Jack Lauman wrote:
I need to reformat the output of the 'dates' field from '2009-04-08' to
'Wed. Apr. 8th'. Any help would be appreciated.
You can either do it using mysql date formats (see
http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_date-format)
or something
See the date function
Http://www.php.net/date
Bastien
Sent from my iPod
On Apr 8, 2009, at 21:41, Jack Lauman jlau...@nwcascades.com wrote:
I need to reformat the output of the 'dates' field from '2009-04-08'
to 'Wed. Apr. 8th'. Any help would be appreciated.
Thanks.
---
for ($counter
On Tue, Aug 5, 2008 at 4:33 PM, Ben Miller [EMAIL PROTECTED]wrote:
Figured there had to be an easier way. Thank you so much.
-Original Message-
From: Simcha Younger [mailto:[EMAIL PROTECTED]
Sent: Tuesday, August 05, 2008 2:48 PM
To: php-db@lists.php.net
Subject: RE: [PHP-DB] Date
Select * FROM ... WHERE DAYOFWEEK(datecol)=7
-Original Message-
From: Ben Miller [mailto:[EMAIL PROTECTED]
Sent: Tuesday, August 05, 2008 8:10 PM
To: PHP. DB Mail List
Subject: [PHP-DB] Date Translation in MySQL
I'm looking for a quick and simple way to query a MySQL database by date,
Figured there had to be an easier way. Thank you so much.
-Original Message-
From: Simcha Younger [mailto:[EMAIL PROTECTED]
Sent: Tuesday, August 05, 2008 2:48 PM
To: php-db@lists.php.net
Subject: RE: [PHP-DB] Date Translation in MySQL
Select * FROM ... WHERE DAYOFWEEK(datecol)=7
Something like this should work.
$today = mktime(0, 0, 0, date(m), date(d), date(Y));
$tomorrow = mktime(0, 0, 0, date(m), date(d) + 1, date(Y));
$sql = SELECT COUNT(*) FROM table WHERE regdate BETWEEN {$today} AND
{$tomorrow};
$thismonth = mktime(0, 0, 0, date(m), 1, date(Y));
$nextmonth =
www.php.net/date will show you all the possibilities of formatting the date
bastien
Date: Sat, 22 Dec 2007 17:54:07 +0600
From: [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] date format problem
my problem is with date format
From: rDubya [EMAIL PROTECTED]
Thanks for the help so far guys!!
Not helping though. I have the date contained in the database as timestamp
(-MM-DD HH:MM:SS).
Do you really need to pull events from the database which are not in your
range of interest? This will only slow down your
From: Instruct ICC [EMAIL PROTECTED]
And while not trusting your indexing, rewrite short_date as:
My short_date rewrite was also wrong. So it looks like you will have to
learn those offsets for this function if you do it on the PHP side. But you
could also do it on the MySQL side.
This DID work, but I recently switched hosting companies...
Is the new server in a different country with a different date format
/ time zone?
Just a thought ;-\
On 9/7/07, Instruct ICC [EMAIL PROTECTED] wrote:
From: Instruct ICC [EMAIL PROTECTED]
And while not trusting your indexing,
WOW!! Thanks for all the help guys!! And Instruct ICC.. you're
solution for pulling the events did work.. but.. it turns out that
the solution was actually much simpler than I thought:
The old mysql database (once again, not sure what version) stored the
date as MMDDHHMMSS. The new
From: rDubya [EMAIL PROTECTED]
My problem is that I have events dated for Sep 2007 and on, and yet
they all come up as being on Dec 7 to 9, 2006.. any ideas?
rDubya
How about having MySQL only return the events you are interested in?
SELECT yourEventFields FROM theTable
WHERE
TO_DAYS(
It's much better to use add_date instead of to_days since mysql isn't
smart enough to do it for you.
Such as:
SELECT yourEventFields FROM theTable
WHERE theEventDate BETWEEN now() AND date_add(now(), INTERVAL 21 DAYS;
This way mysql will calc the now() and date_add and will essentially
convert
Argh, make sure you add the closing paren for the date_add since I
forgot it.
Mike...
