT/whatever. May 29, 2002 isn't a MySQL timestamp, so I'm guessing you have a textual field type. The lesson of all this is: Convert dates to either unix or database-native date format before storing them in the database. Things like May 29, 2002 are useless in a database. At this point I'd recommend running a quick script to strtotime() all your dates and then re-write them to a new field that's in a proper format. miguel -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] date problem
-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 afraid I still can't work it out so some advice/help would be greatly appreciated :-) - -- Nick Wilson // www.explodingnet.com -BEGIN PGP SIGNATURE- Version: GnuPG v1.0.6 (GNU/Linux) iD4DBQE8y++SHpvrrTa6L5oRAh3xAJ9lfFdB1VW3KUHCWxnnksN+Uw950QCYpfg6 OCdApC0Nh/8hi8MYQY3OTQ== =FnlM -END PGP SIGNATURE- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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: 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 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 afraid I still can't work it out so some advice/help would be greatly appreciated :-) - -- Nick Wilson // www.explodingnet.com -BEGIN PGP SIGNATURE- Version: GnuPG v1.0.6 (GNU/Linux) iD4DBQE8y++SHpvrrTa6L5oRAh3xAJ9lfFdB1VW3KUHCWxnnksN+Uw950QCYpfg6 OCdApC0Nh/8hi8MYQY3OTQ== =FnlM -END PGP SIGNATURE- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
-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 that mysql function... - -- Nick Wilson // www.explodingnet.com -BEGIN PGP SIGNATURE- Version: GnuPG v1.0.6 (GNU/Linux) iD8DBQE8y/S9HpvrrTa6L5oRAs9cAJ4jcd8dukhfdaJIla/irYNvnX2ZRwCgoIhU WjS6nkU1UM44crfOGWlBxRA= =RYIM -END PGP SIGNATURE- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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 know about that mysql function... Can we say, RTFM :) -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Everything is possible. Pass the word. -- Rita Mae Brown, Six of One */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] date problem
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 this would work: $qid = db_query(SELECT DATE_FORMAT(TimeStamp, '%e %b %y') AS DispDate FROM sometable); That will extract your timestamp as DD MMM from memory - have a look at the MySQL manual under the chapter on selects and date/time formatting and you'll get the complete list of '%x' options you can use. CYA, Dave -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] date problem
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 into my db? i tried using a hidden field with a foramtted gmt date value, but its not working... any help thanks - eoghan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] date problem
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 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 into my db? i tried using a hidden field with a foramtted gmt date value, but its not working... any help thanks - eoghan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Date Problem
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? Thanks MH -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] date problem
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 started after daylight savings time changed. I hope someone has an idea of what's wrong and how to fix it. As for now, I'm adding time() - 3600 * 8 to all my scripts right now. Steve -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] date problem: some months have 5 weeks... how I discover which ones?
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
Re: [PHP] date problem: some months have 5 weeks... how I discover which ones?
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. 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 months have 5 weeks... how I discover which ones? 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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] date problem: some months have 5 weeks... how I discoverwhich ones?
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 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Yet there are 10 unaccounted days which is 1 and 3/7 more weeks. Jeremy Jeremy Brand :: Sr. Software Engineer :: 408-245-9058 :: [EMAIL PROTECTED] http://www.JeremyBrand.com/Jeremy/Brand/Jeremy_Brand.html for more - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - "LINUX is obsolete" -- Andy Tanenbaum, January 29th, 1992 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - http://www.JEEP-FOR-SALE.com/ -- I need a buyer Get your own Free, Private email at http://www.smackdown.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 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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] date problem: some months have 5 weeks... how I discoverwhichones?
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 Th Fr Sa
123 4 5 6 - 1st week
7 89 10 11 12 13 - 2nd week
14 15 16 17 18 19 20 - 3rd week
21 22 23 24 25 26 27 - 4th week
28 29 30 31 - 5th week
Well, get the timestamp of the first of the month:
$ts = mktime(0,0,0,1,1,2001);
then add 7 days worth of secnds to it until the month is no longer
January:
while ((int)date('m',$ts) == 1)
{
//do stuff for this week
$ts += (60*60*24*7);
}
So january is a month that has 5 weeks for a newspaper, because if I would
put an ad on a tuesday, january has 5 tuesdays.. got it?
How do I do to calculate that?
I hope this help. It should give you the general idea. I don't
have time to write your whole application. ;-) Hopefully I gave you
enough to get you going in the right direction.
Ultimately you are going to need to cutomize this for what you exactly
want to do.
Jeremy
Jeremy Brand :: Sr. Software Engineer :: 408-245-9058 :: [EMAIL PROTECTED]
http://www.JeremyBrand.com/Jeremy/Brand/Jeremy_Brand.html for more
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