Hi All,
I have a database with a bunch of dates in it. I want to count the number of
entries for each year and then display the year and the count.
i.e.
YearCount
200622
200518
200414
200322
This is what I have tried but just not quite getting it.
$query select count(date)
Beauford wrote:
$query select count(date) as count, YEAR(date) as thisyear from stats group
^ = needed here
by thisyear;
-Stut
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Yea, I just figured this out. When I cut and pasted I must have overwrote
the =.
Thanks
-Original Message-
From: Stut [mailto:[EMAIL PROTECTED]
Sent: January 4, 2007 4:27 PM
To: Beauford
Cc: PHP
Subject: Re: [PHP] Date problems
Beauford wrote:
$query select count(date
);
Satyam
- Original Message -
From: Mace Eliason [EMAIL PROTECTED]
To: php-general@lists.php.net
Sent: Sunday, April 09, 2006 4:14 AM
Subject: [PHP] Date problems
Hi,
I am having troubles adding 7 days to the current date. I have been
reading through php.net date() and this is what I
Satyam wrote:
Timestamps are stored as seconds from a certain date. The base date
differ depending on the platform, Windows use 1/1/1980 the rest
1/1/1970, but still seconds. 7 days are 7*24*60*60. Just add that much
to a timestamp. It helps having a constant such as:
define
Hi,
I am having troubles adding 7 days to the current date. I have been
reading through php.net date() and this is what I have come up with but
it doesn't work
$today = date('m/d/Y');
$nextweek = date('m/d/Y',mktime(date(m), date(d)+7, date(Y)));
if I echo the above variables they are the
Mace Eliason wrote:
Hi,
I am having troubles adding 7 days to the current date. I have been
reading through php.net date() and this is what I have come up with but
it doesn't work
$today = date('m/d/Y');
$nextweek = date('m/d/Y',mktime(date(m), date(d)+7, date(Y)));
if I echo the above
Rasmus Lerdorf wrote:
Mace Eliason wrote:
Hi,
I am having troubles adding 7 days to the current date. I have been
reading through php.net date() and this is what I have come up with
but it doesn't work
$today = date('m/d/Y');
$nextweek = date('m/d/Y',mktime(date(m), date(d)+7, date(Y)));
$date1 = 12/12/2001;
$date1 = date(D M j Y, strtotime($date1));
print $date1.br;
When I execute the code above I get the following
output:
Tue Dec 11 2001 instead Wed Dec !2 2001
Why does this happen and how can I get the Dec 12
output? Thanks
sounds like it might have something to do with leap year.
Jim Lucas
- Original Message -
From: toni baker [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, February 04, 2002 12:03 PM
Subject: [PHP] date problems
$date1 = 12/12/2001;
$date1 = date(D M j Y, strtotime($date1
PM
Subject: [PHP] date problems
$date1 = 12/12/2001;
$date1 = date(D M j Y, strtotime($date1));
print $date1.br;
When I execute the code above I get the following
output:
Tue Dec 11 2001 instead Wed Dec !2 2001
Why does this happen and how can I get the Dec 12
output? Thanks
Hi,
I'm having a bit of a weird problem with the date() function. I am using
PostgreSQL to store guestbook entries and updating and displaying them with
Apache 1.3x and PHP4 on OpenBSD. The date information is being stored
correctly in PostgreSQL, but I'm using the date() function to format
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