Mike Gohlke wrote:
It's much better to use add_date instead of to_days since mysql isn't
smart enough to do it for you.
Such as:
SELECT yourEventFields FROM theTable
WHERE theEventDate BETWEEN now() AND
Thanks for the help so far guys!!
Not helping though. I have the date contained in the database as timestamp
(-MM-DD HH:MM:SS). The problem is that not only is it not displaying
events, but if I alter my code so that it displays ALL events, it shows the
events for the last year, and those
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 =
Manoj Singh wrote:
I am developing a site in php implementing the concept of rss feeds. For
that i want to convert the standard date into RFC822 format.
If any one have idea about it, please help me.
Go to http://php.net/date and search the page for RFC822. If you need
further help read the
A simple associate array would work as well, although not quite as elegent.
$months_arr = array(January=01, February=02...);
-Original Message-
From: Mark Bomgardner [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 16, 2006 12:35 PM
To: Php-Db
Subject: [PHP-DB] Date Conversion
PHP 4.4/MySQL
[mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 16, 2006 1:29 PM
To: [EMAIL PROTECTED]; 'Php-Db'
Subject: RE: [PHP-DB] Date Conversion
A simple associate array would work as well, although not quite as elegent.
$months_arr = array(January=01, February=02...);
-Original Message-
From: Mark
This works perfect, Bastien!
Many thanks.
Gerry
On 3/13/06, Bastien Koert [EMAIL PROTECTED] wrote:
select * from table where date_format(date_field, '%Y-%m') = '2006-02'
bastien
From: Gerry Danen [EMAIL PROTECTED]
To: php-db@lists.php.net
CC: [EMAIL PROTECTED]
Subject: [PHP-DB] Date
select * from table where date_format(date_field, '%Y-%m') = '2006-02'
bastien
From: Gerry Danen [EMAIL PROTECTED]
To: php-db@lists.php.net
CC: [EMAIL PROTECTED]
Subject: [PHP-DB] Date question
Date: Sun, 12 Mar 2006 20:44:13 -0700
While I am rebuilding my crashed laptop (the machine that
For example, you have table `logs` with `datelog` field and you want to
select dates
that match 2006-02. You can try this select statement:
SELECT * FROM `logs` WHERE MONTH(datelog)='02' and YEAR(datelog)='2006'
Hope that helps.
LJ Regalado
Gerry Danen wrote:
While I am rebuilding my crashed laptop (the machine that had all my
intelligence), I started thinking about a select statement I need.
I have log info in a table and want to extract it on a monthly basis. The
date field is in -mm-dd format. What's a good way to select
?php
echo date(Y-m-d H:i:s,strtotime(90 minutes ago));
?
bastien
From: Ron Piggott (PHP) [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] Date Time 90 minutes ago
Date: Thu, 19 Jan 2006 16:57:33 -0500
Would someone be able to help me with the
$date_90_minutes_ago = date('m/d/Y',mktime()-(60*90));
$time_90_minutes_ago = date('h:i:s',mktime()-(60*90));
60 seconds * 90 minutes.
=C=
|
| Cal Evans
| http://www.calevans.com
|
| We get our best customers from referrals.
| We would appreciate you referring any of your
| friends or co-workers
Bastien's example is probably the quickest and easiest. I just wanted to point
out that you can use math within the mktime() function as well in case the
relative date/time you need isn't right now.
$month = 1;
$day = 19;
$year = 2006;
$hour = 17;
$minute = 08;
$second = 05;
echo date(Y-m-d
You got the right idea, but you're making it more complicated than it needs to
be.
your $sDate after using explode() is going to contain an array. strtotime
doesn't take an array, it takes a string.
