Re: [PHP-DB] Date problem
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 = getdate();
$sql=select * from events where event_end'.$today.' order by event_start
Asc ;
Thanks
denis
How is your date formatted in the database.
Compare that to the format returned by getdate(), then consider using
date('Y-m-d'). That's assuming your MySQL date is stored as -mm-dd.
Nonetheless, this should set you on the right road.
It would also have helped your diagnosis if you echoed your SQL statement.
Cheers - Miles Thompson
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Re: [PHP-DB] Date problem again ;-)
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 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 and date - is THAT the problem? That it needs to be an actual timestamp? If that is the case I'm not sure how to approach this because the data MUST be in the above format as it comes from a central DB every night. All I need to do is display the current date plus also the previous x amount of days in between. I'm using PHP 4 with MySQL 4.0.22 (Not the latest MySQL I know, sigh). Chris -- Sincerely, Forest Liu() -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date problem: data is current as of yesterday
-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 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 how to fix it (in VBA or SQL Server I would just say, date()-1: $today = getdate(); $month = $today['month'] ; $mday = $today['mday'] -1; $year = $today['year']; echo Data is current as of b$month $mday, $year/bbr; -- The mktime() function is your friend for this kind of date arithmetic. For example, this is one possible way to do what you want: $yesterday = mktime(12, 0, 0, $today['mon'], $today['mday']-1, $today['year']); echo Data is current as of b.date(F j, Y, $yesterday); (Note the use of time 12:00:00 to avoid daylight savings oddities!) The examples on the date() and mktime() manual pages may suggest other possibilities to you. Cheers! Mike - Do you Yahoo!? New and Improved Yahoo! Mail - 100MB free storage! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date problem: data is current as of yesterday
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 this is happening, but I can't find out how to fix it (in VBA or SQL Server I would just say, date()-1: $today = getdate(); $month = $today['month'] ; $mday = $today['mday'] -1; $year = $today['year']; echo Data is current as of b$month $mday, $year/bbr; you can do the same thing with php. I'd use a timestamp and subtract 86400 (24 hours of seconds) $yesterday_at_this_time = date(Y-m-d H:i:s, time() - 86400); hth Jeff - Do you Yahoo!? New and Improved Yahoo! Mail - 100MB free storage! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problem in MySQL DB
From: Angelo Zanetti [EMAIL PROTECTED] This might be slightly off topic but hopefully someone can help. I have a field that is a varchar and I stored dates in it. But now I want to change the type of the column to date, but I have a problem that the formats differ: my format: mm/dd/ mySQL format: -mm-dd So can I either: a. change the format of the date format or b. do i have to write a function that changes my format to the mySQL before changing the column type?? You need to do option (b). Since this is a PHP list, I'd recommend strtotime() and date() to do the formatting. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] date problem in MySQL DB
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($date) {
list($dateDay, $dateMonth, $dateYear) = explode(/, $date);
if ((is_numeric($dateDay)) (is_numeric($dateMonth))
(is_numeric($dateYear))) {
if (!checkdate($dateMonth, $dateDay, $dateYear)) {
return false;
} else {
return($dateYear./.$dateMonth./.$dateDay);
}
} else {
return false;
}
}
function dateCheckForm($date) {
list($dateYear, $dateMonth, $dateDay) = explode(-, $date);
if ((is_numeric($dateDay)) (is_numeric($dateMonth))
(is_numeric($dateYear))) {
return($dateDay./.$dateMonth./.$dateYear);
} else {
return false;
}
}
Many Thanks
Brett
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Re: [PHP-DB] date problem in MySQL DB
Brett King wrote:
Hi Angelo
Yes you will have to reformat and he is something that may help you.
