I have 28 unique events, one event per day. Events repeat after 28. is 1. event again.
Lets say first event is 1970-01-01. Event 28 is 1970-01-28.
Today is 2015-10-16, how do I find out what event should be today? How do I find out what event should be for any given date? (just dates > 1970-01-01 and < 9999-12-31).
I was thinking of converting date to timestamp, then do modulo 28*24*60*60, and then convert it back to date, but I'm afraid that there might be some margin. Any ideas?
You may use DateTime::diff function as follow:
$datetime1 = date_create('1970-01-28');
$datetime2 = date_create('2015-10-16');
$interval = date_diff($datetime2, $datetime1);
/// Get total number of days
$days = $interval->format('%a');
/// Get the remainder of the division by the number of events
$eventNumber = $days % 28;
Another solution is to use unix timestamp to get date differences in terms of seconds. And since 01-01-1970 is zero by definition, all you have to do is:
$today = mktime(); // Or $today = mktime(12,0,0,10,16,2015);
$mod = floor($today/86400) % 28;
Related
How exactly is this done? There's so many questions on stack-overflow about what I'm trying to do; However all of the solutions are to edit the MYSQL Query, and I need to do this from within PHP.
I read about the strtotime('-30 days') method on another question and tried it, but I can't get any results. Here's what I'm trying:
$current_date = date_create();
$current_date->format('U');
... mysql code ...
$transaction_date = date_create($affiliate['Date']);
$transaction_date->format('U');
if($transaction_date > ($current_date - strtotime('-30 days'))) {
} else if(($transaction_date < (($current_date) - (strtotime('-30 days'))))
&& ($transaction_date > (($current_date) - (strtotime('-60 days'))))) {
}
Effectively, I'm trying to sort all of the data in the database based on a date, and if the database entry was posted within the last 30 days, I want to perform a function, then I want to see if the database entry is older than 30 days, but not older than 60 days, and perform more actions.
This epoch math is really weird, you'd think that getting the epoch of the current time, the epoch of the data entry, and the epoch of 30 and 60 days ago would be good enough to do what I wanted, but for some reason it's not working, everything is returning as being less than 30 days old, even if I set the date in the database to last year.
No need to convert to unix timestamp, you can already compare DateTime objects:
$current_date = data_create();
$before_30_day_date = date_create('-30 day');
$before_60_day_date = date_create('-60 day');
$transaction_date = date_create($affiliate['Date']);
if ($transaction_date > $before_30_day_date) {
# transation date is between -30 day and future
} elseif ($transaction_date < $before_30_day_date && $transaction_date > $before_60_day_date) {
# transation date is between -60 day and -30 day
}
This creates (inefficiently, see my comment above) an object:
$current_date = date_create(date("Y-m-d H:i:s"));
From which you try to subtract an integer:
if($transaction_date > ($current_date - strtotime('-30 days'))) {
which is basically
if (object > (object - integer))
which makes no sense.
you're mixing the oldschool time() system, which deals purely with unix timestamps, and the newer DateTime object system, which deals with objects.
What you should have is
$current_date = date_create(); // "now"
$d30 = new DateInterval('P30D'); // 30 days interval
$transaction_date = date_create($affiliate['Date']);
if ($transaction_date > ($current_date->sub($d30)) { ... }
You might consider DatePeriod class, which in essence gives you the ability to deal with a seires of DateTime objects at specified intervals.
$current_date = new DateTime();
$negative_thirty_days = new DateInterval::createFromDateString('-30 day');
$date_periods = new DatePeriod($current_date, $negative_thrity_days, 3);
$thirty_days_ago = $date_periods[1];
$sixty_day_ago = $date_periods[2];
Here you would use $thirty_days_ago, $sixty_days_ago, etc. for your comparisons.
Just showing this as alternative to other options (which will work) as this is more scalable if you need to work with a larger number of interval periods.
I have stored date field at DB.
In PHP, i am getting that field and converted into date.
I want to compare that time with current time. If that difference is above 60 minutes. It will return some value.
I dont know how to write logic for that
$lastUpdatedField = $rows_fetch['lastUpdatedTime'];
$lastUpdatedDate = new DateTime($lastUpdatedField);
$nowDate = new DateTime(date('y-m-d h:m:s'));
I have old date&time is in $lastUpdatedDate variable, and current time is in $nowDate.
How to compare these two
$interval = $nowDate->diff($lastUpdatedDate);
echo $interval->h;
DateDiff: http://www.php.net/manual/en/datetime.diff.php
DateInterval: http://www.php.net/manual/en/class.dateinterval.php
Had The same problem earlier its actually quit simple
heres the piece where you declare your variables
$lastUpdateddate = new DateTime($lastUpdatedField);
$nowDate = new DateTime(date('y-m-d h:m:s'));
Then you have to convert them to second - format so that you can do math with them
To do that use strtotime
$Diff = strtotime($lastUpdatedDate) - strtotime($nowDate);
Then just check to see if the difference in time is more then 60 minutes,
So devide by 60 seconds to get minutes and by 60 to get hours
if ($diff/60/60 <= 1){
//do your thing here
{
First convert the current time and old time to one unit like Unix timestamp passing it through strtotime(). Then differentiate both the timestamp to get the difference between two times.
