|
|
 |
XVIII. Dates et heures- Table des mati�res
- checkdate -- Valide une date/heure.
- date -- Formate une date/heure locale
- getdate -- Retourne la date/heure
- gettimeofday -- Retourne l'heure actuelle
- gmdate -- Formate une date/heure GMT/CUT.
- gmmktime -- Retourne le timestamp UNIX d'une date GMT.
- gmstrftime --
Formate une date/heure GMT/CUT en fonction des param�trages locaux.
- localtime -- Lit l'heure locale
- microtime --
Retourne le timestamp UNIX actuel avec microsecondes.
- mktime --
Retourne le timestamp UNIX d'une date.
- strftime --
Formate une date/heure locale avec les options locales.
- strtotime --
Transforme un texte anglais en timestamp
- time --
Retourne le timestamp UNIX actuel.
User Contributed Notes
Dates et heures |
  |
[email protected]
30-Jan-2000 07:06 |
|
If you need to convert a string to a timestamp, versions 3.0.12 and up
include strtotime($AString) which uses a yacc-generated parser to generate
a timestamp from your user input. It assumes that times are broken up by
":", and that dates use the "/" delimiter and are
entered in American (MM/DD/YY) format. am/pm and 24 hour formats are
accepted for times.
If the string passed to the function only
indicates the time (for example, 12:00am) the timestamp that is generated
will contain that time, on the day that the timestamp was
generated.
Dates without a time are assumed to be 12:00am. There may
be odd behavior related to time-zones, as under certain circumstances the
creation of the date and the printing of the date may each assume that the
date is in UTC, causing a double correction.
One other thing about
strtotime- if your entry is invalid, it'll return -1 to indicate failure.
|
|
[email protected]
18-Feb-2000 04:17 |
|
If you have a date format that's in days (with the decimal part being
fractional) you just need to find the number of days between the beginning
of the pc date and the epoch (set the clock that works off of pc time to
1/1/70 to see how many days it is), subtract that number from the pc date,
and multiply by the number of seconds in a day (60*60*24). That should
work.
|
|
[email protected]
21-Feb-2000 05:04 |
|
The quick and easy to convert a mysql timestamp is to use the
UNIX_TIMESTAMP(field_name) function in mysql when you query the data. It
converts a 14-digit mysql timestamp to a standard unix timestamp.
|
|
|
25-Apr-2000 12:44 |
|
Regarding the postings regarding MySQL timestamps, please be aware that
there are several MySQL date and time types, as documented at
So,
in reference to the above posts, nic_lee's algorhythm works on the MySQL
TIMESTAMP type, while jmat's function works on the MySQL DATETIME type --
they are NOT the same!
|
|
[email protected]
09-Jul-2000 03:11 |
|
To get the number of days between two dates Get the UNIX timestamp for the
two days, using time for the current timestamp and mktime to generate
others. Since timestamps are in seconds from Epoch time you can simply
subtract the two and Multiply by 60 * 60 * 24 to get the number of days.
|
|
[email protected]
02-Aug-2000 07:00 |
|
To get the calendar week use:
strftime("%W",mktime(0,0,0,$month,$day,$year)); This returns the
week number fram a date... -Philip
|
|
[email protected]
23-Aug-2000 01:36 |
|
This is a little procedure to make time loop under a start time, end time
and an interval,I have done the same type thing with the date.
where
you see:
echo "Time: $timeh:$timem
\n";
that is my
command you can of course call your own functions with the vars,$timeh
& $timem.
$timeh = "9"; # Inital Start Time
(hours part)
$timem = "30"; # Inital Start Time (mins
part)
$end_timeh = "17"; # End Time
(hours)
$end_timem = "30"; # End Time
(mins)
$interval = "5"; # Interval
(mins)
$exit_called = "0";
while
($exit_called == "0")
{
# Pad is a my pre-deffined
function add a '0' infront of a var if need be for proper formating.
$timem = pad($timem);
$timeh = pad($timeh);
#
command goes here for execution with the time
echo "Time:
$timeh:$timem
\n";
if($timeh >=
$end_timeh)
{
$exit_called = "1";
}
else
{
# Increment Counters
$timem =
$timem + $interval;
if($timem >=
"60")
{
$answer = ($timem / 60);
$round_val
= round($answer);
if($round_val >
$answer)
{
$round_val = ($round_val -
1);
}
$timeh = ($timeh +
$round_val);
$round_val = ($round_val * 60);
$timem =
($timem - $round_val);
}
}
}
|
|
[email protected]
29-Aug-2000 01:29 |
|
Please note that the calendar week algorithm to apply is country-specific.
