PHP minus one week Sunday - php

I'm a newbie in both php and codeigniter.
I have a function which is used to label the x-axis of a chart of weekly sums of kilometers rode on a bicycle. It works (or is supposed to work) by counting back from today's date either zero Sundays or 1, 2, 3, etc as passed by an array.
I am using PHP Version 5.6.19, and this is my code:
In the Model:
public function week_start($date, $counter){
$tstring = strtotime($date);
$wknumber = ($counter + 1);
$start = strtotime("-$wknumber weeks sunday", $tstring);
return array(date('M d', $start));}
In the Controller:
$datenumbers = array(0, 1, 2, 3, 4, 5, 6, 7);
$dateResult = $this->bike_model->week_start(date("Y/m/d"), $datenumber);
In the view:
for ($i = 0; $i < 8; $i++){
echo "<text class = \"axis\" y = \"357\" x = \"";
echo 26 + ($i*70);
echo "\" dy = \".35em\">";
echo $dates_for_table[$i];}
I have had it up for few days and it has been showing Sunday July 3 as the first number on on my x-axis, as it should have. But today, it is Sunday and I expect it to now say Sunday July 10 as my most recent week. However, it is still showing last Sunday!
So my question is, is there a bug here or am I misunderstanding how '- one week Sunday' should work?
My goal is to have my function with zero as the argument return the most recent week start--either the current Sunday if it is Sunday or the past Sunday. I have found a workaround by adding an if else to the function checking if it is Sunday.
if ($tstring == strtotime('this Sunday')){
$wknumber = $counter;}
else{
$wknumber = ($counter + 1);}
However, this feels clunky and I'm looking for a more elegant way. I also want to understand what went wrong with "- one week Sunday" since I thought I had understood how it was working.
Thank you!

Here is how I did something similar using the object oriented DateTime style:
$startDate = new DateTime('NOW'); // can pass date string if not today
# reset start date to last Sunday or today if it is Sunday:
$startDate->sub(new DateInterval('P'.($startDate->format('w')).'D'));
# for debugging, output first Sunday:
echo $startDate->format('Y-m-d H:i:s') . " => full date and time (check for timezone issues) \n\n";
$weeks = 8;
for ($i = 0; $i < $weeks; $i++) {
# don't decrement first pass; sub 1 week on subsequent:
$startDate->sub(new DateInterval('P'. (($i) ? 1 : 0) .'W'));
echo $startDate->format('Y-m-d') . "\n";
};
You should be able to adapt for your purpose.
If you want to get fancy, you can also use DatePeriod instead of using sub within a loop:
$period = new DatePeriod(
$startDate,
DateInterval::createFromDateString('-1 week'),
7
);
foreach ($period as $date) {
echo $date->format('Y-m-d') . "\n";
}

Related

how to manipulate back date and exclude saturday and sunday?

How to manipulate the date and exclude saturday and sunday?. The objective is, I need to create a cron job that will run and execute on datas that were created 5 days ago,"BUT", saturday and sunday shouldn't be included in that 5 days period.
here's what I have so far
$numdays = 5;
$today = strtotime('now');
$start = date('Y-m-d',strtotime('-'.$numdays.' day',$today));
echo $start;
if you try to run my code snippet above, it will show you the exact date 5 days ago 2016-02-10. But that one doesn't "exclude" saturday and sunday in the computation. it should be be 2016-02-08. So how to do that?
You can use PHP's date week of day, there are several versions, here is one using N:
<?php
$current = new DateTime();
$interval = new DateInterval('P1D');
$x = 5;
while ($x > 1) {
// Check if day of week is not saturday/sunday (1 => Monday ... 7 -> Sunday)
if ($current->format('N') >= 6) {
$x++;
}
$current->sub($interval);
$x--;
}
echo $current->format('Y-m-d') . PHP_EOL;
Example Run.
You can get a whole week and discard the weekends, keeping the furthest element in the array as a result.
$days = array_filter(array_map(function ($daysBack) {
$date = new \DateTimeImmutable("$daysBack days ago", new \DateTimeZone('UTC'));
return (!in_array($date->format('N'), [6, 7])) ? $date : null;
}, Range(1, 7)));
$fiveWorkingDaysAgo = end($days);