$difftime = strtotime(date('Y-m-d H:i:s')) - strtotime($rows_fetch['lastUpdatedTime']);
Then convert the difference to days as follows :
$days=$difftime/24*60*60;
Once you get the days you can get the minutes from it as below to compare to meet your need.
$timediff = $days * 24 * 60;
I want to add an x number of week days (e.g. 48 weekday hours) to the current timestamp. I am trying to do this using the following
echo (strtotime('2 weekdays');
However, this doesn't seem to take me an exact 48 hours ahead in time. For example, inputting the current server time of Tuesday 18/03/2014 10:47 returns Thursday 20/03/2014 00:00. using the following function:
echo (strtotime('2 weekdays')-mktime())/86400;
It can tell that it's returning only 1.3 weekdays from now.
Why is it doing this? Are there any existing functions which allow an exact amount of weekday hours?
Given you want to preserve the weekdays functionality and not loose the hours, minutes and seconds, you could do this:
$now = new DateTime();
$hms = new DateInterval(
'PT'.$now->format('H').'H'.
$now->format('i').'M'.
$now->format('s').'S'.
);
$date = new DateTime('2 weekdays');
$date->add($hms);//add hours here again
The reason why weekday doesn't add the hours is because, if you add 1 weekday at any point in time on a monday, the next weekday has to be tuesday.
The hour simply does not matter. Say your date is 2014-01-02 12:12:12, and you want the next weekday, that day starts at 2014-01-03 00:00:00, so that's what you get.
My last solution works though, and here's how: I use the $now instance of DateTime, and its format method to construct a DateInterval format string, to be passed to the constructor. An interval format is quite easy: it starts with P, for period, then a digit and a char to indicate what that digit represents: 1Y for 1 Year, and 2D for 2 Days.
However, we're only interested in hours, minutes and seconds. Actual time, which is indicated using a T in the interval format string, hence we start the string with PT (Period Time).
Using the format specifiers H, i and s, we construct an interval format that in the case of 12:12:12 looks like this:
$hms = new DateInterval(
'PT12H12M12S'
);
Then, it's a simple matter of calling the DateTime::add method to add the hours, minutes and seconds to our date + weekdays:
$weekdays = new DateTime('6 weekdays');
$weekdays->add($hms);
echo $weekdays->format('Y-m-d H:i:s'), PHP_EOL;
And you're there.
Alternatively, you could just use the basic same trick to compute the actual day-difference between your initial date, and that date + x weekdays, and then add that diff to your initial date. It's the same basic principle, but instead of having to create a format like PTXHXMXS, a simple PXD will do.
Working example here
I'd urge you to use the DateInterface classes, as it is more flexible, allows for type-hinting to be used and makes dealing with dates just a whole lot easier for all of us. Besides, it's not too different from your current code:
$today = new DateTime;
$tomorrow = new DateTime('tomorrow');
$dayAfter = new DateTime('2 days');
In fact, it's a lot easier if you want to do frequent date manipulations on a single date:
$date = new DateTime();//or DateTime::createFromFormat('Y-m-d H:i:s', $dateString);
$diff = new DateInterval('P2D');//2 days
$date->add($diff);
echo $date->format('Y-m-d H:i:s'), PHP_EOL, 'is the date + 2 days', PHP_EOL;
$date->sub($diff);
echo $date->format('Y-m-d H:i:s'), PHP_EOL, 'was the original date, now restored';
Easy, once you've spent some time browsing through the docs
I think I have found a solution. It's primitive but after some quick testing it seems to work.
The function calculates the time passed since midnight of the current day, and adds it onto the date returned by strtotime. Since this could fall into a weekend day, I've checked and added an extra day or two accordingly.
function weekDays($days) {
$tstamp = (strtotime($days.' weekdays') + (time() - strtotime("today")));
if(date('D',$tstamp) == 'Sat') {
$tstamp = $tstamp + 86400*2;
}
elseif(date('D',$tstamp) == 'Sun') {
$tstamp = $tstamp + 86400;
}
return $tstamp;
}
Function strtotime('2 weekdays') seems to add 2 weekdays to the current date without the time.
If you want to add 48 hours why not adding 2*24*60*60 to mktime()?
echo(date('Y-m-d', mktime()+2*24*60*60));
The currently accepted solution works, but it will fail when you want to add weekdays to a timestamp that is not now. Here's a simpler snippet that will work for any given point in time:
$start = new DateTime('2021-09-29 15:12:10');
$start->add(date_interval_create_from_date_string('+ 3 weekdays'));
echo $start->format('Y-m-d H:i:s'); // 2021-10-04 15:12:10
Note that this will also work for a negative amount of weekdays:
$start = new DateTime('2021-09-29 15:12:10');
$start->add(date_interval_create_from_date_string('- 3 weekdays'));
echo $start->format('Y-m-d H:i:s'); // 2021-09-24 15:12:10
I have a table which shows the time since a job was raised.