$monthName = date(F, strtotime($row_events['Sdate']));
$monthNumber = date(m,
use timestamp column type and populate it by
$date = strtottime(date(r));
then when you want to display it
$date = date('r',$row['datefieldname']);
Bastien
From: Ron Piggott [EMAIL PROTECTED]
Reply-To: Ron Piggott [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] DATE(r)
You need to use:
date('Y-m-d H:i:s');
It's in the comments at:
http://www.php.net/date
Jordan
On Sep 9, 2005, at 12:52 PM, Ron Piggott wrote:
Question:
I am trying to for the first time create a table with a column that is
defined as datetime
I wanted to populate that column with the
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 there
To view the terms under which this email is distributed, please go to
http://disclaimer.leedsmet.ac.uk/email.htm
On 16 December 2004 06:00, neil wrote:
Hi
I am needing to convert a d/m/y date such as 30/11/2004 into the
format that mysql can use ie. 2004-11-20
If I try the following:
On Thursday 16 December 2004 14:00, neil wrote:
I am needing to convert a d/m/y date such as 30/11/2004 into the format
that mysql can use ie. 2004-11-20
If I try the following:
$testdate=30/11/2004;
echo date(Y-m-d, strtotime($testdate));
the result is - 2006-06-11
I can't find any
Thank you for all your help.
Among all the variations I found this to be the clearest:
list($d,$m,$y) = explode(/,$testdate);
$mysqldate = date(Y-m-d, mktime(0,0,0,$m,$d,$y));
But I also thought the use of split instead of explode so you could nominate
multiple delimiters was good.
eg.
How are you trying to convert the date?
bastien
From: Bomgardner, Mark A [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Date Question
Date: Wed, 27 Oct 2004 11:17:16 -0500
I am having trouble converting a date from mm/dd/ to -mm-dd on a
user form. I know there was post about
Bomgardner, Mark A wrote:
I am having trouble converting a date from mm/dd/ to -mm-dd on a
user form. I know there was post about this, but I keep getting an
error message when I try to search the archives.
I have looked at the manual, but I am not finding what I am looking for.
echo
-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
How can I query a MySQL table to get the latest results from a date field?
Basically, I am inserting several records at a time at the end of each
week.
I want to have a page that displays the results for the last week only.
The date format in the field is -MM-DD
if you want the latest row
] [mailto:[EMAIL PROTECTED]
Sent: 25 June 2004 12:15
To: Tom Chubb
Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date Select
How can I query a MySQL table to get the latest results from a date field?
Basically, I am inserting several records at a time at the end of each
week.
I want
Content-Transfer-Encoding: 7bit
Subject: RE: [PHP-DB] Date help needed
One thing he wanted which I didn't know how to do (Javascript I guess which
I don't know much about) was to preload a database of email address, and as
he started to type an email address it would do a sort of auto-complete
You could loop through the weeks and put those 3 specifically in:
$days = array();
for($i = 0; $i 365; $i +=7) {
$days[] = strtotime('next Monday', strtotime('+ '.$i.' days'));
$days[] = strtotime('next Friday', strtotime('+ '.$i.' days'));
$days[] = strtotime('next Sunday', strtotime('+
Hi there,
Just got back and tried it and it works perfectly, thank you so much for
your help, you've got me out of a bind here ;-)
Chris
You could loop through the weeks and put those 3 specifically in:
$days = array();
for($i = 0; $i 365; $i +=7) {
$days[] = strtotime('next Monday',
A drop down with 365 days !?!? Isn't that a little big?
I have a problem, I currently have some code which populates a dropdown
box
- this code gives me every day for the next x amount of days (EG: a years
worth of days), however what I really need to be able to do, is to find a
way to
Hi there,
A drop down with 365 days !?!? Isn't that a little big?
Actually it's Fridays, Sundays and Tuesdays for 2 years (365 was an example)
it's for an Admin for a client, and he asked that it be in a dropdown box
and he's the boss, so he gets what he wants :-)
One thing he wanted which I
SELECT
IF( B.Booking_End_Date != -00-00,
DATE_FORMAT( B.Booking_End_Date, %Y-%m-%d ),
N/A
) AS Booking_End_Date
FROM Bookings AS B, etc.
HTH
Ignatius
_
- Original Message -
From: Shaun [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent:
try the WHERE clause.
SELECT name, address, phone, date FROM usertable WHERE date =
'-00-00'
And please, look at the mysql documentation, or at least the tutorial.