calculate sundays between two dates

I want to calculate all Sunday's between given two dates. I tried following code. It works fine if days are less but if i enter more days. It keeps processing and Maximum execution time exceeds i changed the time but it even keeps processing even execution time is 200sec.
code is
<?php
$one="2013-01-01";
$two="2013-02-30";
$no=0;
for($i=$one;$i<=$two;$i++)
{
$day=date("N",strtotime($i));
if($day==7)
{
$no++;
}
}
echo $no;
?>
please help.
John Conde's answer is correct, but here is a more efficient and mathy solution:
$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');
$days = $start->diff($end, true)->days;
$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);
echo $sundays;
Let me break it down for you.
$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');
First, create some DateTime objects, which are powerful built-in PHP objects meant for exactly this kind of problem.
$days = $start->diff($end, true)->days;
Next, use DateTime::diff to find the difference from $start to $end (passing true here as the second parameter ensures that this value is always positive), and get the number of days between them.
$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);
Here comes the big one - but it's not so complicated, really. First, we know there is one Sunday for every week, so we have at least $days / 7 Sundays to begin with, rounded down to the nearest int with intval.
On top of that, there could be a Sunday in a span of time less than a week; for example, Friday to Monday of the next week contains 4 days; one of them is a Sunday. So, depending on when we start and end, there could be another. This is easy to account for:
$start->format('N') (see DateTime::format) gives us the ISO-8601 day of the week for the start date, which is a number from 1 to 7 (1 is Monday, 7 is Sunday).
$days % 7 gives us the number of leftover days that don't divide evenly into weeks.
If our starting day and the number of leftover days add up to 7 or more, then we reached a Sunday. Knowing that, we just have to add that expression, which will give us 1 if it's true or 0 if it's false, since we're adding it to an int value.
And there you have it! The advantage of this method is that it doesn't require iterating over every day between the given times and checking to see if it's a Sunday, which will save you a lot computation, and also it will make you look really clever. Hope that helps!
<?php
$no = 0;
$start = new DateTime('2013-01-01');
$end = new DateTime('2013-04-30');
$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt)
{
if ($dt->format('N') == 7)
{
$no++;
}
}
echo $no;
See it in action
Here is a solution if you want the Sundays in a specific date range.
function dateRange($begin, $end, $interval = null)
{
$begin = new DateTime($begin);
$end = new DateTime($end);
$end = $end->modify('+1 day');
$interval = new DateInterval($interval ? $interval : 'P1D');
return iterator_to_array(new DatePeriod($begin, $interval, $end));
}
/* define date range */
$dates = dateRange('2018-03-01', '2018-03-31');
/* define weekdays */
$weekends = array_filter($dates, function ($date) {
$day = $date->format("N");
return $day === '6' || $day === '7';
});
/* weekdays output */
foreach ($weekends as $date) {
echo $date->format("D Y-m-d") . "</br>";
}
/* define sundays */
$sundays = array_filter($dates, function ($date) {
return $date->format("N") === '7';
});
/* sundays output */
foreach ($sundays as $date) {
echo $date->format("D Y-m-d") . "</br>";
}
/* define mondays */
$mondays = array_filter($dates, function ($date) {
return $date->format("N") === '1';
});
/* mondays output */
foreach ($mondays as $date) {
echo $date->format("D Y-m-d") . "</br>";
}
Just change the number for any days you want in your output:
Monday = 1
Tuesday = 2
Wednesday = 3
Thursday = 4
Friday = 5
Saturday = 6
Sunday = 7

Getting Dates from Week Numbers - Year Starts from March - Week 1 = March 1st week - PHP

I am trying to write a custom function for getting dates from week numbers but the requirement is that the week numbers are custom i.e. the week1 starts from 1st Week of March and the First Day of each week is Friday
I trying to do this in PHP.
Your help or advice would be really helpful for writing this function.
I have gone through some of functions here but doesnt actually fit my requirement.
Thanking You
This might give you a starting point:
$year = 2012;
$startDate = new \DateTime($year . '-03-01');
$startDate->modify('Friday');
echo $startDate->format('Y-m-d') , PHP_EOL;
$endDate = new \DateTime($year+1 . '-03-01');
$endDate->modify('Friday');
echo $endDate->format('Y-m-d') , PHP_EOL;
$interval = new \DateInterval('P1W');
$weekPeriod = new \DatePeriod ($startDate, $interval, $endDate);
foreach ($weekPeriod as $key => $weekDate) {
echo 'Week #' , $key + 1 , ' starts on ';
echo $weekDate->format('Y-m-d') , PHP_EOL;
}
You can use this to build an array, that you can then use as a lookup

PHP. How to get a bimonthly recurring event?