Hope it does
function dateCheckMysql ($date) {
list($dateDay, $dateMonth, $dateYear) = explode(/, $date);
if ((is_numeric($dateDay)) (is_numeric($dateMonth))
(is_numeric($dateYear))) {
if (!checkdate($dateMonth, $dateDay, $dateYear)) {
return false;
} else {
return($dateYear./.$dateMonth./.$dateDay);
}
} else {
return false;
}
}
function dateCheckForm($date) {
list($dateYear, $dateMonth, $dateDay) = explode(-, $date);
if ((is_numeric($dateDay)) (is_numeric($dateMonth))
(is_numeric($dateYear))) {
return($dateDay./.$dateMonth./.$dateYear);
} else {
return false;
}
}
Many Thanks
Brett
Wow, that's lots of code. I'd recommend something simpler. Like:
$mysqlDate = date('Y-m-d', strtotime($myDate));
$myDate = date('m/d/Y', strtotime($mysqlDate));
No checking involved, just one line of code.
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RE: [PHP-DB] Date problem
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 Pay It Forward mailto:[EMAIL PROTECTED] http://accessingram.com -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 brain has probably fried :-) But here's the code, when I convert it back to d,m,Y it says it's 2004 which it shouldn't. Thanks Chris --- $secsinweek = 604800; $numweeks = $secsinweek * 7; $curdate = date(d-m-Y); // Split the strings into day, month, year list($cd, $cm, $cy) = split(-, $curdate); // Create unix time stamps for the start and end date $now = mktime(0,0,0,$cd,$cm,$cy); // Do some number crunching here // $newnow = $now-$numweeks; echo $now; $converted_date = date(d-m-y,$now); echo br$converted_date;
Re: [PHP-DB] date problem
maybe you should try
if($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 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 message.
Also, what is the CREATE TABLE schema for the date field?
Recommend showing us a little more of the surrounding code.
Have you tried adding a debug ECHO $date immediately before the IF
statement, to be sure what you're dealing with?
Please advise,
=dn
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RE: [PHP-DB] Date problem
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problem
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 message.
Also, what is the CREATE TABLE schema for the date field?
Recommend showing us a little more of the surrounding code.
Have you tried adding a debug ECHO $date immediately before the IF
statement, to be sure what you're dealing with?
Please advise,
=dn
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RE: [PHP-DB] date problem
if( strcmp($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 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
-
Express yourself with a super cool email address from BigMailBox.com.
Hundreds of choices. It's free!
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RE: [PHP-DB] Date problem with new year
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 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. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
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 over, I had no difficulty subtracting days, etc. Is there a bug in Mysql? PHP that I don't know of.. Please help. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Date problem with new year
Okay, when I try to change the gregorian date to julian, I get
error: call to undefined function: gregoriantojd() in line...
Here's the code I used to replace $datetoget=($datetoget-1);
//from what I understand ,must be in month, day, year format
//so passed today's date
$greg_date=GregorianToJD(1,2,2002);
//subtracted a day
$greg_date=($greg_date-1);
//converted it back to gregorian format
$datetoget=JDToGregorian($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: [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 to Gregorian (or Hebrew ;-). I've not had much success
with MySQL's date functions.
On Wed, 2 Jan 2002, Faye Keesic wrote:
I had no problems with retrieving news stories from previous days before
2002, and haven't changed any of my code. This site has been up for a year,
and I can't recall anything like this happening when 2001 rolled around.
Please excuse the messy code, as this was the first php/db web site I
designed. I took out the formatting stuff... :)
From my php page, I call a function:
get_most_recent_stories(date(Ymd));
- which passes today's date, formatted for for the db
In my code:
- this is the function I call - which in turn calls the function below
(no_yesterdays_stories($datetoget) to subtract a day if there is no news
items for the current date, which then calls the same
get_most_recent_stories() function, but passes a lower date. Ideally,
these will loop until there are stories found for a certain day.
function get_most_recent_stories($datetoget)
{
//Get the headlines, summaries and links to the story
db_connect();
$sql = SELECT * FROM tblStory WHERE (StoryDate=$datetoget);
$results = mysql_query($sql);
if ($results) {
if ($row = mysql_fetch_array($results)) {
echo $row[StorySum];}
while ($row = mysql_fetch_array($results));
}
else {
no_yesterdays_stories($datetoget); } }
}
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);
}
--
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
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 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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