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 =
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 = getdate();
$sql="select * from events where eve
I suppose there is a " submit_time" field in your DB table, so you could:
"SELECT * FROM `tablename` WHERE `submit_time`>=".lastdays(3)
lastdays() is a function you could create using mktime(),time(), and date()
On Tue, 22 Mar 2005 16:28:39 -0500, Chris Payne <[EMAIL PROTECTED]> wrote:
> Hi ther
Hi there everyone,
OK I'm using the following in my query to display entries in the LAST 3
days:
DATE_SUB(CURDATE(),INTERVAL 3 DAY) <= ListingDate
But it doesn't seem to affect the results, the date format in my DB column
(ListingDate - datatype is Date) is 2000-00-00 as in year, month
-Original Message-
From: Karen Resplendo
To: [EMAIL PROTECTED]
Sent: 02/07/04 19:36
Subject: [PHP-DB] Date problem: data is current as of yesterday
The database queries all the sources at night after everyone has gone
home. That means the data was current as of yesterday. This little
accidentally replied only to karen.
> The database queries all the sources at night after everyone has gone
home. That means the data was current as of yesterday. This little snippet
below returns yesterday's date, except that the first day of the month
returns "0" for > the day. Now, I know why
Use strtotime() instead..
http://us2.php.net/manual/en/function.strtotime.php
Regards,
Neal Carmine
Nine Systems Corporation
-Original Message-
From: Karen Resplendo [mailto:[EMAIL PROTECTED]
Sent: Friday, July 02, 2004 12:36 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Date problem
The database queries all the sources at night after everyone has gone home. That means
the data was current as of yesterday. This little snippet below returns yesterday's
date, except that the first day of the month returns "0" for the day. Now, I know why
this is happening, but I can't find out
Brett King wrote:
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($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($dateMonth)) &&
(is_numer
From: "Angelo Zanetti" <[EMAIL PROTECTED]>
> This might be slightly off topic but hopefully someone can help.
> I have a field that is a varchar and I stored dates in it. But now I want
to
> change the type of the column to date, but I have a problem that the
formats
> differ:
>
> my format: mm/dd
Hi guys
This might be slightly off topic but hopefully someone can help.
I have a field that is a varchar and I stored dates in it. But now I want to
change the type of the column to date, but I have a problem that the formats
differ:
my format: mm/dd/
mySQL format: -mm-dd
So can I eithe
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> -Original Message-
> From: Chris Payne [mailto:[EMAIL PROTECTED]
> Sent: Monday, July 14, 2003 3:45 AM
> To: php
> Subject: [PHP-DB] Date problem
Hi there everyone,
I use the below code to grab the current date, convert it to a unix timestamp to do
some bits then it's supposed to change the date back with the new values (IE: current
date - x amount of seconds) but it's not doing it correctly, it's probably DEAD
obvious to everyone as my
maybe you should try
if($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
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
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
> 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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i get $postdate from my mysql database,the problem is that i want to increase that
date by 6 months??
its stored in mysql as date field
if( strcmp($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 data
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
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 $date!="-00-00"
Rehab M.Shouman
-
t;
> --
> Faye Keesic
> Computer Programmer Analyst/Web Page Design
>
>
> > From: Rick Emery <[EMAIL PROTECTED]>
> > Date: Wed, 2 Jan 2002 09:05:42 -0600
> > To: [EMAIL PROTECTED]
> > Subject: RE: [PHP-DB] Date problem with new year
> >
> >
> >
> > > From: [EMAIL PROTECTED]
> > > Date: Wed, 2 Jan 2002 07:55:02 -0800 (PST)
> > > To: Faye Keesic <[EMAIL PROTECTED]>
> > > Cc: [EMAIL PROTECTED]
> > > Subject: Re: [PHP-DB] Date problem with new year
> > >
> > > Ah... the
=JDToGregorian($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
t; }
>
> --
> 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] Date problem with new year
> >
>
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
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 over, I had no difficulty subtracting days, etc.
Is there a bug in Mysql? PHP that I don't know of..
Please help.
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