[PHP-DB] Date formatting question
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 = 0; $counter mysql_num_rows($resultID); $counter++);
while ($row = mysql_fetch_object($resultID))
{
print tr;
print td . $row-dates . /td;
print td . $row-times . /td;
print td . $row-am_pm . /td;
print td . $row-height . /td;
print td . $row-cond . /td;
print /tr;
}
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Re: [PHP-DB] Date formatting question
Jack Lauman wrote:
I need to reformat the output of the 'dates' field from '2009-04-08' to
'Wed. Apr. 8th'. Any help would be appreciated.
You can either do it using mysql date formats (see
http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_date-format)
or something like date('...', strtotime($result-date));
See http://php.net/date and http://php.net/strtotime
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Re: [PHP-DB] Date formatting question
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 = 0; $counter mysql_num_rows($resultID); $counter++);
while ($row = mysql_fetch_object($resultID))
{
print tr;
print td . $row-dates . /td;
print td . $row-times . /td;
print td . $row-am_pm . /td;
print td . $row-height . /td;
print td . $row-cond . /td;
print /tr;
}
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Re: [PHP-DB] Date Translation in MySQL
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 Translation in MySQL 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, or more specifically, by day of the week. My dates are stored in the DB in -MM-DD HH:MM:SS format. Question, is there a built-in for PHP or MySQL that will take this column and return only Saturdays, for example, without having to use PHP to find each of the last X number of Saturdays and then for each Saturday, query the database? In case it matters, I am running on PHP v 4.4.7 and MySQL version 4.1.22. Thanks in advance. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php No virus found in this incoming message. Checked by AVG - http://www.avg.com Version: 8.0.138 / Virus Database: 270.5.12/1592 - Release Date: 05/08/2008 06:03 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Also look at the mysql DATE_FORMAT function -- Bastien Cat, the other other white meat
[PHP-DB] Date Translation in MySQL
I'm looking for a quick and simple way to query a MySQL database by date, or more specifically, by day of the week. My dates are stored in the DB in -MM-DD HH:MM:SS format. Question, is there a built-in for PHP or MySQL that will take this column and return only Saturdays, for example, without having to use PHP to find each of the last X number of Saturdays and then for each Saturday, query the database? In case it matters, I am running on PHP v 4.4.7 and MySQL version 4.1.22. Thanks in advance. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date Translation in MySQL
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, or more specifically, by day of the week. My dates are stored in the DB in -MM-DD HH:MM:SS format. Question, is there a built-in for PHP or MySQL that will take this column and return only Saturdays, for example, without having to use PHP to find each of the last X number of Saturdays and then for each Saturday, query the database? In case it matters, I am running on PHP v 4.4.7 and MySQL version 4.1.22. Thanks in advance. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php No virus found in this incoming message. Checked by AVG - http://www.avg.com Version: 8.0.138 / Virus Database: 270.5.12/1592 - Release Date: 05/08/2008 06:03 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date Translation in MySQL
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 -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, or more specifically, by day of the week. My dates are stored in the DB in -MM-DD HH:MM:SS format. Question, is there a built-in for PHP or MySQL that will take this column and return only Saturdays, for example, without having to use PHP to find each of the last X number of Saturdays and then for each Saturday, query the database? In case it matters, I am running on PHP v 4.4.7 and MySQL version 4.1.22. Thanks in advance. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php No virus found in this incoming message. Checked by AVG - http://www.avg.com Version: 8.0.138 / Virus Database: 270.5.12/1592 - Release Date: 05/08/2008 06:03 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date calculation from MySql table
I want to calculate the registed users today Also total users this week Total users this month Total users this year The Mysql table has a row of INT(11) with time() value inserted. I did something like this $today = strtotime(+1 day) Then $sql = SELECT COUNT(*) FROM table WHERE dateReg = $today; Same with year/months also, only I use strtotime(+1 week) for a week, strtotime(+1 month) for a month, Can someone help me with this calculation? On 4/7/08, Bruno Lustosa [EMAIL PROTECTED] wrote: On Mon, Apr 7, 2008 at 2:42 PM, Dee Ayy [EMAIL PROTECTED] wrote: I was thinking of using output buffering and then making 1 call to utf8_encode, but I think a better question is, how do I stop using utf8_encode completely? If all components are using utf-8, you should have no problems with charsets at all. By all components, I mean: - Script files in utf-8; - Database in utf-8; - Database connection using utf-8; - Content-type header set to utf-8. With all these, you're free of charset hell, and can enjoy the beauty of utf-8 completely without problems. The rendered view I see in Firefox 2.0.0.12 is a question mark ? where the French character should have appeared. If I use utf8_encode, the character appears as it should. Question mark means the character is not utf-8. Check where it comes from. Might be the database or the way you are connecting to it. I don't know much about mysql, I use postgresql. With it, you just have to call pg_set_client_encoding() to make the connection in utf-8 mode, and create database with encoding='unicode' to set up a database using utf-8. Luckily I'm on PHP 4.3.10, so I can't see what mb_check_encoding would report -- if that would even help normally. Shouls upgrade to PHP 5. PHP 4 is way out of date, is not getting updates anymore, and will not even get security bugfixes after august 8th. It's been almost 4 years since PHP 5 was released. http://www.php.net/archive/2007.php Check the PHP 4 end of life announcement. -- Bruno Lustosa [EMAIL PROTECTED] ZCE - Zend Certified Engineer - PHP! http://www.lustosa.net/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- I develop dynamic website with PHP MySql, Let me know about your site -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date calculation from MySql table
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 = mktime(0, 0, 0, date(m) + 1, 1, date(Y));
$sql = SELECT COUNT(*) FROM table WHERE regdate BETWEEN {$thismonth}
AND {$nextmonth};
$thisyear = mktime(0, 0, 0, 1, 1, date(Y));
$nextyear = mktime(0, 0, 0, 1, 1, date(Y) + 1);
$sql = SELECT COUNT(*) FROM table WHERE regdate BETWEEN {$thisyear} AND
{$nextyear};
HOWEVER, consider to use the mysql date functions instead of a unix
timestamp.
A. Joseph wrote:
I want to calculate the registed users today
Also total users this week
Total users this month
Total users this year
The Mysql table has a row of INT(11) with time() value inserted.
I did something like this
$today = strtotime(+1 day)
Then $sql = SELECT COUNT(*) FROM table WHERE dateReg = $today;
Same with year/months also, only I use strtotime(+1 week) for a week,
strtotime(+1 month) for a month,
Can someone help me with this calculation?
On 4/7/08, Bruno Lustosa [EMAIL PROTECTED] wrote:
On Mon, Apr 7, 2008 at 2:42 PM, Dee Ayy [EMAIL PROTECTED] wrote:
I was thinking of using output buffering and then making 1 call to
utf8_encode, but I think a better question is, how do I stop using
utf8_encode completely?
