At 12:26 PM 1/21/2007, Denis L. Menezes wrote:
Dear friends.
I have a date field in mysql called event_end .
I want to run a query to find all records where the event_and is greater
than today's date. I have written the following code. It does not work.
Please point out the mistake.
$today =
I suppose there is a " submit_time" field in your DB table, so you could:
"SELECT * FROM `tablename` WHERE `submit_time`>=".lastdays(3)
lastdays() is a function you could create using mktime(),time(), and date()
On Tue, 22 Mar 2005 16:28:39 -0500, Chris Payne <[EMAIL PROTECTED]> wrote:
> Hi ther
-Original Message-
From: Karen Resplendo
To: [EMAIL PROTECTED]
Sent: 02/07/04 19:36
Subject: [PHP-DB] Date problem: data is current as of yesterday
The database queries all the sources at night after everyone has gone
home. That means the data was current as of yesterday. This little
snipp
accidentally replied only to karen.
> The database queries all the sources at night after everyone has gone
home. That means the data was current as of yesterday. This little snippet
below returns yesterday's date, except that the first day of the month
returns "0" for > the day. Now, I know why
Use strtotime() instead..
http://us2.php.net/manual/en/function.strtotime.php
Regards,
Neal Carmine
Nine Systems Corporation
-Original Message-
From: Karen Resplendo [mailto:[EMAIL PROTECTED]
Sent: Friday, July 02, 2004 12:36 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Date problem: dat
Brett King wrote:
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($date) {
list($dateDay, $dateMonth, $dateYear) = explode("/", $date);
if ((is_numeric($dateDay)) && (is_numeric($dateMonth)) &&
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($date) {
list($dateDay, $dateMonth, $dateYear) = explode("/", $date);
if ((is_numeric($dateDay)) && (is_numeric($dateMonth)) &&
(is_numer
From: "Angelo Zanetti" <[EMAIL PROTECTED]>
> This might be slightly off topic but hopefully someone can help.
> I have a field that is a varchar and I stored dates in it. But now I want
to
> change the type of the column to date, but I have a problem that the
formats
> differ:
>
> my format: mm/dd
> // Do some number crunching here //
>
> $newnow = $now-$numweeks;
>
> echo $now;
>
> $converted_date = date("d-m-y",$now);
>
> echo "$converted_date";
You should be running your $convewrted_date on $newnow, not $now
Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
maybe you should try
if($date!=="-00-00")
{do something}
"Dl Neil" <[EMAIL PROTECTED]> wrote in message
0df801c1cb3d$c947e420$c200a8c0@jrbrown">news:0df801c1cb3d$c947e420$c200a8c0@jrbrown...
> Rehab,
>
> > i have an input field in a form that accept date called $date and in
> databse i made i
> i get $postdate from my mysql database,the problem is that i want
> to increase that date by 6 months??
> its stored in mysql as date field
http://www.mysql.com/doc/D/a/Date_and_time_functions.html
check DATE_ADD :)
hth
pa
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To unsubscribe,
if( strcmp($date,"-00-00") )
{ do something}
-Original Message-
From: its me [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 14, 2002 12:16 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] date problem
i have an input field in a form that accept date called $date and in databse
i made it
Rehab,
> i have an input field in a form that accept date called $date and in
databse i made it of type $date so its defualt is -00-00
>
> the problem is when i say:
> if($date!="-00-00")
> {do something}
>
> he doesn't understand $date!="-00-00"
Please copy-paste the actual error m
unction no_yesterdays_stories($datetoget)
>
> {
> //query has returned no stories from yesterday (holiday, etc.)
> //keep subtracting on day from yesterday's date until there are stories
> found
> $datetoget=($datetoget-1);
> get_most_recent_stories($datetoget);
> }
&g
> >
> > > From: [EMAIL PROTECTED]
> > > Date: Wed, 2 Jan 2002 07:55:02 -0800 (PST)
> > > To: Faye Keesic <[EMAIL PROTECTED]>
> > > Cc: [EMAIL PROTECTED]
> > > Subject: Re: [PHP-DB] Date problem with new year
> > >
> > > Ah... the
=JDToGregorian($greg_date);
> --
> Faye Keesic
> Computer Programmer Analyst/Web Page Design
>
>
> > From: [EMAIL PROTECTED]
> > Date: Wed, 2 Jan 2002 07:55:02 -0800 (PST)
> > To: Faye Keesic <[EMAIL PROTECTED]>
> > Cc: [EMAIL PROTECTED]
> > Subject: Re: [
Faye Keesic <[EMAIL PROTECTED]>
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Date problem with new year
>
> Ah... the "$datetoget = ($datetoget-1)" is the problem.
>
> Convert the Gregorian Ymd date to Julian, subtract 1, then convert the
> Julian date back
> function no_yesterdays_stories($datetoget)
>
> {
> //query has returned no stories from yesterday (holiday, etc.)
> //keep subtracting on day from yesterday's date until there are stories
> found
> $datetoget=($datetoget-1);
> get_most_recent_stories($datetoget);
&g
Did you convert the dates to PHP's "Julian" format before subtracting?
On Wed, 2 Jan 2002, Faye Keesic wrote:
> I have a db site that always displays the most recent occurrence of news
> stories
>
> The last occurrence of stories was last year (2001-12-31) and up to the new
> year rolling ov
ntil there are stories
found
$datetoget=($datetoget-1);
get_most_recent_stories($datetoget);
}
--
Faye Keesic
Computer Programmer Analyst/Web Page Design
> From: Rick Emery <[EMAIL PROTECTED]>
> Date: Wed, 2 Jan 2002 09:05:42 -0600
> To: [EMAIL PROTECTED]
> Subject: RE: [PHP-DB] Da
No bug, mate.
Post your code so that we can help ya...
-Original Message-
From: Faye Keesic [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 02, 2002 8:56 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Date problem with new year
I have a db site that always displays the most recent occu
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