The first week of a year is usually just a partial week (e.g. Wed-Sun),
and the ideas whether this partial week is week 0, week 52 of the previous
year, or week 1 depend on social (and business) conventions which can vary
across countries; sometimes the definition varies according to whether
that partial week has at least 4 days. These conventions may even vary
with time; for example, Germany had its weeks from Sunday to Saturday and
went to a Monday-to-Sunday cycle some time between ten and twenty years
ago.
|
|
[email protected]
07-Sep-2000 10:18 |
|
if you were trying to convert a mysql format date or time string you could
always do it in the select statement by using DATE_FORMAT and
TIME_FORMAT.
mysql date is 2000-09-07
mysql time is
03:13:27
so...
SELECT DATE_FORMAT(date, '%m.%d.%Y') AS
date, TIME_FORMAT(time, '%l.%i %p') AS time
would produce date as
September 7, 2000 and time as 3:13 AM. See
for more date and time formatting specifiers.
|
|
[email protected]
08-Sep-2000 09:02 |
|
There is an an excellent article by Allan Kent on PHP date/time. The full
article can be found at:
|
|
[email protected]
24-Sep-2000 02:53 |
|
Need the week of the month?
Heres a function that should do the
trick:
/*Call the function as:
weekofmonth(year,month,date)
-
either can be int's or strings with leading 0's.
If theres no
arguements today will be the arguements.
Output is in the form
"YYYYmmw" where "w" represents the week of the
month.
*/
function weekofmonth($yr=0,$mn=0,$dd=0) {
if(!($yr>0&$mn>0&$dd>0)) {
$yr=date("Y");$mn=date("m");$dd=date("d");
}
$thisweek =
strftime("%U",mktime(0,0,0,$mn,$dd,$yr));
$firstweekofmonth = strftime("%U", mktime(0, 0, 0,
date("m", mktime(0, 0, 0, $mn, $dd+(6-date("w",
mktime(0, 0, 0, $mn, $dd, $yr))), $yr)), 1,$ yr));
$thisweek =
(abs($thisweek-$firstweekofmonth)>5) ?
$firstweekofmonth:$thisweek;
return date("Y", mktime(0, 0,
0, $mn, $dd+(6-date("w", mktime(0, 0, 0, $mn, $dd, $yr))), $yr))
. date("m", mktime(0, 0, 0, $mn,
$dd+(6-date("w",mktime(0, 0, 0, $mn, $dd, $yr))), $yr)) .
(($thisweek - $firstweekofmonth) + 1);
}
|
|
[email protected]
13-Nov-2000 06:13 |
|
I was trying to make a Month-At-A-Glance and I finally got it to work so I
thought I'd share it too. What you need to get this to work is the
"Day Of The Week Number", i.e., Sunday=1 and the "Number Of
Days in the Month". I also used Allan Kent's Date/Time Column at
PHPBuilder to get the required information. I also added the process I
used to the requried
information.
Enjoy!
<html>
<head>
<title>Month-At-A-Glance</title>
</head>
<body>
<?php
//
*** These are here if you need to calculate the required values ***
//
*** Reference Allan Kent's Column on Dates at
PHPBuilder.com
//$thismonth = mktime($hours, $minutes,$seconds
,$month, $day,$year);
//$firstday = mktime($hours, $minutes,$seconds
,$month,
1,$year);
//$dayofweek=date("D",$thismonth);
//$firstdayofmonth=date("D",$firstday);
//$weekdaynum=array("Sun"=>1,
"Mon"=>2, "Tue"=>3, "Wed"=>4,
"Thu"=>5, "Fri"=>6,
"Sat"=>7);
//$dayNumber=$weekdaynum[$dayofweek];
print
"<table width=80% align=center
border=1>\n";
print"<tr><th>Sun(1)</th>
<th>Mon(2)</th> <th>Tue(3)</th>
<th>Wed(4)</th> <th>Thu(5)</th>
<th>Fri(6)</th> <th>Sat(7)</th>
</tr>";
$daysinmonth=31;//For July 2000 this is 31
days
$daycount=1;
$firstdayofmonth=7;//For July 2000 this is Sat,
or day 7
for ($week=1; $week <= 6;
$week++){
print"<tr>\n";
for($day=1; $day
<=7; $day++){
if( ($day >= $firstdayofmonth) || ($week
>1) ){
print"<td height=40px>";
if ($daycount <= $daysinmonth){
print $daycount;
$daycount++;
}
print" </td>\n";
}
else{
print"<td height=40px>
</td>\n";
}
}
print"\n</tr>\n";
}
print"</table>";
?>
</body>
</html>
|
|
[email protected]
16-Nov-2000 03:58 |
|
need the age of a person or how many years passed since a certain
date?
this should work, also works when the person is born on Feb,
29th
function get_age($birthday) {
if (!ereg
("([0-9]{4})-([0-9]{1,2})-([0-9]{1,2})", $birthday, $regs))
{ return false;} // Geb-Datum nicht korrekt �bergeben
$age =
date("Y") - $regs[1];
if ($regs[2] >
date("m")) {$age--;}
elseif ($regs[2] ==
date("m")) {
if ($regs[3] > date("d"))
{$age--;}
}
return $age;
} // Ende function get_age
|
|
[email protected]
26-Nov-2000 04:38 |
|
Function 'wend'
Pass to: A number representing the week of the year
($wkno).
