I've got to write a loop that should start and end between two times. I know there are many ways to skin this cat, but I'd like to see a real programmers approach to this function.
Essentially I have Wednesday, for instance, that opens at 6:00pm and closes at 10:30pm.
I'm looking to write a loop that will give me a table with all of the times in between those two in 15 minute intervals.
So, I basically want to build a one column table where each row is
6:00pm
6:15pm
7:15pm
etc...
My two variables to feed this function will be the open time and the close time.
Now don't accuse me of "write my code for me" posting. I'll happily give you my hacked solution on request, I'd just like to see how someone with real experience would create this function.
Thanks :)
$start = new DateTime("2011-08-18 18:00:00");
$end = new DateTime("2011-08-18 22:30:00");
$current = clone $start;
while ($current <= $end) {
echo $current->format("g:ia"), "\n";
$current->modify("+15 minutes");
}
Try it on Codepad: http://codepad.org/JwBDOQQE
PHP 5.3 introduced a class precisely for this purpose, DatePeriod.
$start = new DateTime("6:00pm");
$end = new DateTime("10:30pm");
$interval = new DateInterval('PT15M');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $time) {
echo $time->format('g:ia'), PHP_EOL;
}
echo $end->format('g:ia'); // end time is not part of the period
$start = strtotime('2011-08-11 18:00:00');
for ($i = 0; $i < 20; $i++) {
echo date('g:ia', $start + ($i * (15 * 60))), '<br>';
}
I would go with the DateTime functions and increase the time by 15 minutes every loop-turn as long as the current time is lower then the end-time.
EDIT: as user576875 has posted
$start_date = '2019-07-30 08:00:00';
$end_date = '2019-09-31 08:00:00';
while (strtotime($start_date) <= strtotime($end_date)) {
echo "$start_date<br>";
$start_date = date ("Y-m-d H:i:s", strtotime("+1 hours", strtotime($start_date)));
}
Related
Gooooood evening all...
Edit:
I'm trying to write a php script to output the dates I will work for the remainder of the year if given a specific start date.
For example I start on 2020-05-14 and work for 4 days then take 4 days off.
This continues for the remainder of the year.
I would like to output the dates i will work and think it could probably be done using the php for loop, however i've been thinking of how to do this for too long and can't seem to break through the wall.
The start of my code is this:
<?php
$z = date("z",mktime(0,0,0,05,14,2020));
for($i=$z;$i<=365;$i+=4) {
echo("$i<br>");
}
?>
Any advice would be much appreciated on a possible solution.
You're computing the day of the year (e.g. for 2020-05-14, that would be 134), which is fine for letting your script know when to stop.
However, you're wanting to output dates, where you're currently outputting the day of the year instead.
I think there's an easier way to do what you're trying to accomplish.
<?php
$start = new \DateTime("2020-05-14 00:00:00");
$end = new \DateTime($start->format("Y") . "-12-31 00:00:00");
$four = new \DateInterval("P4D");
$one = new \DateInterval("P1D");
$format = 'Y-m-d';
$date = $start;
while ($date <= $end) {
echo $date->format($format), PHP_EOL;
for ($day = 2; $day <= 4; $day++) {
$date->add($one);
echo $date->format($format), PHP_EOL;
}
$date->add($four);
}
I want to loop a date so that every time date is increment by previous date. my code is here. plz reply anyone, thanks in advance
$today = date('Y-m-d');
for($i=1; $i<=4; $i++){
$repeat = strtotime("+2 day",strtotime($today));
echo $rdate = date('Y-m-d',$repeat);
}
I want result as if today is 2016-04-04 than, 2016-04-06, 2016-04-08, 2016-04-10, 2016-04-12.
actually i want to make a reminder date where user enter reminder. lets a user want to add reminder today and want repeat it 5 time after 2days, 3days or what ever he wants, in next comming day. than how i repeat date with for loop.
