Isn't DATEDIFF() a MySQL 4.x function? The server I'm using has 3.x and I
can't upgrade...
- Original Message -
From: Jasper Bryant-Greene [EMAIL PROTECTED]
To: php-general@lists.php.net
Sent: Wednesday, June 29, 2005 7:49 AM
Subject: Re: [PHP] Re: date problem
Mario netMines wrote
Firstly, this shouldn't be in the PHP list, as you're asking for help
with SQL.
Mario netMines wrote:
carrental_from (datetime field)
carrental_to (datetime field)
carrental_price (datetime field) [rates are per hour]
carrental_price shouldn't be a datetime field, as it isn't a datetime
, 2005 4:28 AM
Subject: [PHP] Re: date problem
Firstly, this shouldn't be in the PHP list, as you're asking for help
with SQL.
Mario netMines wrote:
carrental_from (datetime field)
carrental_to (datetime field)
carrental_price (datetime field) [rates are per hour]
carrental_price shouldn't
Mario netMines wrote:
Hi Jasper and thanks for the quick reply.
something tells me it's not a straightforward SQL query that I have to
use here but a logic using PHP and SQL.
Please don't top-post.
It can be done in SQL quite easily, as can many things people use PHP
for. Go to the MySQL
For me, on Windows, it won't work because Windows won't do anything prior to
1970.
On linux, I get 17 as the result. If I change the year to 2000, then I get
08 on both.
John
Shaun [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi,
Why does the following code print '00', surely
From the documentation:
http://ca2.php.net/manual/en/function.mktime.php
Date with year, month and day equal to zero is considered
illegal (otherwise it what be regarded as 30.11.1999, which
would be strange behavior).
I think the point here to think about is that the date(),
Try:
$ts = time();
$i = 0;
while( $i 2 ) {
$day = date(dS, $ts + $i * 86400);
print(td$day/td);
$i++;
}
On Wed, 19 Mar 2003, shaun wrote:
hi,
using date(dS); how can i can increase the days so that it shows
19th 20th 21st
I have tried
while ($i 2){
$day++;
$date_array=explode(-,$newdate);
$day = $date_array[2];
$month = $date_array[1];
$year = $date_array[0];
Or,
$timestamp - strtotime($newdate);
$today = getdate($timestamp);
$month = $today['month'];
$mday = $today['mday'];
$year = $today['year'];
Alexander Tsonev [EMAIL PROTECTED] wrote in
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
Yeh, ive allready looked at that before, but where and when do i use
DATE_FORMAT() ? When im inserting the row or selecting it?
Jome [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi, im making a tab/lyric portal, and for viewing tabs i want to display
the
Yeh, ive allready looked at that before, but where and when do i use
DATE_FORMAT() ? When im inserting the row or selecting it?
Replace your existent query with this one and try:
SELECT
`artist_id`,`title`,`content`,`user_id`,DATE_FORMAT(date,'%d-%m-%Y
%H:%i:%s'),`type`,`views` FROM
Alternativly you could store the dates as UNIX timestamps.
That is what I do. It is then eaiser to do certian things(ie show stuff
released in the last month)
--
JJ Harrison
[EMAIL PROTECTED]
www.tececo.com
Tony Harrison [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL
I tried using UNIX stamps but it dont work, and why the hell does it default
to that date anyway? I thought it was supposed to default to the current
time?
Jj Harrison [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Alternativly you could store the dates as UNIX
I store all dates in unix timestamp format. It's the easiest one to work
with, and it's easy to do things like date + three days, because it's just
a case of adding the right number of seconds to the current stamp.
You don't have to split anything, or get substr()'s of anything... and since
I store all dates in unix timestamp format. It's the easiest one to
work
with, and it's easy to do things like date + three days, because
it's
just
a case of adding the right number of seconds to the current stamp.
You don't have to split anything, or get substr()'s of anything... and
You can include date formatting function in your SQL statement:
DATE_FORMAT(date,format)
http://www.mysql.com/doc/D/a/Date_and_time_functions.html
--
Kind regards,
Yuri.
www.AceHoster.com Quality web hosting
Nick Wilson [EMAIL PROTECTED] ???/ ? ?:
[EMAIL
Mysql?
INSERT INTO table VALUES (NOW());
--
Julio Nobrega.
Um dia eu chego lá:
http://sourceforge.net/projects/toca
Ajudei? Salvei? Que tal um presentinho?
http://www.submarino.com.br/wishlistclient.asp?wlid=664176742884
Eoghan [EMAIL PROTECTED] wrote in message
[EMAIL
you can use the MySql's function:
DATE_ADD(datefield, INTERVAL 3 MONTH)
Mindhunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
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
On Wed, 21 Nov 2001 09:31:33 +0200, [EMAIL PROTECTED]
(Mindhunter) wrote:
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
Mindhunter wrote:
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
date() shoudl give you the time in your timezone, gmdate() should give you
the time in the GMT timezone. I would check your server and make sure the
timezone is correctly set.
--
Chris Lee
[EMAIL PROTECTED]
Steve Tsai [EMAIL PROTECTED] wrote in message news:none...
For reference:
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