Uses: $t - the current time and return value
$dx - the
current day of the week (0-6)
$woy - the current week in the
year
DAY - The number of seconds in a day
WEEK - The number of
seconds in a week
MAXDAY - The max day-of-the-week num.
The
bulk of the work is performed by the line immediately preceeding the
'return $t;'.
First of all the weekending timestamp is defined
for the current week '((MAXDAY - $dx) * DAY) ' it is then adjusted for the
week of the year that you wish to determine weekending date for by the
'(($wkno - $woy) * WEEK)' term.
Returns: A timestamp ($t) for
the weekending (a Saturday) relating to the week number passed the date is
then generated in the main program using a line like..... '$wkending =
date("d/m/y",(wend($wk)));'
function
wend($wkno)
{
define ("DAY",
"86400");
define ("WEEK",
"604800");
define ("MAXDAY",
"6");
$t = time();
$dx = date("w");
$woy =
exec('date +%U');
$t += (((MAXDAY - $dx) * DAY) + (($wkno - $woy) *
WEEK));
return $t;
}
|
|
[email protected]
13-Dec-2000 02:19 |
|
You can use Mysql
function:
<?php
$data_time="1998-12-31
23:59:59";
$connect_id=mysql_connect('localhost');
$query_id=mysql_query("SELECT
DATE_ADD('$data_time',INTERVAL 1
YEAR)",$connect_id);
$data_time=mysql_result($query_id,0);
mysql_close($connect_id);
echo
$data_time;
?>
|
|
[email protected]
11-Jan-2001 05:00 |
|
I had some problems with dates between mySQL and PHP. PHP had all these
great date functions but I wanted to store a usable value in my database
tables. In this case I was using TIMESTAMP(14) <or
'YYYYMMDDHHMMSS'>.
This is perhaps the easiest way I have found to
pull the PHP usable UNIX Datestamp from my mySQL datestamp stored in the
tables:
Use the mySQL UNIX_TIMESTAMP() function in your SQL
definition string. i.e.
$sql= "SELECT field1, field2,
UNIX_TIMESTAMP(field3) as your_date
FROM your_table
WHERE field1 = '$value'";
The query will return a temp
table with coulms "field1" "Field2"
"your_date"
The "your_date" will be formatted
in a UNIX TIMESTAMP! Now you can use the PHP date() function to spew out
nice date formats.
Sample using above $sql:
20010111002747 =
Date Stored on mySQL table (TIMESTAMP(14))
979172867 = value returned
as your_date in sql stmt (UNIX_TIMESTAMP)
if we use $newdate =
date("F jS, Y -- g:ia", $row["your_date"]);
--(after fetching our array from the sql results of
course)--
echo "$newdate"; --Will
produce:
January 11th, 2001 -- 12:27am
Hope this helps
someone out there!
|
|
[email protected]
31-Jan-2001 10:44 |
|
in jmat's example above
function mysql_to_epoch ($datestr) {
list($year,$month,$day,$hour,$minute,$second) =
split("([^0-9])",$datestr);
return
date("U",mktime($hour,$minute,$second,$month,$day,$year));
}
If your $datestr has quotes around it you will need to add
a placeholder to the list.
Ex:
list($junk,$year,$month,$day,$hour,$minute,$second) =
split("([^0-9])",$datestr);
|
|
[email protected]
07-Feb-2001 11:23 |
|
Some general date functions.
