Re: [PHP-DB] SELECT date query

[Bastien Koert]

i tend to take the approach of

$next_wed = date(Y-m-d, strtotime(next wednesday));


Bastien


From: Niel Archer [EMAIL PROTECTED]
Reply-To: php-db@lists.php.net
To: php-db@lists.php.net
Subject: Re: [PHP-DB] SELECT date query
Date: Sat, 07 Oct 2006 05:49:36 +0100

Hi Ron

I've made the assumption that if today is Wednesday, you still want next
Wednesday.
Try this:

$offset = array(3,2,1,7,6,5,4);
$date = explode(-, date(Y-n-j));
$ToDay = DayOfWeek($date[0], $date[1], $date[2]);
$NextWed = date(Y-n-j, time() + ($offset[$ToDay] * 24 * 60 * 60));

// Returns a digit in range 0-6. 0 = Sunday, 6 = Saturday
function DayOfWeek($Year, $Month, $Day)
{
$t = array(0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4);
$Year -= $Month  3;
return ($Year + ($Year / 4) - ($Year / 100) + ($Year / 400) +
$t[$Month - 1] + $Day) % 7;
}


Niel

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Re: [PHP-DB] SELECT date query

[Hodicska Gergely]

Hi!


You can make this easier with date('w').
 $correction = array(3, 2, 1, 7, 6, 5, 4);
 list($year, $month, $day, $dayOfWeek) = explode('|', date('Y|m|d|w'));
 echo date (Y.m.d, mktime 
(0,0,0,$month,$day+$correction[$dayOfWeek],$year));



Regards,
Felhő


Niel Archer wrote:

Hi Ron

I've made the assumption that if today is Wednesday, you still want next
Wednesday.
Try this:

$offset = array(3,2,1,7,6,5,4);
$date = explode(-, date(Y-n-j));
$ToDay = DayOfWeek($date[0], $date[1], $date[2]);
$NextWed = date(Y-n-j, time() + ($offset[$ToDay] * 24 * 60 * 60));

// Returns a digit in range 0-6. 0 = Sunday, 6 = Saturday
function DayOfWeek($Year, $Month, $Day)
{
$t = array(0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4);
$Year -= $Month  3;
return ($Year + ($Year / 4) - ($Year / 100) + ($Year / 400) +
$t[$Month - 1] + $Day) % 7;
}


Niel



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Re: [PHP-DB] SELECT date query

[Niel Archer]

Hi

 You can make this easier with date('w').

Doh, that'll teach me to code at 5 am.  I knew there was a better way,
but couldn't think of it, the sound of my bed calling was too
distracting.

Niel

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[PHP-DB] SELECT date query

[Ron Piggott (PHP)]

I am wondering if someone would help me write a SELECT date query ...

Weekly mailings go out every Wednesday.  I am setting up an
administration function and table to store the mailing name, PDF to be
contained within the mailing and the date for it to be used.

The SELECT query I want to create is for the next 12 records which are
going to be used to be displayed.  The first record would be next
Wednesday (not October 11th, but calendar wise based on when the script
ran) and then the following 11 Wednesdays.

SELECT * FROM christian_discipleship WHERE created_for_date =
'$next_wednesday' ORDER BY created_for_date ASC LIMIT 12

I am not sure how to generate the value for $next_wednesday

Any ideas?

Ron

Re: [PHP-DB] SELECT date query

[Niel Archer]

Hi Ron

I've made the assumption that if today is Wednesday, you still want next
Wednesday.
Try this:

$offset = array(3,2,1,7,6,5,4);
$date = explode(-, date(Y-n-j));
$ToDay = DayOfWeek($date[0], $date[1], $date[2]);
$NextWed = date(Y-n-j, time() + ($offset[$ToDay] * 24 * 60 * 60));

// Returns a digit in range 0-6. 0 = Sunday, 6 = Saturday
function DayOfWeek($Year, $Month, $Day)
{
$t = array(0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4);
$Year -= $Month  3;
return ($Year + ($Year / 4) - ($Year / 100) + ($Year / 400) +
$t[$Month - 1] + $Day) % 7;
}


Niel

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