--
Marcjon
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Shannon Doyle wrote:
My question, how do I get the date entered into the form add 35days to it
and then include that into the same sql query as the first one. Or do I have
to use a second sql query? If the second query how would I get the date and
add 35days??
INSERT INTO table (date1, date2)
From: Angelo Zanetti [EMAIL PROTECTED]
This might be slightly off topic but hopefully someone can help.
I have a field that is a varchar and I stored dates in it. But now I want
to
change the type of the column to date, but I have a problem that the
formats
differ:
my format: mm/dd/
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($date) {
list($dateDay, $dateMonth, $dateYear) = explode(/, $date);
if ((is_numeric($dateDay)) (is_numeric($dateMonth))
Brett King wrote:
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($date) {
list($dateDay, $dateMonth, $dateYear) = explode(/, $date);
if ((is_numeric($dateDay)) (is_numeric($dateMonth))
This might not be the best place for this but here goes.
You are correct. At the very least, you should probably be using the
general PHP list, not the database one.
I want to create a dropdown list with a date range of
Not to be obstinate, but it looks like you already have. But, assuming
have a look at the PEAR date package.
http://pear.php.net/package/Date
_
- Original Message -
From: OpenSource [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Sunday, November 02, 2003 11:13 PM
Subject: [PHP-DB] date function
Hi guys,
This might not be the best
1) Why do you send this to a DB list?
2) Try seeing the Date class of PEAR (PEAR::Date).
El Dom 02 Nov 2003 19:13, OpenSource escribió:
Hi guys,
This might not be the best place for this but here goes.
I want to create a dropdown list with a date range of
On Saturday 06 September 2003 06:01, Darryl wrote:
I have some php code that pulls from the mysql database. Here it is:
?php
mysql_connect(wildcat.osborneindustries.com, webuser,
webpass);
$mymonth = date('n');
$cyear = date('Y');
$query = SELECT name,hdate FROM
I have a field in my mysql db wich is a timestamp with the format
mmddhhmmss
and I would like to display it like:
dd/mm/
http://www.mysql.com/doc/en/Date_and_time_functions.html#IDX1333
SELECT DATE_FORMAT(timestamp_col_name, '%d/%m/%Y') FROM table_name
--
PHP Database
snip
// Do some number crunching here //
$newnow = $now-$numweeks;
echo $now;
$converted_date = date(d-m-y,$now);
echo br$converted_date;
/snip
You should be running your $convewrted_date on $newnow, not $now
Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
you could change mysql to a few different date formats, but i don't _think_
that is one of them.
Why not just take care of a format change on the way out with DATE_FORMAT.
Also, if you store the date as one of the defualt date[time] values then
you have tons of functions that you can run against
Hi David,
Try this one:
?
function formatDate($val)
{
setlocale (LC_ALL, '');
$arr = explode(-, $val);
return strftime (%A %e %B %Y, mktime (0, 0, 0, $arr[1], $arr[2],
$arr[0]));
}
echo formatDate($DatePick);
?
Regards,
Frank
- Original Message -
From: David Shugarts [EMAIL
At 05:23 PM 6/10/2003 -0400, David Shugarts wrote:
I have a simple need to reformat a variable coming in as $DatePick, brought
forward from a MySQL select in the format:
2003-11-18
I need to convert it to the format
November 18, 2003
try the dice-n-slice method. use substr and
I have a simple need to reformat a variable coming in as $DatePick,
brought
forward from a MySQL select in the format:
2003-11-18
I need to convert it to the format
November 18, 2003
$f_date = date('F d, Y',strtotime($db_date));
---John Holmes...
--
PHP Database Mailing List
This worked perfectly, is very simple for me to use, and I would never have
come upon it in 10,000 years of trying. Thanks!
Thanks also to everyone who offered ideas.
$f_date = date('F d, Y',strtotime($db_date));
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit:
function datetoymd($date){
$dateymd = date('Y-m-d',$date);
return $dateymd;
}
this function when output gives me the date 1970-01-01 (date of the unix
timestamp start) so, ehm, why!?
Because you're not passing a valid Unix timestamp or whatever your passing
is empty.