function sub($timestamp,
$seconds,$minutes,$hours,$days,$months,$years) {
$mytime =
mktime(1+$hours,0+$minutes,0+$seconds,1+$months,1+$days,1970+$years);
return
$timestamp - $mytime;
}
function add($timestamp,
$seconds,$minutes,$hours,$days,$months,$years) {
$mytime =
mktime(1+$hours,0+$minutes,0+$seconds,1+$months,1+$days,1970+$years);
return
$timestamp + $mytime;
}
function dayOfWeek($timestamp)
{
return
intval(strftime("%w",$timestamp));
}
function
daysInMonth($timestamp) {
$timepieces =
getdate($timestamp);
$thisYear =
$timepieces["year"];
$thisMonth =
$timepieces["mon"];
for($thisDay=1;checkdate($thisMonth,$thisDay,$thisYear);$thisDay++);
return
$thisDay;
}
function firstDayOfMonth($timestamp)
{
$timepieces = getdate($timestamp);
return
mktime( $timepieces["hours"],
$timepieces["minutes"],
$timepieces["seconds"],
$timepieces["mon"],
1,
$timepieces["year"]);
}
function
monthStartWeekDay($timestamp) {
return
dayOfWeek(firstDayOfMonth($timestamp));
}
function
weekDayString($weekday) {
$myArray = Array( 0 =>
"Sun",
1 => "Mon",
2 =>
"Tue",
3 => "Wed",
4 =>
"Thu",
5 => "Fri",
6 =>
"Sat");
return $myArray[$weekday];
}
function
stripTime($timestamp) {
$timepieces =
getdate($timestamp);
return mktime( 0,
0,
0,
$timepieces["mon"],
$timepieces["mday"],
$timepieces["year"]);
}
function
getDayOfYear($timestamp) {
$timepieces =
getdate($timestamp);
return
intval($timepieces["yday"]);
}
function
getYear($timestamp) {
$timepieces = getdate($timestamp);
return
intval($timepieces["year"]);
}
function
dayDiff($timestamp1,$timestamp2) {
$dayInYear1 =
getDayOfYear($timestamp1);
$dayInYear2 =
getDayOfYear($timestamp2);
return ((getYear($dayInYear1)*365 +
$dayInYear1) -
(getYear($dayInYear2)*365 +
$dayInYear2));
}
hope they are usefull to you.
-
Vincent
|
|
[email protected]
18-Feb-2001 09:39 |
|
To use dates and times in ODBC-compliant databases (including MySQL), you
can use the timestamp format: {ts 'yyyy-mm-dd hh:mm:ss'}. {ts
'1899-12-30 00:00:00'} is the earliest date and time that can be inserted
into a database using this format. This is perhaps the most portable and
supported syntax out there by most databases (however, a good database
will support several methods for date/time objects).
|
|
[email protected]
17-Apr-2001 10:42 |
|
Some useful functions:
/* dayofweek() will return the day of the
week a given date falls. 0=Sunday, 1=Monday, etc. */
function
dayofweek($day,$month,$year) {
/* Check date for validity
*/
if (!checkdate($month,$day,$year))
return -1;
$a=(int)((14-$month) / 12);
$y=$year-$a;
$m=$month + (12*$a) - 2;
$retval=($day + $y + (int)($y/4) - (int)($y/100) + (int)($y/400) +
(int)((31*$m)/12)) % 7;
return $retval;
}
/*
phpdow_mod is a mod function that deals with negative numbers properly,
used by the nthDayOfMonth function below. */
function
phpdow_mod($a,$b) {
if ($a <= 0)
return
(int)phpdow_mod($b-abs($a),$b);
else
return (int)($a%$b);
}
/*
*
nthDayOfMonth($n,$dow,$month,$year) will compute the Nth day of the
given
* month. For example, the first Monday in April 2001.
*
* Parameters:
* $n - the Nth day you want, i.e. 2 for 2nd
* $dow - The day of week you want, 0=Sunday, 1=Monday, etc.
[0-6]
* $month - The month you want [1-12]
* $year - The full
year [like 2001]
*
* Returns:
* The date, in the month you
passed, that fits the criteria.
* or..
* -1 = invalid
date
* -2 = There is no Nth day of that month, like no 5th
Tuesday
* of the specified month
*/
function
nthDayOfMonth($n,$dow,$month,$year) {
/* Check the date
*/
if ($month > 12)
return -1;
if ($dow > 6)
return -1;
/*
Valid Nth day, should be no more than 5 */
if (($n <= 0) ||
($n > 5))
return -1;
$retval =
(7*$n)-6+phpdow_mod($dow-dayofweek(1,$month,$year),7);
/* Sanity
check */
if (!checkdate($month,$retval,$year))
return -2;
return $retval;
}
|
|
[email protected]
18-May-2001 04:52 |
|
This function will return the unix-timestamp of the specified
date.
function dateToTime($date, $blnUs = false)
{
$blnUs
? list($m,$d,$y)=split('[.-/,]',$date) :
list($d,$m,$y)=split('[.-/,]',$date);
return
mktime(0,0,0,$m,$d,$y);
}
The string containing the date can
be of any digit-based format:
d.m.yy, dd/m/yyyy,....
If the
2nd parameter is set to true the function also handles us - formatted
dates (m-d-yyyy, mm.d.yyyy,...)
|
|
[email protected]
22-Jun-2001 01:57 |
|
For add 30 day to a date from a form.
