Find day light saving time for a given year using php - php

i would like to know is there a anyway we can find the DST(day light saving time) for the given year.

For the United States daylight savings time begins the Second Sunday of March and ends on the First Sunday of November. So...
<?php
$remove_hour = strtotime("Second Sunday March 0");
$add_hour = strtotime("First Sunday November 0");
$time = time();
if( $time >= $remove_hour && $time < $add_hour )
{
var_dump("Lost an hour");
}
else
{
var_dump("Gained an hour");
}
?>
Although as far as I know this doesn't change every year? Not sure what you mean by that?

You can use the I parameter in the standard date function to get a 0/1 flag if DST.
PHP Manual Date

Related

Daylight savings calculation in Php

In my application I have varchar field in mysql for 365 days in particular format i.e. '0000-04-12' (12th of April). Right now I am having a function that should check if particular date from field falls in daylight saving zone of Calgary-Canada, which ranges from 2:00 pm Second Sunday of March to 2:00 pm First Sunday of November. and returns true or false accordingly.
<?php
/*
#param $time form Database e.g. 0000-04-12
*/
function isDaylightSaving($time){
// to replace 0000 in start with current year
$today = substr_replace($time,date('Y'),0,4);
$date = new DateTime($today);
$week = $date->format('W');
// TO DO
// find date falls between 2:00 pm Second Sunday of March to 2:00 pm First Sunday of November.
return ;
}
?>
Right now its returning the week with respect to particular year but I need to check week with respect to month to perform particular logic, which I am unable to get.Thanks.
Try this:
if($time > date('Y-m-d',strtotime('second sunday of march')) &&
$time < date('Y-m-d',strtotime('first sunday of november'))){
return true;
}else{
return false;
}
strtotime() is a great function.
The example given by #Ezenhis sounds good, but lacks of some elements.
Extending it:
$low = new DateTime('#'.strtotime('second sunday of march'));
$high = new DateTime('#'.strtotime('first sunday of november'));
if($date > $low && $date < $high) {
// we're good
}
You shouldn't compare strings from date functions. Using DateTime is also a better way to handle dates/times in PHP since PHP 5.0.

How to set the "first day of the week" to Thursday in PHP

I want to set the first day of the week to Thursday (not Sunday or Monday), because it's the company's cut-off date.
I already have a code to determine the current week number of a date but it starts in Sunday or Monday.
How to modify these to my preference?
function findweek($date) {
$monthstart=date("N",strtotime(date("n/l/Y",strtotime($date))));
$newdate=(date("j",strtotime($date))+$monthstart)/7;
$ddate=floor($newdate);
if($ddate != $date) {
$ddate++;
}
return $ddate;
}
http://php.net/manual/en/datetime.formats.relative.php says that as of PHP version 5.6.23, 7.0.8 "Weeks always start on monday. Formerly, sunday would also be considered to start a week." That said, is your problem that the number of weeks returned might be incorrect depending on whether today falls on or before Thursday of the current week? Maybe try something like this:
$date = new DateTime();
$week = intval($date->format('W'));
$day = intval($date->format('N'));
echo $day < 4 ? $week-1 : $week;
If subtracting 1 isn't the answer you could play around with addition/subtraction, comparing the result with the actual answer you know to be true until you get the right formula. Hope this helps!
This should work.
function findweek($date, $type = "l") {
$time = strtotime($date);
return date($type, mktime(0, 0, 0, date("m", $time) , date("d", $time)-date("d", $time)+1, date("Y", $time)));
}
echo findweek('2015-09-16');

Display the 1st, 2nd, 3rd, 4th, & 5th date of the current week in php

Ok, So im trying to figure out the best way in php to write (or print) the 1st, 2nd, 3rd, 4th, & 5th date of the current week (starting with Monday) in php.
For example... Using the week of the 4th through the 8th of February...
I would need a script for the 1st day of the week (starting with Monday) displayed as February 04, 2013 and then I would need the same script four more times to display Tues, Wed, Thurs, & Fri...
All-together I would end up with 5 scripts or one script that I could copy and manipulate to work the way I need it to...
Also, If you could tell me how to do the same for Saturday and Sunday that would be greatly appreciated as-well.
If there is anything that you do not understand, please let me know and I will try my hardest to clarify...
Thanks in advance!
First you need to get the "current" week's Monday. To do this, I would suggest calling date("N") and see if it's 1. If it is, then now is the Monday you want. Otherwise last monday is. Pass that to strtotime to get the timestamp corresponding to the first monday of the week. Then repeatedly add 24 hours (24*3600 seconds) to get each day.
$startofweek = date("N") == 1 ? time() : strtotime("last monday");
for($i=0; $i<5; $i++) {
$day = $startofweek + 24*3600*$i;
echo "Day ".($i+1).": ".date("d/M/Y",$day)."<br />";
}
You can use strtotime with special parameters. like:
$time = strtotime('monday this week');
$time = strtotime('today');
$time = strtotime('next monday');
$time = strtotime('previous monday');
Same goes for every other days of the week

How to get previous month and year relative to today, using strtotime and date?