I'm trying many approaches but then I get stuck half way.
Let's say order was created today. I need to display when the next recurring order will happen. So I have order created June 13, 2012. Then I have set the schedule to bimonthly recurring order, every 1st of month. How to calculate when the next recurring order will happen? The answer is August 1st.
If someone can outline an approach it would be very useful, it doesn't have to be code. This is what I have so far...
// first, get starting date
$start_date_month = date('m', strtotime($start_date));
// get this year
$this_year = date('Y');
// if this month is december, next month is january
$this_month = date('m', $timestamp_month);
if($this_month == 12){
$next_month = 1;
// if this month is not december add 1 to get next month
}else{
$next_month = $this_month + 1;
}
// get array of months where recurring orders will happen
$months = array();
for ($i=1; $i<=6; $i++) {
$add_month = $start_date_month+(2*$i); // 2, 4, 6, 8, 10, 12
if($add_month == 13){$add_month = 1;$year = $this_year+1;}
elseif($add_month == 14){$add_month = 2;$year = $this_year+1;}
elseif($add_month == 15){$add_month = 3;$year = $this_year+1;}
elseif($add_month == 16){$add_month = 4;$year = $this_year+1;}
elseif($add_month == 17){$add_month = 5;$year = $this_year+1;}
elseif($add_month == 18){$add_month = 6;$year = $this_year+1;}
elseif($add_month == 19){$add_month = 7;$year = $this_year+1;}
elseif($add_month == 20){$add_month = 8;$year = $this_year+1;}
else{$year = $this_year;}
echo $what_day.'-'.$add_month.'-'.$year.'<br />';
$months[] = $add_month;
}
echo '<pre>';
print_r($months);
echo '</pre>';
I don't want to simply find what's the date in two months from now. Let's say order created June 1. Next recurring order is August 1. Then let's say now, today is September 1st, but next recurring order is October 1st. See my dilemma?
Just take the current month, so since it's June, we get 6. 6 mod 2 == 0. Next month is July, we get 7. 7 mod 2 == 1.
So just check if current month % 2 == (first month % 2).
Then just check if it's the 1st of the month.
In PHP modulus is defined with the percentage symbol.
$month = date('n');
$createdMonth = 6;
if($month % 2 == $createdMonth % 2){
// stuff
}
You might find the library called When useful for this (I'm the author).
Here is code which will get you the next 2 recurring monthly dates (from todays date):
include 'When.php';
$r = new When();
$r->recur(new DateTime(), 'monthly')
->count(2)
->interval(2) // every other month
->bymonthday(array(1));
while($result = $r->next())
{
echo $result->format('c') . '<br />';
}
// output
// 2012-08-01T13:33:33-04:00
// 2012-10-01T13:33:33-04:00
Taking this a step further, you likely only want to find the 2 first business days:
include 'When.php';
$r = new When();
$r->recur(new DateTime(), 'monthly')
->count(2)
->interval(2) // every other month
->byday(array('MO', 'TU', 'WE', 'TH', 'FR')) // week days only
->bymonthday(array(1, 2, 3)) // the first weekday will fall on one of these days
->bysetpos(array(1)); // only return one per month
while($result = $r->next())
{
echo $result->format('c') . '<br />';
}
// output
// 2012-08-01T13:33:33-04:00
// 2012-10-01T13:33:33-04:00
Also note, the code is currently under a rewrite -- it works well but it is a little confusing and not well documented.
strtotime to the rescue:
<?php
date_default_timezone_set('Europe/London');
$d = new DateTime('2012-01-31');
$d->modify('first day of +2 months');
echo $d->format('r'), "\n";
?>
Let's say you want the next six orders:
$order_date = '6/13/2012';
$start = date('Y-m-01', strtotime($order_date));
$order_count = 6;
$future_orders = array();
$next = strtotime('+2 months', strtotime($start));
while(count($future_orders) < $order_count){
$future_orders[] = date('m/d/Y',$next);
$next = strtotime('+2 months', $next);
}
This can, obviously, be improved upon, but it should get you started ...
I got this:
$today = new DateTime();
$target_date = $today->modify("first day of +2 months");
echo "Your event is on " . $target_date->format("d/m/Y") . "!";