Try this:
<?php
$today = date('Y-m-d');
for($i=1; $i<=4; $i++)
{
$repeat = strtotime("+2 day",strtotime($today));
$today = date('Y-m-d',$repeat);
echo $today;
}
Output:
2016-04-06
2016-04-08
2016-04-10
2016-04-12
The easiest way is what answer
aslawin
The below example is to go through the date
$begin = new DateTime($check_in);
$end = new DateTime($check_out);
$step = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($begin, $step, $end);
foreach ($period as $dt)
{
<sample code here>
}
You can try this:
$today = date('Y-m-d');
for($i=1; $i<=8; $i++){
if($i%2 == 0){
$repeat = strtotime("+$i day",strtotime($today));
echo $rdate = date('Y-m-d',$repeat);
}
}
Result:
2016-04-06
2016-04-08
2016-04-10
2016-04-12
In this example, you can use $i%2 == 0 with limit <= 8
Use a for loop with base 2, then directly output your dates:
for( $i=2; $i<9; $i=$i+2 )
{
echo date('Y-m-d', strtotime( "+ $i days" )) . PHP_EOL;
}
Result:
2016-04-06
2016-04-08
2016-04-10
2016-04-12
actually i want to make a reminder date where user enter reminder.
lets a user want to add reminder today and want repeat it 5 time after
2days, 3days or what ever he wants, in next comming day. than how i
repeat date with for loop.
I'll help with the above. First of all I will just say I have a huge personal preference towards the DateTime object over simply using date it's more flexible and a hell of a lot more readable in my opinion, so when working with dates I would always suggest using that over date()
So here is some Code:
$date = new DateTime(); // Pretend this is what the User entered. We got it via $_POST or something.
$interval = 2; // Repeat x times at y day intervals. (Not including the initial)
$repeatAmount = 2; // Repeat the reminder x times
for ($i = 0; $i <= $repeatAmount; ++$i) {
echo $date->format('d/m/Y');
$date->modify('+'. $interval .' day');
}
$date = new DateTime()Imagine this is the date the user entered, this is our starting point, our first reminder will at this time.
$interval and $repeatAmount are the interval in days, i.e. I want this to every 2 days and the amount of times you want it to repeat, in our example 2.
for ($i = 0; $i <= $repeatAmount; ++$i) { We want to loop as many times as the user says they want to repeat. Little note ++$i tends to be a very minor performance boost over $i++ in some scenarios, so it is usually better to default to that unless you specifically need to use $i++
echo $date->format('d/m/Y'); Simply print out the date, i'll let you handle the reminder logic.
$date->modify('+' . $interval . ' day'); Increment the dateTime object by the interval that the user has asked for, in our case increment by 2 days.
Any questions let me know.
I want to check between two user-specified times everyday and not run some function call (i.e. "Do Not Disturb").
For example, a user set a "Do Not Disturb" time block between 10:00pm to 6:00am (next day).
FYI, no days/dates are being specified by the end-user, ONLY times. This will run consistently everyday, 7 days a week.
So between 10pm-6am (next day), any function call is ignored. This is what I've written up so far:
$now = time(); // or $now = strtotime('11:00pm'); to simulate time to test
$start = strtotime('10:00pm');
$end = strtotime('6:00am +1 day');
// alternative time block
//$start = strtotime('10:00am');
//$end = strtotime('11:00am');
//debug
//echo date('r', $now) . '<br>' . date('r', $start) . '<br>' . date('r', $end) . '<br><br>';
if($start > $now || $now > $end) {
echo 'disturb';
} else {
echo 'do not disturb';
}
But this doesn't seem to work, because once you reach midnight, it's a new day, but the $end variable is already a day ahead.
I tried putting it a day behind, but then the issue is that the value of $end ends up being lower than the value of $start, which isn't correct.
I also tried adding a day to the $now variable whenever the time reaches midnight, but the issue w/ that is, what if the $start and $end times are within the same day?
What am I missing here?
Apparently you're trying to build some kind of calendar functionality here.
If you use strtotime('10:00pm'); this will change to the timestamp of the next day after midnight.
So you need to give the variable a date
$start = strtotime('2015-02-26 10:00pm');
$end = strtotime('2015-02-27 6:00am');
Not sure how you store these time blocks, but ideally they would be stored in a database table.
If it's every day the same you could do:
$now = time(); // or $now = strtotime('11:00pm'); to simulate time to test
$start = strtotime('10:00pm');
$end = strtotime('6:00am'); // without the +1 day
if($start > $end) {
if($start > $now && $now > $end) {
echo 'disturb';
} else {
echo 'do not disturb';
}
}else{
if($now < $start || $now > $end) {
echo 'disturb';
} else {
echo 'do not disturb';
}
}
That's a nice question actually,
You can use the the relatively new object oriented way of dealing with times.