I need to get previous month and year, relative to current date.
However, see following example.
// Today is 2011-03-30
echo date('Y-m-d', strtotime('last month'));
// Output:
2011-03-02
This behavior is understandable (to a certain point), due to different number of days in february and march, and code in example above is what I need, but works only 100% correctly for between 1st and 28th of each month.
So, how to get last month AND year (think of date("Y-m")) in the most elegant manner as possible, which works for every day of the year? Optimal solution will be based on strtotime argument parsing.
Update. To clarify requirements a bit.
I have a piece of code that gets some statistics of last couple of months, but I first show stats from last month, and then load other months when needed. That's intended purpose. So, during THIS month, I want to find out which month-year should I pull in order to load PREVIOUS month stats.
I also have a code that is timezone-aware (not really important right now), and that accepts strtotime-compatible string as input (to initialize internal date), and then allows date/time to be adjusted, also using strtotime-compatible strings.
I know it can be done with few conditionals and basic math, but that's really messy, compared to this, for example (if it worked correctly, of course):
echo tz::date('last month')->format('Y-d')
So, I ONLY need previous month and year, in a strtotime-compatible fashion.
Answer (thanks, #dnagirl):
// Today is 2011-03-30
echo date('Y-m-d', strtotime('first day of last month')); // Output: 2011-02-01
Have a look at the DateTime class. It should do the calculations correctly and the date formats are compatible with strttotime. Something like:
$datestring='2011-03-30 first day of last month';
$dt=date_create($datestring);
echo $dt->format('Y-m'); //2011-02
if the day itself doesn't matter do this:
echo date('Y-m-d', strtotime(date('Y-m')." -1 month"));
I found an answer as I had the same issue today which is a 31st. It's not a bug in php as some would suggest, but is the expected functionality (in some since). According to this post what strtotime actually does is set the month back by one and does not modify the number of days. So in the event of today, May 31st, it's looking for April-31st which is an invalid date. So it then takes April 30 an then adds 1 day past it and yields May 1st.
In your example 2011-03-30, it would go back one month to February 30th, which is invalid since February only has 28 days. It then takes difference of those days (30-28 = 2) and then moves two days past February 28th which is March 2nd.
As others have pointed out, the best way to get "last month" is to add in either "first day of" or "last day of" using either strtotime or the DateTime object:
// Today being 2012-05-31
//All the following return 2012-04-30
echo date('Y-m-d', strtotime("last day of -1 month"));
echo date('Y-m-d', strtotime("last day of last month"));
echo date_create("last day of -1 month")->format('Y-m-d');
// All the following return 2012-04-01
echo date('Y-m-d', strtotime("first day of -1 month"));
echo date('Y-m-d', strtotime("first day of last month"));
echo date_create("first day of -1 month")->format('Y-m-d');
So using these it's possible to create a date range if your making a query etc.
If you want the previous year and month relative to a specific date and have DateTime available then you can do this:
$d = new \DateTimeImmutable('2013-01-01', new \DateTimeZone('UTC'));
$firstDay = $d->modify('first day of previous month');
$year = $firstDay->format('Y'); //2012
$month = $firstDay->format('m'); //12
date('Y-m', strtotime('first day of last month'));
strtotime have second timestamp parameter that make the first parameter relative to second parameter. So you can do this:
date('Y-m', strtotime('-1 month', time()))
if i understand the question correctly you just want last month and the year it is in:
<?php
$month = date('m');
$year = date('Y');
$last_month = $month-1%12;
echo ($last_month==0?($year-1):$year)."-".($last_month==0?'12':$last_month);
?>
Here is the example: http://codepad.org/c99nVKG8
ehh, its not a bug as one person mentioned. that is the expected behavior as the number of days in a month is often different. The easiest way to get the previous month using strtotime would probably be to use -1 month from the first of this month.
$date_string = date('Y-m', strtotime('-1 month', strtotime(date('Y-m-01'))));
I think you've found a bug in the strtotime function. Whenever I have to work around this, I always find myself doing math on the month/year values. Try something like this:
$LastMonth = (date('n') - 1) % 12;
$Year = date('Y') - !$LastMonth;
date("m-Y", strtotime("-1 months"));
would solve this
Perhaps slightly more long winded than you want, but i've used more code than maybe nescessary in order for it to be more readable.
That said, it comes out with the same result as you are getting - what is it you want/expect it to come out with?
//Today is whenever I want it to be.
$today = mktime(0,0,0,3,31,2011);
$hour = date("H",$today);
$minute = date("i",$today);
$second = date("s",$today);
$month = date("m",$today);
$day = date("d",$today);
$year = date("Y",$today);
echo "Today: ".date('Y-m-d', $today)."<br/>";
echo "Recalulated: ".date("Y-m-d",mktime($hour,$minute,$second,$month-1,$day,$year));
If you just want the month and year, then just set the day to be '01' rather than taking 'todays' day:
$day = 1;
That should give you what you need. You can just set the hour, minute and second to zero as well as you aren't interested in using those.
date("Y-m",mktime(0,0,0,$month-1,1,$year);
Cuts it down quite a bit ;-)
This is because the previous month has less days than the current month. I've fixed this by first checking if the previous month has less days that the current and changing the calculation based on it.
If it has less days get the last day of -1 month else get the current day -1 month:
if (date('d') > date('d', strtotime('last day of -1 month')))
{
$first_end = date('Y-m-d', strtotime('last day of -1 month'));
}
else
{
$first_end = date('Y-m-d', strtotime('-1 month'));
}
If a DateTime solution is acceptable this snippet returns the year of last month and month of last month avoiding the possible trap when you run this in January.
function fn_LastMonthYearNumber()
{
$now = new DateTime();
$lastMonth = $now->sub(new DateInterval('P1M'));
$lm= $lastMonth->format('m');
$ly= $lastMonth->format('Y');
return array($lm,$ly);
}
//return timestamp, use to format month, year as per requirement
function getMonthYear($beforeMonth = '') {
if($beforeMonth !="" && $beforeMonth >= 1) {
$date = date('Y')."-".date('m')."-15";
$timestamp_before = strtotime( $date . ' -'.$beforeMonth.' month' );
return $timestamp_before;
} else {
$time= time();
return $time;
}
}
//call function
$month_year = date("Y-m",getMonthYear(1));// last month before current month
$month_year = date("Y-m",getMonthYear(2)); // second last month before current month
function getOnemonthBefore($date){
$day = intval(date("t", strtotime("$date")));//get the last day of the month
$month_date = date("y-m-d",strtotime("$date -$day days"));//get the day 1 month before
return $month_date;
}
The resulting date is dependent to the number of days the input month is consist of. If input month is february (28 days), 28 days before february 5 is january 8. If input is may 17, 31 days before is april 16. Likewise, if input is may 31, resulting date will be april 30.
NOTE: the input takes complete date ('y-m-d') and outputs ('y-m-d') you can modify this code to suit your needs.