---John Holmes...
ehmmm what kind of dates do you use, maybe dates of birth?
when using date of unix timestamp you can't have dates befor 1970 so if
those timestamps represents an older date you can't use timestamps!
-mark-
-Oorspronkelijk bericht-
Van: David Rice [mailto:[EMAIL PROTECTED]
I have a form which displays 3 dropdowns, day, month and year, I have
it
displaying the default of the dropdown for month no problem, but how
can I
get it to select the current date (day) in the dropdown, but show 31
days
nomatter how many days are in the month? This is important even
though
Hi there,
Thank you John, you are a lifesaver :-)
Chris
I have a form which displays 3 dropdowns, day, month and year, I have
it
displaying the default of the dropdown for month no problem, but how
can I
get it to select the current date (day) in the dropdown, but show 31
days
Sorry, I forgot to add the part at the bottom where I am calling the
function
=
?
function tips($weekstart){
$start = date('Ymd',strtotime($weekstart));
$query = SELECT * FROM Rota WHERE date = $start and date = ($start +
Professionals. Get your copy
today. http://www.phparch.com/
-Original Message-
From: David Rice [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 05, 2003 12:34 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] date functions (generates parse error)
Sorry, I forgot
I am looking for a way to take a date stored in a mysql database... and
find
out the date seven days later.
how would i do this?!
too easy...
SELECT date_column + INTERVAL 7 DAY FROM your_table ...
---John Holmes...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe,
Good stuff here.
http://www.mysql.com/doc/en/Date_and_time_functions.html
Check out the SELECT DATE_ADD section. That might be what you're looking
for.
HTH,
Rich
-Original Message-
From: David Rice [mailto:[EMAIL PROTECTED]
Sent: Tuesday, March 04, 2003 9:36 AM
To: [EMAIL
I want to use it in this function that i am creating (it's for a
resteraunt
automated tips system, to work out how much tips each staff member is
entitled to.
function tips($weekstart){
/* JUST BELOW HERE IS WHERE I
Here is the whole code of my function
Whenever i run it, it say's there is a parse error on line 6, can't see what
is the problem
the format of $weekstart (as it is stored in the Database) is -MM-DD
=
?
function
Here is the whole code of my function
Whenever i run it, it say's there is a parse error on line 6, can't see
what
is the problem
the format of $weekstart (as it is stored in the Database) is -MM-DD
=
?
function
go to mysql.com en look for date
-Original Message-
From: Jorge Miguel Fonseca Martins [mailto:[EMAIL PROTECTED]
Sent: vrijdag 28 februari 2003 16:23
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Date
How can a make a query that lists all the fields where a datefield as
the date of tomorow?
when retrieving data from the db and displaying it on the
page by creation date how can i reverse the order so that the most
recent entry appears first? here's what i have now:
?
require(config.php);
$obj = mysql_db_query($dbname,select * from ads order by
createdate);
?
You
Simple MySQL.
After your ORDER BY clause use either ASC (I think this is the default) or
DESC. I'm pretty sure this'll work on date fields to produce the results you
want.
Your query will end up looking like:
...order by createdate DESC
-Original Message-
From: Addison Ellis
thank you... this worked. best, addison
Simple MySQL.
After your ORDER BY clause use either ASC (I think this is the default) or
DESC. I'm pretty sure this'll work on date fields to produce the results you
want.
Your query will end up looking like:
...order by createdate DESC
-Original
I am working on an app that needs to post information to a website
based on date. The tricky part for me is that the date is a range that
spans either 2 or 3 days. I want the web page to dynamically populate
this
information based on a query. The query will look something like this:
the difference of 2 or 3 days?
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, February 04, 2003 10:04 AM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date Range Question...
I am working on an app that needs to post information to a website
strftime()
offers a lot of formatting options for dates.
Hope it helps.
RUBANOWICZ Lisa wrote:
Hi All, I have a date format of -MM-DD in MySQL and am showing it on a PHP page. However I want to show it as
2 February, 2003
or 2 February
Can someone please help me. The date will not
]
Subject: Re: [PHP-DB] Date format in
MySQL
02/03/2003
09:32 AM
$query = mysql_query(SELECT UNIX_TIMESTAMP(fieldname) AS date WHERE id = '1';);
$array = mysql_fetch_array($query);
$mydate = date(j F, Y,$array[date]);
Change fieldname and the where clause, and that should work.