If all components are using utf-8, you should have no problems with
charsets at all. By all components, I mean:
- Script files in utf-8;
- Database in utf-8;
- Database connection using utf-8;
- Content-type header set to utf-8.
With all these, you're free of charset hell, and can enjoy the beauty
of utf-8 completely without problems.
The rendered view I see in Firefox 2.0.0.12 is a question mark ?
where the French character should have appeared. If I use
utf8_encode, the character appears as it should.
Question mark means the character is not utf-8. Check where it comes
from. Might be the database or the way you are connecting to it. I
don't know much about mysql, I use postgresql. With it, you just have
to call pg_set_client_encoding() to make the connection in utf-8 mode,
and create database with encoding='unicode' to set up a database
using utf-8.
Luckily I'm on PHP 4.3.10, so I can't see what mb_check_encoding would
report -- if that would even help normally.
Shouls upgrade to PHP 5. PHP 4 is way out of date, is not getting
updates anymore, and will not even get security bugfixes after august
8th. It's been almost 4 years since PHP 5 was released.
http://www.php.net/archive/2007.php
Check the PHP 4 end of life announcement.
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http://www.lustosa.net/
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[PHP-DB] date format problem
my problem is with date format mysql support -mm-dd but my client not use to enter date in -mm-dd format he use to in dd-mm-yyy format how can it possible to input date in dd-mm- format
RE: [PHP-DB] date format problem
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 mysql support -mm-dd but my client not use to enter date in -mm-dd format he use to in dd-mm-yyy format how can it possible to input date in dd-mm- format _ Use fowl language with Chicktionary. Click here to start playing! http://puzzles.sympatico.msn.ca/chicktionary/index.html?icid=htmlsig -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problems
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 processing time. Instead,
you could be looping over valid events and not using your incorrect
check_date function.
If you insist upon using a check_date function on the PHP side (which you
claimed to have worked in the past), on the format -MM-DD HH:MM:SS where
the first Y is at index 0, then substr($mysql_timestamp, 4, 2) is not the
month MM, it is -M. You need 5,2. Your other offsets are also wrong.
Rather than debugging your check_date function, you should just pull the
info you actually need from the database (even to the point of not selecting
*, and instead naming specific fields if you really don't need all fields),
and remove your check_date function. Then your while loop will loop over
only valid events.
Also, although this doesn't affect the running of your code, your variable
names show a misunderstanding of what parameters you are passing and
receiving.
$sql = SELECT...;
$resultResource = mysql_query($sql);// or just $result = mysql_query($sql);
if($result != false){
while($event_data = mysql_fetch_assoc($result)) {
//Do something with a valid event instead of checking if it is valid
now on the PHP side.
}
}
Too bad you are not on MySQL 5, because the SQL query could have been even
closer to the phrasing you posted to the list it is between now and three
weeks from now, similar to Mike's reply as:
$sql = SELECT * FROM `EVENTS`
WHERE `event_city` = ' . $city . '
AND `theEventDate` BETWEEN NOW() AND DATE_ADD( NOW(), INTERVAL 3 WEEK)
ORDER BY `theEventDate` ASC;
But for your MySQL version, WEEK is unavailable and you need INTERVAL 21 DAY
(watch the unit versus the expr; DAY not DAYS -- WEEK not WEEKS)
http://dev.mysql.com/doc/refman/4.1/en/date-and-time-functions.html#function_date-add
http://dev.mysql.com/doc/refman/4.1/en/comparison-operators.html
But if you are not familiar with BETWEEN or DATE_ADD, then the way I posted
is another way to skin this cat.
$sql = SELECT * FROM `EVENTS`
WHERE `event_city` = ' . $city . '
AND TO_DAYS( `theEventDate` ) = TO_DAYS( NOW() )
AND TO_DAYS( `theEventDate` ) = TO_DAYS( NOW() ) + 21
ORDER BY `theEventDate` ASC;
http://dev.mysql.com/doc/refman/4.1/en/date-and-time-functions.html#function_to-days
And while not trusting your indexing, rewrite short_date as:
function short_date ($mysql_timestamp) {
return date(D M j, $mysql_timestamp);
}
or this is probably a 1-liner in the while loop:
while...//I'm already using valid events due to my query
... date(D M j, $event_data['theEventDate']) ...
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Re: [PHP-DB] date problems
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. _ Get a FREE small business Web site and more from Microsoft® Office Live! http://clk.atdmt.com/MRT/go/aub0930003811mrt/direct/01/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problems
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, 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. _ Get a FREE small business Web site and more from Microsoft(r) Office Live! http://clk.atdmt.com/MRT/go/aub0930003811mrt/direct/01/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- Graham -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problems
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 database stores the date as -MM-DD HH:MM:SS. All I had to do was adjust my code to pull only the values and none of the delimeters (ie. -, , and :). DUH!!! But once again, thanks you guys for all the help rDubya On 9/7/07, Instruct ICC [EMAIL PROTECTED] wrote: 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. _ Get a FREE small business Web site and more from Microsoft(r) Office Live! http://clk.atdmt.com/MRT/go/aub0930003811mrt/direct/01/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] date problems
I'm having a problem with dates in php and mysql. I run a site that
promotes dated events and concerts and has the information for each
stored in a mysql database with the timestamp field.
Here is the function that checks the date of the event to ensure it is
between now and three weeks from now (only events in this time period
are displayed on a page, with all events being displayed on another
page)
function check_date ($mysql_timestamp, $days) {
$timestamp = mktime(0, 0, 0, substr($mysql_timestamp, 4, 2),
substr($mysql_timestamp, 6, 2), substr($mysql_timestamp, 0, 4));
$event_day = date(z, $timestamp);
$event_year = date(Y, $timestamp);
$actual_day = date(z);
$actual_year = date(Y);
while ($event_year $actual_year) {
$event_day = $event_day + 365;
$event_year--;
}
if (($event_day - $actual_day) = $days $event_day =
$actual_day) { return TRUE; }
else { return FALSE; }
}
Then, to display the events that fill this criteria, there is this
code for the date:
function short_date ($mysql_timestamp) {
$stimestamp = mktime(0, 0, 0, substr($mysql_timestamp, 4, 2),
substr($mysql_timestamp, 6, 2), substr($mysql_timestamp, 0, 4));
$sformatted_date = date(D M j, $stimestamp);
return $sformatted_date;
}
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
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RE: [PHP-DB] date problems
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( theEventDate ) = TO_DAYS( NOW() ) AND TO_DAYS( theEventDate ) = TO_DAYS( NOW() ) + 21 I like theEventDate to be in the format -MM-DD http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_to-days _ A place for moms to take a break! http://www.reallivemoms.com?ocid=TXT_TAGHMloc=us -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problems
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 them to static values. If there's an index on theEventDate it will be used. Mike... Instruct ICC wrote: 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( theEventDate ) = TO_DAYS( NOW() ) AND TO_DAYS( theEventDate ) = TO_DAYS( NOW() ) + 21 I like theEventDate to be in the format -MM-DD http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_to-days _ A place for moms to take a break! http://www.reallivemoms.com?ocid=TXT_TAGHMloc=us -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problems
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 date_add(now(), INTERVAL 21 DAYS; This way mysql will calc the now() and date_add and will essentially convert them to static values. If there's an index on theEventDate it will be used. Mike... Instruct ICC wrote: 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( theEventDate ) = TO_DAYS( NOW() ) AND TO_DAYS( theEventDate ) = TO_DAYS( NOW() ) + 21 I like theEventDate to be in the format -MM-DD http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_to-days _ A place for moms to take a break! http://www.reallivemoms.com?ocid=TXT_TAGHMloc=us -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date problems
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 upcoming, as all being in the first
couple weeks of December, 2006.