// These are unix epoch times...
$raised = 1360947684;
$now = 1361192598;
$difference = 244914;
$difference needs to exclude any time outside of business hours (ex, 9-5 and weekends).
How could I tackle this?
The thing you have to do are 3 in numbers.
You take your start date and calculate the rest time on this day (if it is a business day)
You take your end date and calulate the time on this day and
you take the days in between and multiply them with your business hours (just those, that are business days)
And with that you are done.
Find a little class attached, which does those things. Be aware that there is no error handling, time zone settings, daylight saving time, ...
input:
start date
end date
output:
difference time in seconds
adjustable constants:
Business hours
Days that are not business days
Very bad idea, but I had no choice because I'm on php 5.2
<?php
date_default_timezone_set('Asia/Seoul');
$start = 1611564957;
$end = 1611670000;
$res = 0;
for($i = $start; $i<$end; $i++){
$h = date("H", $i);
if($h >= 9 && $h < 18){
//echo date("Y-m-d H:i:s", $i) . "<br>";
$res = $res + 1;
}
}
echo $res;
Use DateTime.
Using UNIX time for this is slightly absurd, and you would have to literally remake DateTime.
Look up relative formats where you can specify the hour on the day, e.g.
$date = new DateTime($raised);
$business_start = new DateTime("09:00"); // 9am today
$business_end = new DateTime("17:00"); // 5pm today
The rest is for you to work out.
Oh, and instead of start/end, you could probably use DateInterval with a value of P8H ("period 8 hours")
The problem with using timestamps directly is that you are assigning a context to a counter of seconds. You have to work backwards from the times you want to exclude and work out their timestamps beforehand. You might want to try redesigning your storage of when a job is raised. Maybe set an expiry time for it instead?
I am developing a quiz site and there is time for x min to answer the quiz. So when user clicks on start quiz link the starttime (current time at this instant) is recored in session. Also the endtime (start_time+ 30 min) is recorded in session and every time he submits a answer the current time is compared with the quiz end time. Only if the current time is less than end_time the answer should be accepted.
How can I get the currentdatetime?
How can I add x minutes to current this datetime?
How can I compare (<=) datetime ?
I think we should use date time. Is it right?
PHP measures time as seconds since Unix epoch (1st January 1970). This makes it really easy to work with, since everything just a single number.
To get the current time, use: time()
For basic maths like adding 30 minutes, just convert your interval into seconds and add:
time() + 30 * 60 // (30 * 60 ==> 30 minutes)
And since they're just numbers, just do regular old integer comparison:
$oldTime = $_SESSION['startTime'];
$now = time();
if ($now < $oldTime + 30 * 60) {
//expired
}
If you need to do more complicated things like finding the date of "next tuesday" or something, look at strtotime(), but you shouldn't need it in this case.
use php builtin functions to get time:
<?php
$currentTimeStamp = time(); // number of seconds since 1970, returns Integer value
$dateStringForASpecificSecond = date('Y-m-d H:i:s', $currentTimeStamp);
?>
for your application that needs to compare those times, using the timestamp is more appropriate.
<?php
$start = time();
$end = $start + (30 * 60); // 30 minutes
$_SESSION['end_time'] = $end;
?>
in the page where the quiz is submitted:
<?php
$now = time();
if ( $now <= $_SESSION['end_time'] ) {
// ok!
}
?>
Use the time() function to get a UNIX timestamp, which is really just a large integer.
The number returned by time() is the number of seconds since some date (like January 1, 1970), so to add $x minutes to it you do something like (time() + ($x*60)).
Since UNIX timestamps are just numbers, you can compare them with the usual comparison operators for numbers (< <= > >= ==)
time() will give you the current time in seconds since 1/1/1970 (an integer), which looks like it should be good.
To add x minutes, you'd just need to add x*60 to that, and you can compare it like any other two integers.
Source: http://us3.php.net/time
This is an old question but I wanted to provide an answer based on the PHP 5.2 DateTime class which I feel is much easier to use and much more versatile than any previous functions.
So how can i get the currentdatetime?
You can create a new DateTime object like this:
$currentTime = new DateTime();
But at this point, $currentTime is a datetime object and must be converted to a string in order to store it in a database or output it.
$currentTime = $currentTime->format('Y-m-d H:i:s');
echo $currentTime;
Outputs 2014-05-10 21:14:06
How can i add x minutes tocurrent this datetime?
You can add x minutes with the modify method:
$currentTime = new DateTime();
$addedMinutes = $currentTime->modify('+10 minutes');
echo $addedMinutes;
Outputs 2014-05-10 21:24:06
How can i comapare (<=) datetime ?
With the DateTime class, you can not only easily compare datetime objects, you can get the difference between them.
$currentTime = new DateTime('2014-05-10 21:14:06');
$addDays = $currentTime->modify('+10 days');
To compare
if ($currentTime >= $addDays) {
//do something//
}
$diffTime = new DateTime('2014-05-10 21:14:06');
$diff = $addDays->diff($diffTime);
$diff = $diff->format('There are %d days difference.');
echo $diff;
Outputs There are 10 days difference.