Page1
<INPUT
TYPE='text' NAME='codea' maxlength='4' size=4 value=''>-
<INPUT TYPE='text' NAME='codeb' maxlength='2' size=2 value=''>-
<INPUT TYPE='text' NAME='codec' maxlength='2' size=2
value=''>
Page2
<?
$encoded_date =
"$codea-$codeb-$codec";
$poll =
"$codea/$codeb/$codec";
$encoded_datc = date("Y m
d",$data + 2592000);
echo " $encoded_datc ";
?>
|
|
[email protected]
17-Jul-2001 01:42 |
|
For those who want portable timestamps between ODBC-compliant datasources,
use the following format:
{ts '1890-12-30 00:00:00'}
(The
above timestamp is associated as the "zero-base" in many
datasources)
So, to use the above timestamp in a query, do
something like this:
INSERT INTO MyTable (Table_ID,
Table_TS)
VALUES (1, {ts '1890-12-30 00:00:00'})
If you just
want to use the date, use this:
{d '1890-12-30'}
Or just
the time:
{t '00:00:00'}
That should ease someone's
pain. Instead of running a query against the database for a timestamp
(which is generally slow), you can generate one real-time. For
instance:
echo "{ts '".gmdate("Y-m-d h:i:s",
time())."'}";
It is just a matter of modifying the
above code for whatever queries you need to
perform.
MacTruck
(BTW, if you can't tell, those
are braces '{}' rather than parenthesis '()' surrounding the
timestamp...they are also known as "escape characters" in ODBC
terminology).
|
|
[email protected]
31-Jul-2001 12:07 |
|
In addition to [email protected]'s example. I needed a quick simple
function that added 30 days to a date for a PHP Site running
3.0.7.
I came up with,
//
-----------------------------
// -- Add thirty days --
//
-----------------------------
$timenow = time();
$plusthirtydays = date("d-m-Y", $timenow + 2592000);
//
-----------------------------
$plusthirtydays now contains your
date, 30 days later than current.
You can change the timestamp if
you wish to increase the number of days etc. And you can select whether to
add or subtract.
Alternatively, I believe strtotime() is a good
function to do this job, available in 3.0.12
Its not much, but I
like it :),
Amadiere
|
|
[email protected]
16-Aug-2001 11:20 |
|
Simple addDays function: add $days to a typical $datetime value (see
return).
function addDays($datetime, $days) {
$dd =
strtotime($datetime);
$dd2 = $dd + 3600*24*$days;
return
date("Y-m-d H:i:s", $dd2);
}
|
|
[email protected]
19-Aug-2001 01:06 |
|
[Editor's note: a simpler and more efficient solution is to use the
UNIX_TIMESTAMP(fieldname) function in MySQL when performing a SELECT
query. This is documented in the MySQL manual.]
Ever use the
mysql column type of "timestamp"? It crams all the digits fo
the date/time togeter and makes it hard to deal with. Here's a
solution:
#####################################################################
//
myts_date is identical to date except that it takes a
// mysql
timestamp instead of epoch seconds for an argument
// written by
William Janoch - [email protected]
//
// I write
"clever little" code that never
// complains when the sky
isn't falling.
// /* see also "chicken little" code
*/
function myts_date($format,$mytimestamp)
{
$month = substr($mytimestamp,4,2);
$day =
substr($mytimestamp,6,2);
$year =
substr($mytimestamp,0,4);
$hour =
substr($mytimestamp,8,2);
$min =
substr($mytimestamp,10,2);
$sec =
substr($mytimestamp,12,2);
$epoch =
mktime($hour,$min,$sec,$month,$day,$year);
$date = date
($format, $epoch);
return
$date;
}
#####################################################################
|
|
[email protected]
31-Aug-2001 05:21 |
|
For all of us who speak Spanish, here is a function that adds a
determinated amount of days to a given date... the format for the incoming
and the outgoing date is: DD-MM-YYYY
-----------------------
Para
todos quienes hablamos espa�ol, he aqu� una funci�n que agrega una
determinada cantidad de d�as a una fecha dada... el formato tanto para la
fecha de entrada como para la de salida es: DD-MM-YYYY
function
agregaDias($fecha, $dias) {
$t1 =
mktime(0,0,0,substr($fecha,3,2),substr($fecha,0,2),substr($fecha,6,4));
$t2 = $t1 + 3600*24*$dias;
return
date("d-m-Y", $t2);
}
|
|
[email protected]
15-Sep-2001 09:50 |
|
PM or AM = A
Seconds = s
Minuts = i
Hours = h
Year = Y
Month = F
Month (16th) = dS
Day = 1
|
|
[email protected]
09-Oct-2001 08:06 |
|
Simple way to calculate the days until something (looks like one of the
entries above suggest multiplying instead of dividing!). This'll return a
nice number formatted to two (2) decimal places.