Here is the script in action: http://www.clubandpub.ca/lobby/?city=1
The events SHOULD display on the right column, under where it says events.
If you click to the GUIDE page, the calendar should be highlighted on the
dates where there are events, and the events in the coming weeks (as well as
concerts) should display below that (under their respective headings).
This DID work, but I recently switched hosting companies as the one I was
with has become increasingly unreliable after the company changed hands. My
new server uses PHP 5 server, with MySQL 4.1 (I'm not 100% sure what the old
was.. I think it was PHP 4, and while I think it was the same MySQL version,
it could have been an earlier one).
Cheers!
rDubya
also, here is the code that actually displays the events:
?php
$db = mysql_connect(xx, user, pw) or error(mysql_error);
mysql_select_db(database, $db) or error(mysql_error);
$eventquery = mysql_query(SELECT * FROM `EVENTS` WHERE `event_city` =
'$city' ORDER BY `date` ASC, $db);
while($event_data = mysql_fetch_assoc($eventquery)) {
if (check_date($event_data[date], 21) == TRUE) {
?
it then displays the event information of the events within that time period
(pulled from the database, with the date being displayed using the
short_date function posted earlier)
On 9/6/07, Mike Gohlke [EMAIL PROTECTED] wrote:
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 date_add(now(), INTERVAL 21 DAYS;
This way mysql will calc the now() and date_add and will essentially
convert them to static values. If there's an index on theEventDate it
will be used.
Mike...
Instruct ICC wrote:
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( theEventDate ) = TO_DAYS( NOW() )
AND
TO_DAYS( theEventDate ) = TO_DAYS( NOW() ) + 21
I like theEventDate to be in the format -MM-DD
http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_to-days
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[PHP-DB] DATE
Would someone help me with this --- How do I get the results for 2 days before Easter Sunday from this --- what do I have to do to my date statement to figure out Good Friday? $good_Friday = date(Y-m-d, easter_date($current_year));
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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[PHP-DB] Date Conversion in RFC822 format
Hello all, 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. Regards Manoj
Re: [PHP-DB] Date Conversion in RFC822 format
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 rest of that page. -Stut -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date Conversion
PHP 4.4/MySQL 4.0
I am tying to convert a date to put into a database from a string (ie:
January, February) to a numeric value (ie: 01,02). I am taking the
value from a form, which is a drop down menu listing the months.
$sMonth1 = $_POST['Smonth']; returns the money selected from the form.
When I go to format the month from a string to numeric with
date('m',strtotime($sMonth1));
it returns 12, no matter which month I select.
In reading the docs at php.net, date('m',strtotime($sMonth1)); is
correct to reformat from a string to a time format.
What am I missing.
mark bomgardner
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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 Bomgardner [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 16, 2006 12:35 PM
To: Php-Db
Subject: [PHP-DB] Date Conversion
PHP 4.4/MySQL 4.0
I am tying to convert a date to put into a database from a string (ie:
January, February) to a numeric value (ie: 01,02). I am taking the
value from a form, which is a drop down menu listing the months.
$sMonth1 = $_POST['Smonth']; returns the money selected from the form.
When I go to format the month from a string to numeric with
date('m',strtotime($sMonth1));
it returns 12, no matter which month I select.
In reading the docs at php.net, date('m',strtotime($sMonth1)); is
correct to reformat from a string to a time format.
What am I missing.
mark bomgardner
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RE: [PHP-DB] Date Conversion
ADDENDUM:
To Convert your date string, use the associate array:
$months_arr = array(January=01, February
=02...);
$month = $months_arr[February];
* Replace $months_arr[February] with $_POST[Month];
// Output
$month = 01;
Ralph Brickley
-Original Message-
From: Ralph Brickley [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 Bomgardner [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 16, 2006 12:35 PM
To: Php-Db
Subject: [PHP-DB] Date Conversion
PHP 4.4/MySQL 4.0
I am tying to convert a date to put into a database from a string (ie:
January, February) to a numeric value (ie: 01,02). I am taking the
value from a form, which is a drop down menu listing the months.
$sMonth1 = $_POST['Smonth']; returns the money selected from the form.
When I go to format the month from a string to numeric with
date('m',strtotime($sMonth1));
it returns 12, no matter which month I select.
In reading the docs at php.net, date('m',strtotime($sMonth1)); is
correct to reformat from a string to a time format.
What am I missing.
mark bomgardner
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Re: [PHP-DB] Date question
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 question Date: Sun, 12 Mar 2006 20:44:13 -0700 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 those dates that match 2006-02, for example. I apologize if the solution should be staring me in the face, but all my favorites and help files are toast, until I can restore some of them.
