PHP Adding 15 minutes to Time value - php

I have a form that receives a time value:
$selectedTime = $_REQUEST['time'];
The time is in this format - 9:15:00 - which is 9:15am. I then need to add 15 minutes to this and store that in a separate variable but I'm stumped.
I'm trying to use strtotime without success, e.g.:
$endTime = strtotime("+15 minutes",strtotime($selectedTime)));
but that won't parse.

Your code doesn't work (parse) because you have an extra ) at the end that causes a Parse Error. Count, you have 2 ( and 3 ). It would work fine if you fix that, but strtotime() returns a timestamp, so to get a human readable time use date().
$selectedTime = "9:15:00";
$endTime = strtotime("+15 minutes", strtotime($selectedTime));
echo date('h:i:s', $endTime);
Get an editor that will syntax highlight and show unmatched parentheses, braces, etc.
To just do straight time without any TZ or DST and add 15 minutes (read zerkms comment):
$endTime = strtotime($selectedTime) + 900; //900 = 15 min X 60 sec
Still, the ) is the main issue here.

Though you can do this through PHP's time functions, let me introduce you to PHP's DateTime class, which along with it's related classes, really should be in any PHP developer's toolkit.
// note this will set to today's current date since you are not specifying it in your passed parameter. This probably doesn't matter if you are just going to add time to it.
$datetime = DateTime::createFromFormat('g:i:s', $selectedTime);
$datetime->modify('+15 minutes');
echo $datetime->format('g:i:s');
Note that if what you are looking to do is basically provide a 12 or 24 hours clock functionality to which you can add/subtract time and don't actually care about the date, so you want to eliminate possible problems around daylights saving times changes an such I would recommend one of the following formats:
!g:i:s 12-hour format without leading zeroes on hour
!G:i:s 12-hour format with leading zeroes
Note the ! item in format. This would set date component to first day in Linux epoch (1-1-1970)

strtotime returns the current timestamp and date is to format timestamp
$date=strtotime(date("h:i:sa"))+900;//15*60=900 seconds
$date=date("h:i:sa",$date);
This will add 15 mins to the current time

To expand on previous answers, a function to do this could work like this (changing the time and interval formats however you like them according to this for function.date, and this for DateInterval):
(I've also written an alternate form of the below function here.)
// Return adjusted time.
function addMinutesToTime( $time, $plusMinutes ) {
$time = DateTime::createFromFormat( 'g:i:s', $time );
$time->add( new DateInterval( 'PT' . ( (integer) $plusMinutes ) . 'M' ) );
$newTime = $time->format( 'g:i:s' );
return $newTime;
}
$adjustedTime = addMinutesToTime( '9:15:00', 15 );
echo '<h1>Adjusted Time: ' . $adjustedTime . '</h1>' . PHP_EOL . PHP_EOL;

get After 20min time and date
function add_time($time,$plusMinutes){
$endTime = strtotime("+{$plusMinutes} minutes", strtotime($time));
return date('h:i:s', $endTime);
}
20 min ago Date and time
date_default_timezone_set("Asia/Kolkata");
echo add_time(date("Y-m-d h:i:sa"),20);

In one line
$date = date('h:i:s',strtotime("+10 minutes"));

You can use below code also.It quite simple.
$selectedTime = "9:15:00";
echo date('h:i:s',strtotime($selectedTime . ' +15 minutes'));

Current date and time
$current_date_time = date('Y-m-d H:i:s');
15 min ago Date and time
$newTime = date("Y-m-d H:i:s",strtotime("+15 minutes", strtotime($current_date)));

Quite easy
$timestring = '09:15:00';
echo date('h:i:s', strtotime($timestring) + (15 * 60));

Related

PHP adding exact weekdays to a timestamp

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

In an LDAP 'lastlogon' lookup how do I decipher the result?

I'm using a PHP script to grab data from Active Directory using LDAP..
When I get the user values for 'lastlogon' I get a number like 129937382382715990
I've tried to figure out how to get the date/time from this but have no idea, can anybody help?
Read this comment on the PHP: LDAP Functions page.
All of them are using "Interval" date/time format with a value that represents the number of 100-nanosecond intervals since January 1, 1601 (UTC, and a value of 0 or 0x7FFFFFFFFFFFFFFF, 9223372036854775807, indicates that the account never expires): https://msdn.microsoft.com/en-us/library/ms675098(v=vs.85).aspx
So if you need to translate it from/to UNIX timestamp you can easily calculate the difference with:
<?php
$datetime1 = new DateTime('1601-01-01');
$datetime2 = new DateTime('1970-01-01');
$interval = $datetime1->diff($datetime2);
echo ($interval->days * 24 * 60 * 60) . " seconds\n";
?>
The difference between both dates is 11644473600 seconds. Don't rely on floating point calculations nor other numbers that probably were calculated badly (including time zone or something similar).
Now you can convert from LDAP field:
<?php
$lastlogon = $info[$i]['lastlogon'][0];
// divide by 10.000.000 to get seconds from 100-nanosecond intervals
$winInterval = round($lastlogon / 10000000);
// substract seconds from 1601-01-01 -> 1970-01-01
$unixTimestamp = ($winInterval - 11644473600);
// show date/time in local time zone
echo date("Y-m-d H:i:s", $unixTimestamp) ."\n";
?>
This is the number 100-nanosecond ticks since 1 January 1601 00:00:00 UT.
System time article in Wikipedia can give you more details.
What about this:
$timeStamp = 129937382382715990;
echo date('Y-m-d H:i:s', $timeStamp);
EDIT ------
I just tried the following and noticed that this method wont work unless the clock on your machine is set 10 years in the future. Below is the code I used to prove the above pretty much useless unless you do more processing maybe..
$time = time();
echo date('Y-m-d H:i:s', $time);
echo "<br />";
$timeStamp = 129937382382715990;
echo date('Y-m-d H:i:s', $timeStamp);
In my case I'm using Pentaho. With a Modified Javascript value you can convert the values, lastLogon is the column I wanna convert from data stream:
calendar = java.util.Calendar.getInstance();
calendar.setTime(new Date("1/1/1601"));
base_1601_time = calendar.getTimeInMillis();
calendar.setTime(new Date("1/1/1970"));
base_1970_time = calendar.getTimeInMillis();
ms_offset = base_1970_time - base_1601_time;
calendar.setTimeInMillis( lastLogon / 10000 - ms_offset); //lastLogon is a column from stream
var converted_AD_time = calendar.getTime(); // now just add this variable 'converted_AD_time' to the 'Fields' as a show in the image below