I'll link you some info as I don't have time to write an entire example
http://php.net/manual/en/datetime.diff.php
http://php.net/manual/en/class.datetime.php
http://php.net/manual/en/class.dateinterval.php
specifically from the docs :
<?php
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
?>
Hope it helps
I would convert to DateTime() objects instead. Then you won't get any issues with days ending.
// obviously you'll need to feed in the date as well so
// that might involve some refactoring
$now = new DateTime();
$start = new DateTime('2015-02-26 10:00');
$end = new DateTime('2015-02-27 06:00');
Now you can compare as before.
If you don't know the date and your users are only specifying time, you might need to add the date dynamically. These are just for example.
Edit: to cope with unknown days, you could dynamically generate after grabbing today:
$today = new DateTime();
$start = new DateTime($today->format('Y-m-d') . ' 10:00');
$end = new DateTime($today->format('Y-m-d') . ' 06:00');
$end->add(new DateInterval('P1D'));
Do you know what the problem is by looking at the code?
I would be happy if you helped me:
list($from_day,$from_month,$from_year) = explode(".","27.09.2012");
list($until_day,$until_month,$until_year) = explode(".","31.10.2012");
$iDateFrom = mktime(0,0,0,$from_month,$from_day,$from_year);
$iDateTo = mktime(0,0,0,$until_month,$until_day,$until_year);
while ($iDateFrom <= $iDateTo) {
print date('d.m.Y',$iDateFrom)."<br><br>";
$iDateFrom += 86400;
}
Date of writing the same problem 2 times
October (31) for writing 2 times in history draws the ends October 30th: (
27.09.2012
28.09.2012
...
26.10.2012
27.10.2012
[[28.10.2012]]
[[28.10.2012]]
29.10.2012
30.10.2012
Your problem is because you have set time to 00:00:00, set it to 12:00:00. That is because the Daylight saving time.
Stop using date() function, use Date and Time classes.
Solution (PHP >= 5.4):
$p = new DatePeriod(
new DateTime('2012-09-27'),
new DateInterval('P1D'),
(new DateTime('2012-10-31'))->modify('+1 day')
);
foreach ($p as $d) {
echo $d->format('d.m.Y') . "\n";
}
Solution (PHP < 5.4)
$end = new DateTime('2012-10-31');
$end->modify('+1 day');
$p = new DatePeriod(
new DateTime('2012-09-27'),
new DateInterval('P1D'),
$end
);
foreach ($p as $d) {
echo $d->format('d.m.Y') . "\n";
}
You have daylight savings time issues. Adding seconds from one timestamp to another is prone to problems around these sorts of edge conditions (leap days can be problematic is well), You should get in the habit of using PHP's DateTime and DateInterval objects. It makes working with dates a snap.
$start_date = new DateTime('2012-09-27');
$end_date = new DateTime('2012-10-31');
$current_date = clone $start_date;
$date_interval = new DateInterval('P1D');
while ($current_date < $end_date) {
// your logic here
$current_date->add($date_interval);
}
My idea for solving this would be something like this;
$firstDate = "27.09.2012";
$secondDate = "31.10.2012";
$daysDifference = (strtotime($secondDate) - strtotime($firstDate)) / (60 * 60 * 24);
$daysDifference = round($daysDifference);
for ($i = 0; $i <= $daysDifference; $i++)
{
echo date("d.m.Y", strtotime('+'.$i.' day', strtotime($firstDate))) . "<BR>";
}
This should solve your problem and be much easier to read (imho). I've just tested the code, and it outputs all dates and no doubles. It also saves you from all the daylight savings inconsistencies.
I don't know where you're from, but it's likely you're hitting daylight saving changeover in your timezone (it's Nov 4th where I live - exactly one week after Oct 28th). You can not rely on a day being exactly 86400 seconds long.
If you loop incrementing with mktime, you should be fine:
list($from_day,$from_month,$from_year) = explode(".","27.09.2012");
list($until_day,$until_month,$until_year) = explode(".","31.10.2012");
$iDateFrom = mktime(0,0,0,$from_month,$from_day,$from_year);
$iDateTo = mktime(0,0,0,$until_month,$until_day,$until_year);
while ($iDateFrom <= $iDateTo)
{
print date('d.m.Y',$iDateFrom)."<br><br>";
$from_day = $from_day + 1;
$iDateFrom = mktime(0,0,0,$from_month,$from_day,$from_year);
}
Even though $from_day will likely be going well over 31, mktime will make the math conversion for you. (ie 32 days in a 31 day month = day 1 of the next month)
EDIT: sorry, I had the incrementation in the wrong place.