RE: [PHP-DB] Date question
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 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 those dates that match 2006-02, for example. I apologize if the solution should be staring me in the face, but all my favorites and help files are toast, until I can restore some of them. TIA. Gerry -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date question
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 those dates that match 2006-02, for example. I apologize if the solution should be staring me in the face, but all my favorites and help files are toast, until I can restore some of them. TIA. Gerry
Re: [PHP-DB] Date question
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
Re: [PHP-DB] Date question
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 those dates that match 2006-02, for example. I apologize if the solution should be staring me in the face, but all my favorites and help files are toast, until I can restore some of them. Which database are you using? Mysql has this: http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html Postgresql has: http://www.postgresql.org/docs/8.1/static/functions-datetime.html I'm sure others have the same sort of functionality. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date Time 90 minutes ago
Would someone be able to help me with the DATE command syntax to know what the date and time was 90 minutes ago? I am trying to assign these values into two variables: $date_90_minutes_ago $time_90_minutes_ago I am not sure how to handle midnight where if the time is 00:10:00 ninety minutes earlier is a day before. Thanks. Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date Time 90 minutes ago
?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 command syntax to know what the date and time was 90 minutes ago? I am trying to assign these values into two variables: $date_90_minutes_ago $time_90_minutes_ago I am not sure how to handle midnight where if the time is 00:10:00 ninety minutes earlier is a day before. Thanks. Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Time 90 minutes ago
$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=
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Ron Piggott (PHP) wrote:
Would someone be able to help me with the DATE command syntax to know
what the date and time was 90 minutes ago? I am trying to assign these
values into two variables:
$date_90_minutes_ago
$time_90_minutes_ago
I am not sure how to handle midnight where if the time is 00:10:00
ninety minutes earlier is a day before. Thanks. Ron
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RE: [PHP-DB] Date Time 90 minutes ago
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; $second = 05; echo date(Y-m-d H:i:s, mktime($hour, $minute - 90, $second, $month, $day, $year)); It will even adjust for leap years I believe. You can add/subtract/etc any of those items and it's smart enough to figure out what the correct resulting date/time would be. -TG = = = Original message = = = ?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 command syntax to know what the date and time was 90 minutes ago? I am trying to assign these values into two variables: $date_90_minutes_ago $time_90_minutes_ago I am not sure how to handle midnight where if the time is 00:10:00 ninety minutes earlier is a day before. Thanks. Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php ___ Sent by ePrompter, the premier email notification software. Free download at http://www.ePrompter.com. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date Formatting Question
I am trying to format the month portion of a date that I am trying to pull from MySQL to be placed into a drop down menu to modify the date. I have tried several ways and none seem to be working. I am pulling the date out of MySQL with: $sDate = explode(-, $row_events['Sdate']); And then attempting to insert each portion of the array into a drop down menu with: echo select name=Smonth; echo option selected$sDate[1]/option; which is where I am running into the problem. I pull out the month as 2 digit numeric 01, 02, 03 etc., but I want it displayed as January, February, March, etc., I have tried the following with no success: Date(F,strtotime($sDate)); Strftime(%B:,$sDate); Date(F,$sDate); I would use MySQL to format the date, but I have three date fields to modify and it would be easier to do it in PHP Any pointers would be appreciated. Mark Bomgardner Technology Specialist KLETC -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Formatting Question
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'])); $monthNumber = date(m, strtotime($row_events['Sdate'])); // or n if you want 1 instead of 01 for January echo select name='sMonth'\n; for ($i = 1; $i = 12; $i++) // using date() below to get month name, day and year irrelevant $selectMonthText = date(F, mktime(0, 0, 0, $i, 1, 2000)); if ($i == $monthNumber) $selected = SELECTED; else $selected = ; echo option value='$i'$selected$selectMonthText/option\n; echo /select\n; -TG = = = Original message = = = I am trying to format the month portion of a date that I am trying to pull from MySQL to be placed into a drop down menu to modify the date. I have tried several ways and none seem to be working. I am pulling the date out of MySQL with: $sDate = explode(-, $row_events['Sdate']); And then attempting to insert each portion of the array into a drop down menu with: echo select name=Smonth; echo option selected$sDate[1]/option; which is where I am running into the problem. I pull out the month as 2 digit numeric 01, 02, 03 etc., but I want it displayed as January, February, March, etc., I have tried the following with no success: Date(F,strtotime($sDate)); Strftime(%B:,$sDate); Date(F,$sDate); I would use MySQL to format the date, but I have three date fields to modify and it would be easier to do it in PHP Any pointers would be appreciated. Mark Bomgardner Technology Specialist KLETC ___ Sent by ePrompter, the premier email notification software. Free download at http://www.ePrompter.com. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] DATE(r)
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 date(r) command. date(r) on my web site gives this response: Fri, 9 Sep 2005 13:32:19 -0400 How may I manipulate date(r) to give a format which is compatable to the column type -00-00 00:00:00 -MM-DD 24:59:59 Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] DATE(r)
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)
Date: Fri, 9 Sep 2005 12:52:59 -0500
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 date(r) command.
date(r) on my web site gives this response:
Fri, 9 Sep 2005 13:32:19 -0400
How may I manipulate date(r) to give a format which is compatable to the
column type
-00-00 00:00:00
-MM-DD 24:59:59
Ron
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Re: [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 date(r) command.
date(r) on my web site gives this response:
Fri, 9 Sep 2005 13:32:19 -0400
How may I manipulate date(r) to give a format which is compatable
to the
column type
-00-00 00:00:00
-MM-DD 24:59:59
Ron
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[PHP-DB] Date problem again ;-)
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
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 conversions
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: $testdate=30/11/2004; echo date(Y-m-d, strtotime($testdate)); the result is - 2006-06-11 strtotime() is, unfortunately, a little American biased and only recognises the mm/dd/ numeric format (using slashes). So the above is returning the equivalent of the 11th day of the 30th month of 2004! You can use other PHP functions, as has been suggested, but you might also investigate the mySQL date formatting functions, as I believe they work perfectly well on input dates as well as output ones. Cheers! Mike - Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning Information Services, JG125, James Graham Building, Leeds Metropolitan University, Headingley Campus, LEEDS, LS6 3QS, United Kingdom Email: [EMAIL PROTECTED] Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] date conversions
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: $testdate=30/11/2004; echo date(Y-m-d, strtotime($testdate)); the result is - 2006-06-11 I can't find any other function apart from strtotime to do this. Any ideas? Thanks Neil -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date conversions
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 other function apart from strtotime to do this. Use the string functions to manipulate it into the required format, explode() is one approach. If you're desperate search the archives, hundreds of variations of code to do this have been posted in the past. But it might be quicker to write your own than to hit upon the appropriate keywords to search for. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* If practice makes perfect, and nobody's perfect, why practice? */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date conversions
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.
list($d,$m,$y) = split('[/.-]', $testdate);
Another variation was to hardcode the date results instead of using the date
function which is very simple:
eg.
$mysqldate = $y.'-'.$m.'-'.$d;
Sorry for asking a question that is obviously asked often.
Interesting comment on 'normal'. I suspect that numerically more of the
world uses dmy than mdy. There is a lot more of the world outside the US
than in it.
Thanks again for all your help.
Neil
[EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Yeah, this is my problem with relyinig on strtotime(). If you don't give
it a format that it knows, it's going to give you random results (well, not
random, but undesireable). Seems like more of a crutch that leaves too
much chance of error for my taste. I prefer to be a little more explicit:
$testdate=30/11/2004;
list($day,$month,$year) = explode(/,$testdate);
echo date(Y-m-d, mktime(0,0,0,$month,$day,$year));
Try that out. mktime() produces a serial date/time just like strtotime()
but you have a little more control over what it's producing and subsequently
what gets piped into date(). This exact example is what I used once before
when arguing against using strtotime(). Most people are going to use a
'normal' format that strtotime() likes, but the format you're using
(european standard?) and just using the numbers is the one big instance that
strtotime() breaks.