How to get weekend day in this week by any given date?

for example, get the weekend of today.
If you're looking for e.g. the next or previous saturday (or any other weekday for that matter) strtotime is your friend.
$prev = strtotime('-1 saturday');
$next = strtotime('saturday');
var_dump($prev, $next);
It's worth noting that strtotime is quite an expensive function, so multiple calculations will noticiably add to your execution time. A good compromise is using it to get a starting point and using the other date functions to derive further dates.
This is brilliantly easy in PHP:
<?php
echo date( 'Y-m-d', strtotime( 'next Saturday' ) );
?>
get the current date.
get the current day of week. (0=monday, 6 = sunday)
days2weekend = 5 - current day of week
dateadd(currentdate, days, days2weekend)
Your question is a bit a hard to follow, but do you mean "the next weekend" from a certain date?
You could get the the weekday number, and then see how much to add for saturday and sundag? That would look like:
<?php
$dayNR = date('N'); //monday = 1, tuesday = 2, etc.
$satDiff = 6-$dayNR; //for monday we need to add 5 days -> 6 - 1
$sunDiff = $satDiff+1; //sunday is one day more
$satDate = date("Y-m-d", strtotime(" +".$satDiff." days"));
$sunDate = date("Y-m-d", strtotime(" +".$sunDiff." days"));
echo $satDate."\n";
echo $sunDate."\n";
?>
I think the built-in PHP function strtotime should work for you. Eg:
strtotime('next saturday')
http://php.net/manual/en/function.strtotime.php

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