If you want to further munipulate how it looks, just look at the
date
[snip]
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
My big question is about using the
[snip]
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
My big question is about
]
Subject: Re: [PHP-DB] Date math functions...
[snip]
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die
W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 10:32 AM
To: 1LT John W. Holmes; NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date math functions...
[snip]
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said
On Thursday 16 January 2003 00:55, NIPP, SCOTT V (SBCSI) wrote:
Actually this is generating another error. Now, without the single
quotes I am getting the following error:
Column 'StartDate' cannot be null
It looks like for some reason the DATE_ADD is returning a NULL
value.
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES(DATE_ADD($StopDate,
INTERVAL 1 DAY,'',1,2,'$CalendarDetailsID')) or die(mysql_error());
Actually this is generating another error. Now, without
When I use this, I get 12/31/69 as my date.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, November 13, 2002 4:04 PM
To: 'Edward Peloke'; 'Php-Db'
Subject: RE: [PHP-DB] date()
$date = date(m:d:y, $myrow[datefield]);
Will produce: 11:13:02
http
: [PHP-DB] date()
When I use this, I get 12/31/69 as my date.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, November 13, 2002 4:04 PM
To: 'Edward Peloke'; 'Php-Db'
Subject: RE: [PHP-DB] date()
$date = date(m:d:y, $myrow[datefield]);
Will produce: 11:13:02
'; 'Php-Db'
Subject: RE: [PHP-DB] date()
Ok..
I guess it depends on how your date is stored. I always use
unix_timestamps.
If you are too.. then the format I supplied should work as indicated.
Aaron
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: November 18, 2002
18, 2002 2:50 PM
To: 'Php-Db'
Subject: RE: [PHP-DB] date()
it is just a datetime field and the clients use a javascript calander to
pick the date, here is the exact date as it appears in the db.
2002-11-08 00:00:00
Eddie
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED
On Tuesday 19 November 2002 03:26, Aaron Wolski wrote:
Hmm.. I would get it into a unix_timestamp (unless someone can suggest a
reasoning for now doing so).
It really depends on where you want to manipulate the dates. If mostly from
within MySQL then store with DATE, DATETIME TIMESTAMP, if
:
11/18/2002 Subject: RE: [PHP-DB] date()
02:26 PM
@Keane.com To: 'Edward Peloke'
[EMAIL PROTECTED]
cc: 'Php-Db' [EMAIL PROTECTED]
11/18/2002 Subject: RE: [PHP-DB] date
$date = date(m:d:y, $myrow[datefield]);
Will produce: 11:13:02
http://www.php.net/manual/en/function.date.php
Aaron
-Original Message-
From: Edward Peloke [mailto:epeloke;echoman.com]
Sent: November 13, 2002 4:27 PM
To: Php-Db
Subject: [PHP-DB] date()
I have a date field in my
use the php function substr to chop off what you don't need, since all the
dates can come from mysql as -mm-dd or however, just use substr($mydate,
0, 10) or however far out, that truncates the actual $mydate.
-Original Message-
From: [EMAIL PROTECTED] [mailto:epeloke;echoman.com]
If you don't want to change your SQL statement:
echo date ('Y/m/d', strtotime ($myrow['datefield']));
Marco
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On Thursday 14 November 2002 05:27, Edward Peloke wrote:
I have a date field in my mysql db, when I output the data to the screen, I
don't want to see the minutes, just the mmddyy.
DATE fields in MySQL don't have the time (h:m:s).
I can format a date but
can't seem to get it to work
check out the date( ) function in mysql. a quick prelude of it
is.select date(date_col, %m/%d/%Y) from table...this will take
-mm-dd and return mm/dd/
I assume you are trying to insert into a mysql table column that is a date.
insert into table (datecol) values (from_unixtime($new_exp_day));
Example output of from_unixtime:
+---+
| from_unixtime(10) |
+---+
| 2001-09-08 21:46:40 |
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