Hope this helps. You should be able to get whatever date format you need
now.
-TG
*** new email address [EMAIL PROTECTED]
*** old email address [EMAIL PROTECTED] YAY CHAPTER 11! :(
= = = Original message = = =
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:
$testdate=30/11/2004;
echo date(Y-m-d, strtotime($testdate));
the result is - 2006-06-11
I can't find any other function apart from strtotime to do this.
Any ideas?
Thanks
Neil
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[PHP-DB] Date Question
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. Mark A. Bomgardner Technology Specialist KLETC
RE: [PHP-DB] Date Question
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 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. Mark A. Bomgardner Technology Specialist KLETC -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Question
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 date('Y-m-d',strtotime('09/24/2004'));
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[PHP-DB] Date Conversion
Hi all, can someone kindly point me to a resource that converts all kinds of possible date inputs into MySQL format of -MM-DD? example of formats to be converted includes: d/m/yy d/m/ d/mm/yy d/mm/yyy dd/mm/yy dd/mm/yyy d/mmm/yy d/mmm/ dd/mmm/yy dd/mmm/ yy - 2 digit representation of year - ful numeric representation of year m - numeric representation of month, without leading zero mm- numeric representation of month, with leading zero mmm - short textual representation of month d - day of month without leading zero dd - day of month with leadin zero thanx! hwee
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
[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; - Do you Yahoo!? New and Improved Yahoo! Mail - 100MB free storage!
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
[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 to have a page that displays the results for the last week only. The date format in the field is -MM-DD
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 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 - select * from table order by max(date_column) limit 1; if you want rows in a range of dates there are lots o ways, and instead of showing one or two, I'd suggest taking a look at the manual on dates http://dev.mysql.com/doc/mysql/en/Date_and_time_functions.html, and seeing what fits your needs best. Seems like you may want to look at BETWEEN And if you don't want to use those, you are always free to use select * from table where date_column '-mm-dd' and date_column '-mm-dd' order by date_col desc HTH Jeff -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date Select
Jeffrey, Thanks for this, I've just realised that it's always the same no of inserts, so perhaps I should just order by date then my previous sort field and then limit the result to 20? (It's all 20 results that I need displayed) Thanks, Tom -Original Message- From: [EMAIL PROTECTED] [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 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 - select * from table order by max(date_column) limit 1; if you want rows in a range of dates there are lots o ways, and instead of showing one or two, I'd suggest taking a look at the manual on dates http://dev.mysql.com/doc/mysql/en/Date_and_time_functions.html, and seeing what fits your needs best. Seems like you may want to look at BETWEEN And if you don't want to use those, you are always free to use select * from table where date_column '-mm-dd' and date_column '-mm-dd' order by date_col desc HTH Jeff -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date help needed
No, it's actually very easy to do the autocomplete once you get the hang of it. Actually the way I've done it is to populate a multi-select box but you could also use a DIV and write out the values Dump the email addresses as an XML file (generate this dynamically) then use XSLT to read out matching rows on each keyup ... basically you filter the XML file each time till you get down to one value Yes, it's javascript but it works really well as long as you have some control over your client browser (in the case of your boss, probably IE but it can be made to work in mozilla / firefox too) Mail me offlist if you want a working example. Cheers - Neil At 11:07 25/06/2004 +, you wrote: Message-Id: [EMAIL PROTECTED] From: Chris Payne [EMAIL PROTECTED] To: [EMAIL PROTECTED] Date: Thu, 24 Jun 2004 22:53:13 -0400 MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii 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, but have no clue how to go about that so just told him not viable ATM. CaptionKit http://www.captionkit.com : Production tools for accessible subtitled internet media, transcripts and searchable video. Supports Real Player, Quicktime and Windows Media Player. VideoChat with friends online, get Freshly Toasted every day at http://www.fresh-toast.net : NetMeeting solutions for a connected world. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date help needed
Hi there everyone, 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 display this data in the dropdown box but ONLY show 3 days a week, IE: Mondays, Fridays and Sundays, so it would show the dates for each Monday, Friday and Sunday for X amount of days (IE: 365 days in the dropdown). Does anyone have any idea how to do this? I would really appreciate any help, I'd send my sample code only I'm not at my home/work computer ATM. Chris
Re: [PHP-DB] Date help needed
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'));
$days[] = strtotime('next Sunday', strtotime('+ '.$i.' days'));
}
sort($days);
foreach($days as $day) {
echo date('Y-m-d', $day).'br/';
}
(This is not tested, but it *should* work,)
On Thu, 24 Jun 2004 17:07:12 -0400, Chris Payne
[EMAIL PROTECTED] wrote:
Hi there everyone,
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 display this data in the dropdown box but ONLY show 3 days a week,
IE: Mondays, Fridays and Sundays, so it would show the dates for each
Monday, Friday and Sunday for X amount of days (IE: 365 days in the
dropdown).
Does anyone have any idea how to do this? I would really appreciate any
help, I'd send my sample code only I'm not at my home/work computer ATM.
Chris
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RE: [PHP-DB] Date help needed
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) {
$days[] = strtotime('next Monday', strtotime('+ '.$i.' days'));
$days[] = strtotime('next Friday', strtotime('+ '.$i.' days'));
$days[] = strtotime('next Sunday', strtotime('+ '.$i.' days'));
}
sort($days);
foreach($days as $day) {
echo date('Y-m-d', $day).'br/';
}
(This is not tested, but it *should* work,)
On Thu, 24 Jun 2004 17:07:12 -0400, Chris Payne
[EMAIL PROTECTED] wrote:
Hi there everyone,
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 display this data in the dropdown box but ONLY show 3 days a week,
IE: Mondays, Fridays and Sundays, so it would show the dates for each
Monday, Friday and Sunday for X amount of days (IE: 365 days in the
dropdown).
Does anyone have any idea how to do this? I would really appreciate any
help, I'd send my sample code only I'm not at my home/work computer ATM.
Chris
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Re: [PHP-DB] Date help needed
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 display this data in the dropdown box but ONLY show 3 days a week, IE: Mondays, Fridays and Sundays, so it would show the dates for each Monday, Friday and Sunday for X amount of days (IE: 365 days in the dropdown). Does anyone have any idea how to do this? I would really appreciate any help, I'd send my sample code only I'm not at my home/work computer ATM. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date help needed
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 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, but have no clue how to go about that so just told him not viable ATM. Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] date and time problem
Hi guys
I have a script which pulls a date(date default NULL) and a time (time default NULL)
from a MySQL database, now I would like to display that date and time in a 'pretty'
format.
I've been able to show the date nicely with the help of this:
$ntime=strtotime($row['date'] $row['time']);
$ndate=date('D d F Y H:i',$ntime);
This shows the date in a 'nice' format (2004-05-10 20:40:00 is shown as Mon 10 May
2004 00:00), however, it shows the time as 00:00 for all of the games no matter what
the time is. How do I get it to show the date as 20:40 or 20h40 or 8:40 PM?