I have 2 dates. Lets say they look like this.
$start = 2010/12/24;
$end = 2012/01/05;
I query the database to look for visits between these two dates. I find some. I then populate an array called stats.
$stats['2010/12/25'] = 50;
$stats['2010/12/31'] = 25;
...
As you can see, there are days missing. I need to fill the missing dates with a value of zero. I was thinking something like this. (I have pulled day / month / year from start and end dates.
for($y=$start_year; $y <= $end_year; $y++) {
for($m=$start_month; $m <=$end_month; $m++) {
for($d=$start_day; $d <= $end_day; $d++) {
This would work fine for the year however the months and days wouldn't work. If the start day is the 15th. Days 1-14 of each subsequent month would be missed. I could have a solution like this then...
for($y=$start_year; $y <= $end_year; $y++) {
for($m=1; $m <13; $m++) {
$total_days = cal_days_in_month(CAL_GREGORIAN, $m, $y) + 1;
for($d=1; $d <= $total_days; $d++) {
I would then need a bunch of if statements making sure starting and end months and days are valid.
Is there a better way of doing this? Or could this even be done in my mysql query?
Just to demonstrate the power of some of PHP's newer interval handling method (mentioned by pgl in his answer):
$startDate = DateTime::createFromFormat("Y/m/d","2010/12/24",new DateTimeZone("Europe/London"));
$endDate = DateTime::createFromFormat("Y/m/d","2012/01/05",new DateTimeZone("Europe/London"));
$periodInterval = new DateInterval( "P1D" ); // 1-day, though can be more sophisticated rule
$period = new DatePeriod( $startDate, $periodInterval, $endDate );
foreach($period as $date){
echo $date->format("Y-m-d") , PHP_EOL;
}
Does require PHP >= 5.3.0
EDIT
If you need to include the actual end date, then you need to add a day to $endDate immediately before the foreach() loop:
$endDate->add( $periodInterval );
EDIT #2
$startDate = new DateTime("2010/12/24",new DateTimeZone("Europe/London"));
$endDate = new DateTime("2012/01/05",new DateTimeZone("Europe/London"));
do {
echo $startDate->format("Y-m-d") , PHP_EOL;
$startDate->modify("+1 day");
} while ($startDate <= $endDate);
For PHP 5.2.0 (or earlier if dateTime objects are enabled)
If you're using PHP5.3 then Mark Baker's answer is the one to use. If (as you say in your comment) you're still on PHP5.2 something like this should help you:
$startdate = strtotime( '2010/12/24' );
$enddate = strtotime( '2012/01/05' );
$loopdate = $startdate;
$datesArray = array();
while( $loopdate <= $enddate ) {
$datesArray[$loopdate] = 0;
$loopdate = strtotime( '+1 day', $loopdate );
}
It will create an array of the unix timestamp of every date between the start and end dates as the index and each value set to zero. You can then overwrite any actual results you have with the correct values.
$start_date = DateTime::createFromFormat('Y/m/d', '2010/12/24');
$end_date = DateTime::createFromFormat('Y/m/d', '2012/01/05');
$current_date = $start_date;
while($current_date <= $end_date) {
$current_date = $current_date->add(new DateInterval('P1D'));
// do your array work here.
}
See DateTime::add() for more information about this.
$i = 1;
while(date("Y/m/d", strtotime(date("Y/m/d", strtotime($start)) . "+ $i days")) < $end) {
... code here ...
$i++;
}
I would calculate the difference between start and end date in days, iterate on that adding a day to the timestamp on each iteration.
$start = strtotime("2010/12/24");
$end = strtotime("2012/01/05");
// start and end are seconds, so I convert it to days
$diff = ($end - $start) / 86400;
for ($i = 1; $i < $diff; $i++) {
// just multiply 86400 and add it to $start
// using strtotime('+1 day' ...) looks nice but is expensive.
// you could also have a cumulative value, but this was quicker
// to type
$date = $start + ($i * 86400);
echo date('r', $date);
}
I have this bit of horrible code saved:
while (($tmptime = strtotime('+' . (int) $d++ . ' days', strtotime($from))) && ($tmptime <= strtotime($to))) // this code makes baby jesus cry
$dates[strftime('%Y-%m-%d', $tmptime)] = 0;
(Set $from and $to to appropriate values.) It may well make you cry, too - but it sort of works.
The proper way to do it is to use DateInterval, of course.