Please take a look at the following code:
$parts_count = 3;
$diff = strtotime('2013-11-30 00:00:00') - time();
$diff = $diff / $parts_count;
for ($i = 0; $i < $parts_count; $i++) {
echo date('Y-m-d H:m:s', time() + $diff * ($i + 1) );
}
I am trying to divide the difference between the current time and a deadline (2013-11-30 00:00:00) into 3 equal parts.
The result I get is more than interesting though. It is:
2013-11-16 23:11:36
2013-11-23 11:11:48
2013-11-30 00:11:00
As you see, the latest deadline is 11 minutes past the original deadline, which is very weird. Do you have an idea why this happens?
Misspelling in format, should be:
echo date('Y-m-d H:i:s', time() + $diff * ($i + 1) );
Related
This question already has answers here:
How do I find the hour difference between two dates in PHP?
(10 answers)
Closed 5 years ago.
I have datetimes in the following format:
Start: 2017-07-16 20:00
End: 2017-07-16 23:30
Start: 2017-07-18 21:30
End: 2017-07-19 00:30
I need to tell from these intervals how many hours (in 0,5 increments) are spent between 18:00-22:00 and 22:00-06:00 in total for a month.
Thanks in advance for any hint.
The current code I have is this, but I'm not sure if it is covering all timeframe possibilities:
<?php
date_default_timezone_set("UTC");
$start = array(
"2017-07-16 21:30:00",
"2017-07-16 18:00:00"
);
$end = array(
"2017-07-17 00:30:00",
"2017-07-16 18:30:00"
);
$amount_low = 0;
$amount_high = 0;
for($i = 0; $i < sizeof($start); $i++) {
$start_time = date("H:i:s", strtotime($start[$i]));
$end_time = date("H:i:s", strtotime($end[$i]));
$start_date = date("Ymd", strtotime($start[$i]));
$end_date = date("Ymd", strtotime($end[$i]));
// getting chunk before 22:00 if
if(
(strtotime($start[$i]) >= strtotime($start_date . " 18:00") && strtotime($start[$i]) < strtotime($start_date . " 22:00"))
&&
$start_date < $end_date
) {
$interval_low = strtotime($start_date . " 22:00") - strtotime($start[$i]);
$amount_low += ceil($interval_low / 1800) / 2;
}
//amount_high
if(strtotime($start[$i]) > strtotime($start_date . " 22:00") && strtotime($start[$i]) < strtotime($start_date . " 06:00")) {
$interval_high = strtotime($end[$i]) - strtotime($start[$i]); //needs further things
$amount_high += ceil($interval_high / 1800) / 2;
} elseif (strtotime($start[$i]) < strtotime($start_date . " 22:00") && strtotime($end[$i]) > strtotime($start_date . " 22:00")) {
$interval_high = strtotime($end[$i]) - strtotime($start_date . " 22:00");
$amount_high += ceil($interval_high / 1800) / 2;
} else {
$interval_low = strtotime($end[$i]) - strtotime($start[$i]);
$amount_low += ceil($interval_low / 1800) / 2;
}
}
echo $amount_low;
echo "\n$amount_high";
?>
Would this work for you?
$start = strtotime('2017-07-16 20:00');
$end = strtotime('2017-07-16 23:30');
$interval = $end - $start;
$hours = floor($interval / 1800)/2;
echo($hours);
This will display the total hours (in 0,5 increments) between the two dates (rounding down, so if it's 55 minutes, it's 0,5 hours; replace 'floor' with 'ceil' for the opposite).
I believe a part of your solution is found here from Aillyn
You can convert them to timestamps and go from there:
$hourdiff = round((strtotime($time1) - strtotime($time2))/3600, 1);
Dividing by 3600 because there are 3600 seconds in one hour and using round()
to avoid having a lot of decimal places.
$hourdiff = round((strtotime($time1) - strtotime($time2))/3600, 1);
This will give you a rounded hour difference between the two times. You could do something similar, just apply it for each interval for which ever month and adjust your math. There's not really going to be a single PHP function you can call to solve this.
So here is what I need to do:
Get time from database (don't worry about the database stuff for the moment)
Countdown time in days, hours and minutes
When countdown has reached 0, add 7 days to countdown along with adding 1 to an episode count
repeat multiple times until episode count reaches certain number (set by database, again don't worry to much about the database stuff at the moment) then stop countdown and just echo Aired
Basically its going to countdown to the next episode of a TV show's airing time and show what episode number is next, this will carry on until all the episodes have aired.
Here is what I have currently, it works to a degree but will only +7 days/++episode once, after that the countdown will go into negatives. I've tried while loops and some other things but I haven't had too much luck.
$date = "February 12, 2013 5:06 PM";
$date = strtotime($date);
$remaining = $date - time();
$episode = 0;
if ($remaining < 0) {
++$episode;
$remaining = strtotime("+7 day", $date) -time();
}
$days_remaining = floor($remaining / 86400);
$hours_remaining = floor(($remaining % 86400) / 3600);
$mins_remaining = floor(($remaining % 86400 % 3600) / 60);
if ($episode == 3){
echo "Aired";
} else {
echo "$days_remaining:$hours_remaining:$mins_remaining Ep $episode";
}
Any advice is really appreciated, thanks!
Of course it will go to negative.
After $remaining reaches 0, another 7 days are added to it.
if ($remaining < 0) { // remaining: less than 0 days
++$episode; // episode becomes 1
$remaining = strtotime("+7 day", $date) -time(); // remaining: less than 7 days
}
However, after another 7 days, $remaining is still negative and only 7 days are added to it. Since you haven't saved $episode in a database, it is still 0 according to the code.