$today =
time();
$daysuntil = ((($timestamp - $today) / 60) / 60) /
24;
$daysuntil = number_format($daysuntil,2);
|
|
[email protected]
18-Oct-2001 11:24 |
|
A major problem I have run into regarding timestamps is the limitation
implied by the UNIX timestamp range, which starts at 12:00am on 1st
January 1970, and ends some time in 2030 (I think).
Anything outside
this range returns a -1 from the mktime() function.
|
|
florent at kzar.net
25-Oct-2001 06:26 |
|
The Latest date available for timestamp related functions (mktime...)
is:
January 19 th, 2038, 3:23 AM and 27 seconds (or 2147480607 Unix
timestamp).
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[email protected]
04-Nov-2001 11:41 |
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the no of seconds in one day is 86400
so to round a unix timestamp
down to the nearest day, use
$time = floor(time()/86400)*86400
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[email protected]
05-Dec-2001 12:38 |
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For Brazilian users:
Remember that the banking ticket uses a
sequencial date which begins on October 1st, 1997.
Lembre-se que
o fator de vencimento nos boletos bancarios e' uma data sequencial, cujo
dia 1 e' na verdade 7/10/1997.
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[email protected]
22-Jan-2002 02:14 |
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I probably didn't need to write this but it was fun anyway. This function
will work out the difference between two timestamps and process the number
of seconds into a text string broken down into years,weeks,minutes and
seconds. Years aren't accurate since it doesn't take into account leap
years.
function get_time_diff($start,$end){ // Calculates
difference in seconds and returns text string.
$diff =
($end-$start);
if($diff<60){
$sec = $diff;
}
else{
if($diff<3600){
$min = floor($diff/60);
$sec = $diff-(60*$min);
}
else{
if($diff<86400){
$hour = floor($diff/3600);
$min
= floor(($diff-($hour*3600))/60);
$sec =
(($diff-($hour*3600)-(60*$min)));
}
else{
if($diff<604800){
$day = floor($diff/86400);
$hour = floor(($diff-($day*86400))/3600);
$min =
floor(($diff-($day*86400)-($hour*3600))/60);
$sec =
(($diff-($day*86400)-($hour*3600)-(60*$min)));
}
else{
if($diff<31536000){
$week =
floor($diff/604800);
$day =
floor(($diff-($week*604800))/86400);
$hour =
floor(($diff-($week*604800)-($day*86400))/3600);
$min =
floor(($diff-($week*604800)-($day*86400)-($hour*3600))/60);
$sec = (($diff-($week*604800)-($day*86400)-($hour*3600)-(60*$min)));
}
else{
$year =
floor($diff/31536000);
$week =
floor(($diff-($year*31536000))/604800);
$day =
floor(($diff-($year*31536000)-($week*604800))/86400);
$hour
= floor(($diff-($year*31536000)-($week*604800)-($day*86400))/3600);
$min =
floor(($diff-($year*31536000)-
($week*604800)-($day*86400)-($hour*3600))/60);
$sec =
(($diff-($year*31536000)
-($week*604800)-($day*86400)-($hour*3600)
-(60*$min)));
}
}
}
}
}
if($sec !=
''){
if($sec == 1){
$final = $sec . ' second';
}
else{
$final = $sec . ' seconds';
}
}
if($min != ''){
if($min == 1){
$final = $min . ' minute '
. $final;
}
else{
$final = $min . ' minutes ' .
$final;
}
}
if($hour != ''){
if($hour == 1){
$final = $hour . ' hour ' . $final;
}
else{
$final = $hour . ' hours ' . $final;
}
}
if($day !=
''){
if($day == 1){
$final = $day . ' day ' . $final;
}
else{
$final = $day . ' days ' . $final;
}
}
if($week != ''){
if($week == 1){
$final = $week . '
week ' . $final;
}
else{
$final = $week . ' weeks '
. $final;
}
}
if($year != ''){
if($year == 1){
$final = $year . ' year ' . $final;
}
else{
$final = $year . ' years ' . $final;
}
}
return
$final;
}
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[email protected]
30-Jan-2002 12:07 |
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Someone may find this info of some use:
Rules for calculating a leap
year:
1) If the year divides by 4, it is a leap year (1988, 1992,
1996 are leap years)
2) Unless it divides by 100, in which case it
isn't (1900 divides by 4, but was not a leap year)
3) Unless it divides
by 400, in which case it is actually a leap year afterall (So 2000 was a
leap year).
In practical terms, to work out the number of days in X
years, multiply X by 365.2425, rounding DOWN to the last whole number,
should give you the number of days.
The result will never be more
than one whole day inaccurate, as opposed to multiplying by 365, which,
over more years, will create a larger and larger deficit.