If I change the lines to this:
$ntime=strtotime($row[date] $row[time] GMT);
$ndate=date('D d F Y H:i',$ntime);
the only difference is that it now shows Mon 10 May 2004 02:00 (probably because we
are 2 hours ahead of GMT)
Any ideas please?
Thanks
K
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[PHP-DB] Date SELECT with IF
Hi, Is it possible to have a clause in a mysql SELECT statement? I would the query to display the date except where it equals the default 000-00-00 to display n/a or something similar. For example SELECT DATE_FORMAT(B.Booking_End_Date, \%Y-%m-%d\) AS 'Booking End Date' FROM Bookings (IF Booking_End_Date = '-00-00' DISPLAY 'n/a'); I hope this illustrates what I am trying to achieve! Thanks for your help -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date SELECT with IF
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: Thursday, April 15, 2004 7:44 PM Subject: [PHP-DB] Date SELECT with IF Hi, Is it possible to have a clause in a mysql SELECT statement? I would the query to display the date except where it equals the default 000-00-00 to display n/a or something similar. For example SELECT DATE_FORMAT(B.Booking_End_Date, \%Y-%m-%d\) AS 'Booking End Date' FROM Bookings (IF Booking_End_Date = '-00-00' DISPLAY 'n/a'); I hope this illustrates what I am trying to achieve! Thanks for your help -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date SELECT with IF
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
[PHP-DB] Date Manipulation
Hi People, Need some assistance in the following scenario:- Inserting a date into the database that is entered into a form by the site visitor. - This is easy enough. However I now need to use the same date that has been entered by the site visitor add 35 days and then insert into another table. 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?? Cheers, Shannon -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Manipulation
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) VALUES (20040321, 20040321 + INTERVAL 35 DAY) -- ---John Holmes... Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/ php|architect: The Magazine for PHP Professionals www.phparch.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] date problem in MySQL DB
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 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?? TIA Angelo Zanetti Disclaimer This e-mail transmission contains confidential information, which is the property of the sender. The information in this e-mail or attachments thereto is intended for the attention and use only of the addressee. Should you have received this e-mail in error, please delete and destroy it and any attachments thereto immediately. Under no circumstances will the Cape Technikon or the sender of this e-mail be liable to any party for any direct, indirect, special or other consequential damages for any use of this e-mail. For the detailed e-mail disclaimer please refer to http://www.ctech.ac.za/polic or call +27 (0)21 460 3911 -- 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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[PHP-DB] date function
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 -- SELECT name=bdate OPTION value=11/02/2003 SELECTEDToday OPTION value=11/01/2003Yesterday OPTION value=10/31/2003Fri 10/31 OPTION value=10/30/2003Thu 10/30 OPTION value=10/29/2003Wed 10/29 OPTION value=10/28/2003Tue 10/28 OPTION value=10/27/2003Mon 10/27 OPTION value=10/26/2003Sun 10/26 OPTION value=10/25/2003Sat 10/25 OPTION value=10/24/2003Fri 10/24 OPTION value=10/23/2003Thu 10/23 OPTION value=10/22/2003Wed 10/22 OPTION value=10/21/2003Tue 10/21 OPTION value=10/20/2003Mon 10/20 OPTION value=10/19/2003Sun 10/19 OPTION value=10/12/2003Sun 10/12 OPTION value=10/05/2003Sun 10/5 OPTION value=09/28/2003Sun 9/28 OPTION value=09/21/2003Sun 9/21 OPTION value=09/14/2003Sun 9/14 OPTION value=08/31/2003Sun 8/31 OPTION value=08/01/2003Fri 8/1 OPTION value=07/02/2003Wed 7/2 OPTION value=06/02/2003Mon 6/2 OPTION value=05/03/2003Sat 5/3 OPTION value=04/03/2003Thu 4/3 OPTION value=03/04/2003Tue 3/4 OPTION value=02/02/2003Sun 2/2 OPTION value=01/03/2003Fri 1/3 OPTION value=12/04/2002Wed 12/4 /SELECT -- I would like to know i can go about doing that. some examples, or if anyone has a function already the i could use. Thanks in advance,
Re: [PHP-DB] date function
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 that you mean that you want to create this dynamically in PHP starting with today's date, then you need to establish some guidelines or rules for what dates make up the list. I can't really see a pattern in your list (every day for two weeks, then every week for five weeks, then two weeks once, then monthly going back a year) so I can't offer specifics. The answer will probably be found using a for loop and the date function. Larry -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date function
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 place for this but here goes. I want to create a dropdown list with a date range of -- SELECT name=bdate OPTION value=11/02/2003 SELECTEDToday OPTION value=11/01/2003Yesterday OPTION value=10/31/2003Fri 10/31 OPTION value=10/30/2003Thu 10/30 OPTION value=10/29/2003Wed 10/29 OPTION value=10/28/2003Tue 10/28 OPTION value=10/27/2003Mon 10/27 OPTION value=10/26/2003Sun 10/26 OPTION value=10/25/2003Sat 10/25 OPTION value=10/24/2003Fri 10/24 OPTION value=10/23/2003Thu 10/23 OPTION value=10/22/2003Wed 10/22 OPTION value=10/21/2003Tue 10/21 OPTION value=10/20/2003Mon 10/20 OPTION value=10/19/2003Sun 10/19 OPTION value=10/12/2003Sun 10/12 OPTION value=10/05/2003Sun 10/5 OPTION value=09/28/2003Sun 9/28 OPTION value=09/21/2003Sun 9/21 OPTION value=09/14/2003Sun 9/14 OPTION value=08/31/2003Sun 8/31 OPTION value=08/01/2003Fri 8/1 OPTION value=07/02/2003Wed 7/2 OPTION value=06/02/2003Mon 6/2 OPTION value=05/03/2003Sat 5/3 OPTION value=04/03/2003Thu 4/3 OPTION value=03/04/2003Tue 3/4 OPTION value=02/02/2003Sun 2/2 OPTION value=01/03/2003Fri 1/3 OPTION value=12/04/2002Wed 12/4 /SELECT -- I would like to know i can go about doing that. some examples, or if anyone has a function already the i could use. Thanks in advance, -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] date function
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 -- SELECT name=bdate OPTION value=11/02/2003 SELECTEDToday OPTION value=11/01/2003Yesterday OPTION value=10/31/2003Fri 10/31 OPTION value=10/30/2003Thu 10/30 OPTION value=10/29/2003Wed 10/29 OPTION value=10/28/2003Tue 10/28 OPTION value=10/27/2003Mon 10/27 OPTION value=10/26/2003Sun 10/26 OPTION value=10/25/2003Sat 10/25 OPTION value=10/24/2003Fri 10/24 OPTION value=10/23/2003Thu 10/23 OPTION value=10/22/2003Wed 10/22 OPTION value=10/21/2003Tue 10/21 OPTION value=10/20/2003Mon 10/20 OPTION value=10/19/2003Sun 10/19 OPTION value=10/12/2003Sun 10/12 OPTION value=10/05/2003Sun 10/5 OPTION value=09/28/2003Sun 9/28 OPTION value=09/21/2003Sun 9/21 OPTION value=09/14/2003Sun 9/14 OPTION value=08/31/2003Sun 8/31 OPTION value=08/01/2003Fri 8/1 OPTION value=07/02/2003Wed 7/2 OPTION value=06/02/2003Mon 6/2 OPTION value=05/03/2003Sat 5/3 OPTION value=04/03/2003Thu 4/3 OPTION value=03/04/2003Tue 3/4 OPTION value=02/02/2003Sun 2/2 OPTION value=01/03/2003Fri 1/3 OPTION value=12/04/2002Wed 12/4 /SELECT -- I would like to know i can go about doing that. some examples, or if anyone has a function already the i could use. Thanks in advance, -- select 'mmarques' || '@' || 'unl.edu.ar' AS email; - Martín Marqués |[EMAIL PROTECTED] Programador, Administrador, DBA | Centro de Telemática Universidad Nacional del Litoral - -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date Pulldown
Sorry for what is probably going to be a late reply. I'm waiting to pick
someone up at the airport, and the plane was delayed, so I wanted to try my
hand at answering your question.