$episode = 0; // $episode starts at 0 according to the code
if ($remaining < 0) { // remaining: less than -7 days
++$episode; // episode becomes 1
So it continues to count negative and give you $episode == 1.
You should do this instead:
$date = strtotime("February 12, 2013 5:06 PM");
for ($i = 0; $i < 3; $i++) {
$episodes[$i+1] = strtotime("+". 7*$i ." day", $date);
$remainings[$i+1] = $episodes[$i+1] - time();
}
foreach ($remainings as $key => $remaining) {
if ($remaining > 0) {
$episode = $key;
$days_remaining = floor($remaining / 86400);
$hours_remaining = floor(($remaining % 86400) / 3600);
$mins_remaining = floor(($remaining % 86400 % 3600) / 60);
break;
}
}
if (!isset($episode)){
echo "Aired";
} else {
echo "$days_remaining:$hours_remaining:$mins_remaining Ep $episode";
}
How do I get the exact dates of the last 7 days including today in a custom format (dd/mm)?
In the resulting array I would like to get something like (dates are examples only):
1=>11/2 (today minus 7 days)
2=>12/2 (today minus 6 days)
...
7=>17/2 (today)
function getLastNDays($days, $format = 'd/m'){
$m = date("m"); $de= date("d"); $y= date("Y");
$dateArray = array();
for($i=0; $i<=$days-1; $i++){
$dateArray[] = '"' . date($format, mktime(0,0,0,$m,($de-$i),$y)) . '"';
}
return array_reverse($dateArray);
}
Usage:
$arr = getLastNDays(7);
or
$arr = getLastNDays(7, 'd/m/Y');
You can combine the 2 functions date() and strtotime(). for example:
echo date("Y-m-d", strtotime("7 days ago"));
Try:
for ($i=0; $i<7; $i++)
{
echo date("d/m", strtotime($i." days ago")).'<br />';
}
You should be able to work out how to get them in the correct order and into an array :)
Hope that helps
time() gives you the current timestamp.
86400 seconds are one day (60 * 60 * 24).
date() gives you a custom date string.
for ($iDay = 6; $iDay >= 0; $iDay--) {
$aDays[7 - $iDay] = date('d/m', time() - $iDay * 86400);
}
Also see this example.
If you don't want the leading zeros, use 'j/n' as custom date format parameter:
for ($iDay = 6; $iDay >= 0; $iDay--) {
$aDays[7 - $iDay] = date('j/n', time() - $iDay * 86400);
}
Also see this updated example.
=== UPDATE ===
#Dagon's idea to use strtotime() to get the timestamp is great. Here the better solution:
for ($iDay = 6; $iDay >= 0; $iDay--) {
$aDays[7 - $iDay] = date('j/n', strtotime("-" . $iDay . " day"));
}
And an example.
I have a function that creates time intervals between two time marks. The function works but I'm struggling to upgrade from strtotime() and use the DateTime class.
Below is a patch of code I wrote without getting errors
$timestart = new DateTime("14:00:00");
$timestop = new DateTime("20:00:00");
$date_diff = $timestop->diff($timestart);
$time_diff = $date_diff->format('%H');
Next is the entire code untouched. I get DateInterval could not be converted to int erros using the code above. Please kindly advise how to correctly implement the class.
Live example: http://codepad.org/jSFUxAnp
function timemarks()
{
//times are actually retrieved from db
$timestart = strtotime("14:00:00");
$timestop = strtotime("20:00:00");
$time_diff = $timestop - $timestart; //time difference
//if time difference equals negative value, it means that $timestop ends second day
if ($time_diff <= 0)
{
$timestop = strtotime( '+1 day' , strtotime( $row->timestop ) ); //add 1 day and change the timetsamp
$time_diff = $timestop - $timestart;
}
//create interval
$split = 3;
$interval = $time_diff/$split;
//get all timemarks
$half_interval = $interval / 2;
$mid = $timestart + $half_interval;
for ( $i = 1; $i < $split; $i ++) {
//round intervals
$round_mid = round($mid / (15 * 60)) * (15 * 60);
$result .= date('H:i:s', $round_mid) . ", ";
$mid += $interval;
}
$round_mid = round($mid / (15 * 60)) * (15 * 60);
$result .= date('H:i:s', $round_mid);
return $result;
}
outputs 15:00:00, 17:00:00, 19:00:00
Actually you're using DateTime, these are just aliases for creating DateTime instances
The equivalent would look like this:
$timestart = new DateTime("14:00:00");
$timestop = new DateTime("20:00:00");
$date_diff = $timestop->diff($timestart);
$time_diff = $date_diff->format('%H');
So this has to work, I tested it and I got correct results!
i hope some help to me ...
i am trouble with some code IN PHP ,maybe you can see function about this :
I need to check is there on Saturday in one period ($date_start ,$date_end )
so :
input : $date_Start (ex: "2010-07-01")
$date_end (ex: "2010-07-12")
output : total saturday : 2 in your range date
but the way thanks before
JOKONARDI
Efficient implementation with no loops for number of Saturdays in the range:
$d1 = new DateTime("2010-07-01"); //including
$d2 = new DateTime("2010-07-12"); //excluding
$diff = $d2->diff($d1);
$days = $diff->format("%a");
$num = floor($days / 7);
$remaining = $days % 7;
$d = $d1->format("w");
if ($d + $remaining > 6)
$num++;
//result in $num