Finding all weekdays in a month

How do I go about getting all the work days (mon-fri) in a given time period (let's say, today till the end of the next month) ?
If you're using PHP 5.2+ you can use the library I wrote in order to handle date recursion in PHP called When.
With the library, the code would be something like:
$r = new When();
$r->recur(<start date here>, 'weekly')
->until(<end date here>)
->wkst('SU')
->byday(array('MO', 'TU', 'WE', 'TH', 'FR'));
while($result = $r->next())
{
echo $result->format('c') . '<br />';
}
This sample does exactly what you need, in an quick and efficient way.
It doesn't do nested loops and uses the totally awesome DateTime object.
$oDateTime = new DateTime();
$oDayIncrease = new DateInterval("P1D");
$aWeekDays = array();
$sStart = $oDateTime->format("m-Y");
while($oDateTime->format("m-Y") == $sStart) {
$iDayInWeek = $oDateTime->format("w");
if ($iDayInWeek > 0 && $iDayInWeek < 6) {
$aWeekDays[] = clone $oDateTime;
}
$oDateTime->add($oDayIncrease);
}
Try it here: http://codepad.org/wuAyAqnF
To use it, simply pass a timestamp to get_weekdays. You'll get back an array of all the weekdays, as timestamps, for the rest of the current month. Optionally, you can pass a $to argument - you will get all weekdays between $from and $to.
function get_weekdays ($from, $to=false) {
if ($to == false)
$to = last_day_of_month($from);
$days = array();
for ($x = $from; $x < $to; $x+=86400 ) {
if (date('w', $x) > 0 && date('w', $x) < 6)
$days[] = $x;
}
return $days;
}
function last_day_of_month($ts=false) {
$m = date('m', $ts);
$y = date('y', $ts);
return mktime(23, 59, 59, ($m+1), 0, $y);
}
I suppose you could loop through the dates and check the day for each one, and increment a counter.
Can't think of anything else off the top of my head.
Pseudocode coming your way:
Calculate the number of days between now and the last day of the month
Get the current day of the week (i.e. Wednesday)
Based on the current day of the week, and the number of days left in the month, it's simple calculation to figure out how many weekend days are left in the month - it's going to be the number of days remaining in the month, minus the number of Sundays/Saturdays left in the month.
I would write a function, something like:
daysLeftInMonth(daysLeftInMonth, startingDayOfWeek, dayOfWeekToCalculate)
where:
daysLeftInMonth is last day of the month (30), minus the current date (15)
startingDayOfWeek is the day of the week you want to start on (for today it would be Wednesday)
dayOfWeekToCalculate is the day of the week you want to count, e.g. Saturday or Sunday. June 2011 currently has 2 Sunday, and 2 Saturdays left 'til the end of the month
So, your algorithm becomes something like:
getWeekdaysLeft(todaysDate)
...getWeekdaysLeft is something like:
sundaysLeft = daysLeftInMonth(lastDayOfMonth - todaysDate, "Wednesday", "Sunday");
saturdaysLeft = daysLeftInMonth(lastDayOfMonth - todaysDate, "Wednesday", "Saturday");
return ((lastDayOfMonth - todaysDate) - (sundaysLeft + saturdaysLeft));
This code does at least one part you ask for. Instead of "end of next month" it simply works with a given number of days.
$dfrom = time();
$fourweeks = 7 * 4;
for ($i = 0; $i < $fourweeks; $i ++) {
$stamp = $dfrom + ($i * 24 * 60 * 60);
$weekday = date("D", $stamp);
if (in_array($weekday, array("Mon", "Tue", "Wed", "Thu", "Fri"))) {
print date(DATE_RSS, $stamp) . "\n";
}
}
// Find today's day of the month (i.e. 15)
$today = intval(date('d'));
// Define the array that will hold the work days.
$work_days = array()
// Find this month's last day. (i.e. 30)
$last = intval(date('d', strtotime('last day of this month')));
// Loop through all of the days between today and the last day of the month (i.e. 15 through 30)
for ( $i = $today; $i <= $last; $i++ )
{
// Create a timestamp.
$timestamp = mktime(null, null, null, null, $i);
// If the day of the week is greater than Sunday (0) but less than Saturday (6), add the timestamp to an array.
if ( intval(date('w', $timestamp)) > 0 && intval(date('w', $timestamp)) < 6 )
$work_days[] = mktime($timestamp);
}
The $work_days array will contain timestamps which you could use this way:
echo date('Y-m-d', $work_days[0]);
The code above with work in PHP 4 as well as PHP 5. It does not rely on the functionality of the DateTime class which was not available until PHP 5.2 and does not require the use of "libraries" created by other people.

Categories