First I had to figure out what method you used to pick those dates in the
first place. It seems to be...
Today, Yesterday, then 1 day back for a total of 14 days (2 weeks)
Then going 1 week back for a total of 8 weeks (56 days)
Then I'm stumped how you came up with the jump from 9/14 to 8/31 to
8/1 unless the rule is go back 2 weeks then go back 30 days...
Is there a reason the next rule can't be: go 1 month back for a total of 11
months?
Here's a little piece of PHP code which will do just about exactly what
you're looking for, with the rule modification I suggested. Perhaps it
will do the trick? (And, since I'm no guru myself, I welcome comments on
my suggestion). It basically works around mktime and date. Both are
very good functions to accomplish what you're trying to accomplish. Read
up on those and you should be able to handle it.
form name=foo
select name=bdate
?php
output_date_option ( mktime(0,0,0, date(m), date(d), date(Y)) ,
Today , SELECTED );
output_date_option ( mktime(0,0,0, date(m), date(d)-1, date(Y)) ,
Yesterday );
for ($days = 2; $days = 14; $days++) {
output_date_option ( mktime(0,0,0, date(m), date(d)-$days,
date(Y)) );
}
for ($weeks = 3; $weeks = 7; $weeks++) {
output_date_option ( mktime(0,0,0, date(m), date(d)-($weeks*7),
date(Y)) );
}
for ($months = 2; $months = 11; $months++) {
output_date_option ( mktime(0,0,0, date(m)-$months, date(d),
date(Y)) );
}
function output_date_option($current_date, $date_string = , $xparams = ) {
if (!isset($date_string) or $date_string == ) {
$date_string = date ( D m/d , $current_date );
}
echo ( 'OPTION VALUE='.date ( m/d/Y , $current_date ).'
'.$xparams.''.$date_string );
}
?
/select
/form
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[PHP-DB] date function
Greetings,
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');
$query = SELECT name,hdate FROM emp2 where month(hdate)=$mymonth
order by hdate;
$result = mysql_db_query(iweb, $query);
if ($result) {
echo table align=center border=0 cellspacing=5 ;
while ($r = mysql_fetch_array($result)) {
$name = $r[name];
$hyear = date('Y',$r[hdate]);
$timein = $cyear - $hyear;
if ($timein 0) {
echo
trtd$name/ttd$timein/tdtd$cyear/tdtd$hyear/td/tr;}
}
echo /table;
} else {
echo No data.;
}
mysql_free_result($result); ?
I'm trying to figure out years of employment based on hiredate. Its not
working as expected. $timein is always
1969. Date in the database is -MM-DD. What have I screwed up ?
thanks,
Darryl
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Re: [PHP-DB] date function
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');
$query = SELECT name,hdate FROM emp2 where month(hdate)=$mymonth
order by hdate;
$result = mysql_db_query(iweb, $query);
if ($result) {
echo table align=center border=0 cellspacing=5 ;
while ($r = mysql_fetch_array($result)) {
$name = $r[name];
$hyear = date('Y',$r[hdate]);
$timein = $cyear - $hyear;
if ($timein 0) {
echo
trtd$name/ttd$timein/tdtd$cyear/tdtd$hyear/td/tr;}
}
echo /table;
} else {
echo No data.;
}
mysql_free_result($result); ?
I'm trying to figure out years of employment based on hiredate. Its not
working as expected. $timein is always
1969. Date in the database is -MM-DD. What have I screwed up ?
date() in php expects a unix timestamp. Dates in MySQL are NOT stored in unix
timestamp format.
You can perform the calculations directly in MySQL
mysql manual Tutorial Introduction Date Calculations
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[PHP-DB] Date format problem
Hello, 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/ how do I do that? Thank you in advance, Dani --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.502 / Virus Database: 300 - Release Date: 18/7/2003 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Date format problem
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 Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[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
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;
[PHP-DB] date, and time?
Cheers for the help on my last auto inc prob.. I checked out phpmyadmin, and I'm loving it...! cheers all Anyway, I want to know how to store a date like this in a MySQL database: date(F j, Y, g:i a); Resulting in: July 10, 2003, 3:39 am But what do I need to set my database table to, to capture the full data. I can use text, but then I can't sort by date later on Ideas? Cheers list... * The information contained in this e-mail message is intended only for the personal and confidential use of the recipient(s) named above. If the reader of this message is not the intended recipient or an agent responsible for delivering it to the intended recipient, you are hereby notified that you have received this document in error and that any review, dissemination, distribution, or copying of this message is strictly prohibited. If you have received this communication in error, please notify us immediately by e-mail, and delete the original message. ***
Re: [PHP-DB] date, and time?