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[email protected]
30-Jan-2002 12:26 |
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Where possible, use the database built-in time and date functions, as they
often are not affected by the UTS limitation.
For example: I tested
MySQL "dayname" with "1978-08-09" (my birthdate),
which returned "Wednesday", which is correct.
I also tried
"1500-08-09", which returned Thursday. I have no actual way of
verifying this, as no other time functions on my PC go back that far,
though I double-checked it on a Linux server, and that returned the same
day.
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[email protected]
18-Feb-2002 09:31 |
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Some lines about LeapYear and day count of month:
function
mod($a,$b)
{
$x1=(int) abs($a/$b);
$x2=$a/$b;
return $a-($x1*$b);
}
function
IsLeapYear($dt)
{
$y=$dt["year"];
$bulis=((mod($y,4)==0) && ((mod ($y,100)<>0) ||
(mod($y,400)==0)));
return $bulis;
}
function
daycount($dt)
{
$dc_year=$dt["year"];
$dc_month=$dt["mon"];
$dc_day=$dt["mday"];
switch ($dc_month)
{
case
1:
case 3:
case 5:
case 7:
case 8:
case
10:
case 12:
return 31;
break;
case 4:
case
6:
case 9:
case 11:
return
30;
break;
case 2:
if (IsLeapYear($dt)) { return 28; }
else { return 29; };
break;
}
}
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[email protected]
08-Mar-2002 12:37 |
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Warning!
You have to be VERY carefull with things
like:
$same_time_tomorrow = time() + 60*60*24;
It's
because:
1. Mar-24 23:30 + 24 hours = Mar-26 00:30
2. Oct-28 00:30 +
24 hours = Oct-28 23:30
This occurs because of the daylight saving
time shifts.
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[email protected]
11-Mar-2002 04:16 |
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dameFechas($dateINI,$dateEND) in the format "20020311" will
return an array with the dates between that two days.
For
example:
var_dump(dameFechas("20011230","20020102"));
will
return
array(4) {
[0]=>
string(8)
"20020102"
[1]=>
string(8) "20020101"
[2]=>
string(8) "20011231"
[3]=>
string(8)
"20011230"
}
<?
// Bisiesto
function
esBisiesto($y) {
$res=$y%4==0 && $y%100<>0 ||
$y%400==0;
return $res;
}
function numDias($i_ano,$i_mes)
{
switch ($i_mes) {
case 1:
case 3:
case 5:
case
7:
case 8:
case 10:
case 12:
return
31;
break;
case 4:
case 6:
case 9:
case 11:
return 30;
break;
case 2:
if (!esBisiesto($i_ano)) {
return 28; } else { return 29; };
break;
}
}
function
formatoa2($i) {
if (strlen($i)==1)
return
"0".$i;
else
return $i;
}
function
menosmenosuno(&$s_cad)
{
$s_cad--;
$s_cad=formatoa2($s_cad);
}
function
dameFechas($fechaIni,$fechaFin)
{
$a_res=array();
$ini_ano=substr($fechaIni,0,4);
$fin_ano=substr($fechaFin,0,4);
$ini_mes=substr($fechaIni,4,2);
$fin_mes=substr($fechaFin,4,2);
$ini_dia=substr($fechaIni,6,2);
$fin_dia=substr($fechaFin,6,2);
$i=$fin_dia;
while
(($fin_ano.$fin_mes)>=($ini_ano.$ini_mes)) {
while ($fin_mes>=1
&& ($fin_ano.$fin_mes)>=($ini_ano.$ini_mes)) {
while
($fin_dia>=1 &&
($fin_ano.$fin_mes.$fin_dia)>=($ini_ano.$ini_mes.$ini_dia))
{
$a_res[]=$fin_ano.$fin_mes.$fin_dia;
menosmenosuno($fin_dia);
}
menosmenosuno($fin_mes);
$fin_dia=numDias($fin_ano,$fin_mes);
}
$fin_mes=12;
$fin_dia=31;
$fin_ano--;
}
return
$a_res;
}
echo
"<pre>";
var_dump(dameFechas("20020309","20020310"));
echo
"</pre>";
?>
Hope this helps.
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[email protected]
22-Mar-2002 10:39 |
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The editor is going to tell me it's easier to use UNIX_TIMESTAMP (I know
you're watching us ed! ;) ), but I disagree. I often want to just do a
"SELECT * FROM...", and this function makes that
possible.
I also see a guy above has also written basically the
same function. Great minds think alike, or fools seldom differ?
;)
/**
* Convert mysql timestamp to PHP timestamp.
*
* The
format of a mysql date is "yyyymmddhhmmss" whereas
* PHP uses
the (UNIX) format "sssssssssss" (number of seconds
* since
1970). Though it is possible to convert using
* the UNIX_TIMESTAMP
function of mysql, it is sometimes more convenient
* to read in the
timestamp variable just like any other (for example,
* by doing a
"SELECT * FROM mytable"), and convert the timestamp
* field
when it is in the program.
*
* This function simply does that
conversion.
*
* Example:
* $sql_row = mysql_fetch_array( $result
);
* $php_timestamp =
mysql_to_php_timestamp($sql_row[my_timestamp_field]);
* $display_date =
strftime("%d %B %Y %H:%M:%S", $php_timestamp);
*/
function
mysql_to_php_timestamp( $mysql_date )
{
// mysql date looks like
"yyyymmddhhmmss"
$year = substr( $mysql_date, 0, 4 );
$month = substr( $mysql_date, 4, 2 );
$day = substr(
$mysql_date, 6, 2 );
$hour = substr( $mysql_date, 8, 2 );
$min = substr( $mysql_date, 10, 2 );
$sec = substr( $mysql_date,
12, 2 );
// Warning: mktime uses a strange order of
arguments
$php_timestamp = mktime( $hour, $min, $sec, $month, $day,
$year );
return $php_timestamp;
}
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[email protected]
28-Mar-2002 02:21 |
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Corrections about add months of "Some general date functions" in
these notes
function sub($timestamp,
$seconds,$minutes,$hours,$days,$months,$years)
{
$mytime
=
mktime(1+$hours,0+$minutes,0+$seconds,1,1+$days,1970+$years);
$times=$timestamp - $mytime;
if
($months!=0)
{
$year=date('Y',$times)*12+date('n',$times)*1;
$year2=$year-$months;
$year3=floor($year2/12);
$month3=abs($year2-$year3*12);
$times=mktime
(date('H',$times),date('i',$times),date('s',$times),
$month3,date('j',$times),date('Y',$times));
}
return $times;
}
function add($timestamp,
$seconds,$minutes,$hours,$days,$months,$years) {
$mytime
=
mktime(1+$hours,0+$minutes,0+$seconds,1+$months,1+$days,1970+$years);
$times=$timestamp
+ $mytime;
if
($months!=0)
{
$year=date('Y',$times)*12+date('n',$times)*1;
$year2=$year+$months;
$year3=floor($year2/12);
$month3=abs($year2-$year3*12);
$times=mktime
(date('H',$times),date('i',$times),date('s',$times),
$month3,date('j',$times),date('Y',$times));
}
return $times;
}
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[email protected]
31-Mar-2002 08:24 |
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In vincentv's examples, you should use gmmktime instead of mktime for
portability across time zones. For example:
function
DateSub($timestamp, $unit, $amount) {
// Possible $units are:
"hr", "min", "sec",
//
"mon", "day", or "yr"
// $amount should
be an integer
$delta_vars = array("hr"=>0,
"min"=>0,
"sec"=>0,
"mon"=>1,
"day"=>1,"yr"=>1970);
$delta_vars[$unit] += $amount;
$delta =
gmmktime($delta_vars["hr"],
$delta_vars["min"],
$delta_vars["sec"],
$delta_vars["mon"],
$delta_vars["day"],
$delta_vars["yr"]);
return $timestamp - $delta;
}
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emerix at rol dot ro
15-Jul-2002 03:50 |
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The simplest function for formating a timestamp difference.
function
date_diff ( $date1, $date2 ) {
$some=date("z \\d\\a\\y\\s H\\h
i\\m s\\s",$data-7200);
return $some;
}
Where $date1
and $date2 are the timestamps you want.
Ex. date_diff(1024681353,
1023917859);
Will output :
8 days 20h 04m 54s
If you want
smater output you should have a
switch(date("z", $date))
{
...
}
in order to avoid "1 days" ...
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[email protected]
30-Jul-2002 04:59 |
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I wanted to find all records in my database which match the current week
(for a call-back function). I made up this function to find the start and
end of the current week :
function week($curtime)
{
$date_array = getdate (time());
$numdays =
$date_array["wday"];
$startdate =
date("Y-m-d", time() - ($numdays * 24*60*60));
$enddate =
date("Y-m-d", time() + ((7 - $numdays) *
24*60*60));
$week['start'] = $startdate;
$week['end'] =
$enddate;
return $week;
}
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[email protected]
22-Aug-2002 10:42 |
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I have write a library to manipulate date and time as a TDateTime-like
Delphi object.
My new DateTime format is a float value, where the
integer part is the number of day since a given date (1/1/1899), and the
fractional part represents the number of milliseconds since 0h0m0s of this
day.
So it is easy to compare, add, multiply dates, and there is no
1970-2038 limit.
You can convert DateTime number to unix epoch and
mysql DATETIME type.
url is :
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dba_sync