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 your data... http://www.mysql.com/doc/en/Date_and_time_types.html http://www.mysql.com/doc/en/Date_and_time_functions.html hth jeff [EMAIL PROTECTED] sungard.com To: [EMAIL PROTECTED] cc: 07/10/2003 11:02 AM Subject: [PHP-DB] date, and time? Cheers for the help on my last auto inc prob.. I checked out phpmyadmin, and I'm loving it...! cheers all Anyway, I want to know how to store a date like this in a MySQL database: date(F j, Y, g:i a); Resulting in: July 10, 2003, 3:39 am But what do I need to set my database table to, to capture the full data. I can use text, but then I can't sort by date later on Ideas? Cheers list... * The information contained in this e-mail message is intended only for the personal and confidential use of the recipient(s) named above. If the reader of this message is not the intended recipient or an agent responsible for delivering it to the intended recipient, you are hereby notified that you have received this document in error and that any review, dissemination, distribution, or copying of this message is strictly prohibited. If you have received this communication in error, please notify us immediately by e-mail, and delete the original message. *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date Formatting in PHP
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 I am trying to avoid having to mess with the MySQL select statement, as the output is being passed through several documents. So I need to do it within PHP. I tried making a conversion like $DayReport = date (F d, Y, $DatePick); But the value of $DayReport is neither correct nor does it change when $DatePick changes. For instance, in this example, the $DatePick value of 2003-11-18 gets converted to December 31, 1969. TIA Dave Shugarts -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Formatting in PHP
Hi David,
Try this one:
?
function formatDate($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 PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, June 10, 2003 11:23 PM
Subject: [PHP-DB] Date Formatting in PHP
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
I am trying to avoid having to mess with the MySQL select statement, as
the
output is being passed through several documents. So I need to do it
within
PHP.
I tried making a conversion like
$DayReport = date (F d, Y, $DatePick);
But the value of $DayReport is neither correct nor does it change when
$DatePick changes. For instance, in this example, the $DatePick value of
2003-11-18 gets converted to December 31, 1969.
TIA
Dave Shugarts
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Re: [PHP-DB] Date Formatting in PHP
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 concatenation to build $DayReport from the current value of $DatePick. I am trying to avoid having to mess with the MySQL select statement, as the output is being passed through several documents. So I need to do it within PHP. I tried making a conversion like $DayReport = date (F d, Y, $DatePick); But the value of $DayReport is neither correct nor does it change when $DatePick changes. For instance, in this example, the $DatePick value of 2003-11-18 gets converted to December 31, 1969. TIA Dave Shugarts -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Formatting in PHP
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...
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Re: [PHP-DB] Date Formatting in PHP
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));
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[PHP-DB] date to Y-M-D
a function to convert any date to ymd...
function datetoymd($date){
$dateymd = date('Y-m-d',$date);
return $dateymd;
}
this function when output gives me the date 1970-01-01 (date of the unix
timestamp start) so, ehm, why!?
cheers,
dave
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Re: [PHP-DB] date to Y-M-D
function datetoymd($date){
$dateymd = date('Y-m-d',$date);
return $dateymd;
}
this function when output gives me the date 1970-01-01 (date of the unix
timestamp start) so, ehm, why!?
Because you're not passing a valid Unix timestamp or whatever your passing
is empty.
---John Holmes...
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RE: [PHP-DB] date to Y-M-D
ehmmm what kind of dates do you use, maybe dates of birth?
when using date of unix timestamp you can't have dates befor 1970 so if
those timestamps represents an older date you can't use timestamps!
-mark-
-Oorspronkelijk bericht-
Van: David Rice [mailto:[EMAIL PROTECTED]
Verzonden: Tuesday, March 25, 2003 3:42 PM
Aan: [EMAIL PROTECTED]
Onderwerp: [PHP-DB] date to Y-M-D
a function to convert any date to ymd...
function datetoymd($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!?
cheers,
dave
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[PHP-DB] Date in a dropbox box
Hi there everyone, 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 it sounds odd :-) so if it was the 5th of a month, it would show 5 as a default value, but allow you to select 1-4 before that and 6-31 after it. Thanks everyone Chris
RE: [PHP-DB] Date in a dropbox box
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 it
sounds odd :-)
so if it was the 5th of a month, it would show 5 as a default value,
but
allow you to select 1-4 before that and 6-31 after it.
Lots of ways you can do it.
$today = date('d');
for($x=1;$x=31;$x++)
{
echo 'option value=$x' . (($x==$today)?' selected':'') .
'$x/option';
}
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
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Re: [PHP-DB] Date in a dropbox box
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
nomatter how many days are in the month? This is important even
though it
sounds odd :-)
so if it was the 5th of a month, it would show 5 as a default value,
but
allow you to select 1-4 before that and 6-31 after it.
Lots of ways you can do it.
$today = date('d');
for($x=1;$x=31;$x++)
{
echo 'option value=$x' . (($x==$today)?' selected':'') .
'$x/option';
}
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
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Re: [PHP-DB] date functions (generates parse error)
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 +
INTERVAL 6 DAY) ORDER BY staffid;
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
if ( isset ( $tips ) ){
if (isset ( $tips[$row[staffid]] ) ){
$hours = $row[finish] - $row[start];
$tips[$row[staffid]] = $tips[$row[staffid]] + $hours;
}
else{
$tips[$row[staffid]] = $row[finish] - $row[start];
}
}
else{
$tips = array('$row[staffid]' =( $row[finish] - $row[start] ) );
}
}
return $tips;
}
function dbconnect(){
mysql_connect(localhost, filterseveuk, godisadj);
mysql_select_db(filterseveuk);
}
dbconnect();
$date = 2003-03-02;
var_dump(tips($date));
?
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RE: [PHP-DB] date functions (generates parse error)
This is very weird. What are you using to create this code?
If you remove all of the spaces (?) before $query, the code doesn't give
a parse error. It shouldn't give one either way, though, if those are
spaces or a tab before your $query = ... line.
Even if you remove all of the text and only leave ?, ?, and those
spaces, PHP will spit something like:
Notice: Use of undefined constant - assumed ' ' in
c:\inetpub\wwwroot\test.php on line 3
So, the fix is to remove those characters and replace them with spaces
or tabs.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
-Original Message-
From: David Rice [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 05, 2003 12:34 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] date functions (generates parse error)
Sorry, I forgot to add the part at the bottom where I am calling the
function
=
?
function tips($weekstart){
$start = date('Ymd',strtotime($weekstart));
$query = SELECT * FROM Rota WHERE date = $start and date =
($start +
INTERVAL 6 DAY) ORDER BY staffid;
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
if ( isset ( $tips ) ){
if (isset ( $tips[$row[staffid]] ) ){
$hours = $row[finish] - $row[start];
$tips[$row[staffid]] =
$tips[$row[staffid]] +
$hours;
}
else{
$tips[$row[staffid]] = $row[finish] -
$row[start];
}
}
else{
$tips = array('$row[staffid]' =( $row[finish] -
$row[start] ) );
}
}
return $tips;
}
function dbconnect(){
mysql_connect(localhost, filterseveuk, godisadj);
mysql_select_db(filterseveuk);
}
dbconnect();
$date = 2003-03-02;
var_dump(tips($date));
?
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[PHP-DB] date functions
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?! cheers, dave _ Chat online in real time with MSN Messenger http://messenger.msn.co.uk -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