How to convert date into same format?

I want to get the $registratiedag and count a couple of days extra, but I always get stuck on the fact that it needs to be a UNIX timestamp? I did some google-ing, but I really don't get it.
I hope someone can help me figure this out. This is what I got so far.
$registratiedag = $oUser['UserRegisterDate'];
$today = strtotime('$registratiedag + 6 days');
echo $today;
echo $registratiedag;
echo date('Y-m-d', $today);
There's obviously something wrong with the strtotime('$registratiedag + 6 days'); part, because I always get 1970-01-01
You probably want this:
// Store as a timestamp
$registratiedag = strtotime($oUser['UserRegisterDate']);
$new_date = strtotime('+6 days', $registratiedag);
// You'll need to format for printing $new_date
echo date('Y-m-d', $new_date);
// I think you want to compare $new_date against
// today's date. I'd recommend a string comparison here,
// As time() includes the time as well
// time() is implied as the second argument to date,
// But we'll put it anyways just to be clearer
if( date('Y-m-d', $new_date) == date('Y-m-d', time()) ) {
// The dates are equal, do something here
}
else if($new_date < time()) {
// if the new date is earlier than today
}
// etc.
First it converts $registratiedag to a timestamp, then it adds 6 days
EDIT: You probably should change $today to something less misleading like $modified_date or something
try:
$today = strtorime($registratiedag);
$today += 86400 * 6; // seconds in 1 day * 6 days
at least one of your problems is that PHP does not expand variables in single quotes.
$today = strtotime("$registratiedag + 6 days");
//use double quotes and not single quotes when embedding a php variable in a string
If you want to include the value of variable $registratiedag right into the text passed as parameter of strtotime, you have to enclose that parameter with ", not with '.

check time for time difference in 24hr

In php i get the variable $date_time in this format -- 11-01-2010 20:48:25 . This time is GMT time. I have a 2 hour flexibility and if it exceeds 2 hours then i have to reject it. I am set in EST, but i want to do the check based on GMT only so that there is no errors in the time difference. How can i set to GMT in my php code and how do i check for the 2 hours flexible time difference? like for this example it is acceptable for any time between 11-01-2010 18:48:25 and 11-01-2010 22:48:25. Also will it be an issue if $date_time is 11-01-2010 23:48:23?
Clarification
I am doing a $date_time=$_GET['date_time'];. Then i need to check if this new $date_time if within 2 hours range of the current GMT time. if it is in the range, then i will proceed to execute that code, else i will show an error or do something else. I wanted to know how i am going to check this 2 hours range for this $date_time variable.
Here is a way how to convert your time format into a UNIX timestamp:
$date = strptime($date_time, "%m-%d-%Y %T");
$ut = mktime($date['tm_hour'], $date['tm_min'], $date['tm_sec'], 1 + $date['tm_mon'], $date['tm_mday'], 1900 + $date['tm_year']);
$now = time();
if($ut >= $now && $ut <= ($now + 7200)) { // 7200 = 2 * 60 * 60 seconds
// allowed
}
Reference: strptime, mktime, time.
Note: time() always returns the UNIX timestamp in UTC (regardless of time settings). So this assumes that the $date_time timestamp is a GMT time.
Working example (of course you have to provide a valid GMT time for $date_time).
Note 2: If the input time is not in GMT, you can set the timezone with date_default_timezone_set (affects mktime but not time).
Working example (change time and timezone accordingly)
If PHP >= 5.3 (you've got a seriously weird format BTW):
date_default_timezone_set('EST');
$inputtime = DateTime::createFromFormat('m-d-Y H:i:s','11-01-2010 20:48:25',new DateTimeZone("GMT"));
$diff = $inputtime->getTimestamp() - time();
if(abs($diff) > 7200){
//more then 2 hours difference.
}
If you run on PHP > 5.3, you can use DateTime for this :
$my_date = "11-01-2010 20:48:25";
$date = DateTime::createFromFormat('m-d-Y H:i:s', $my_date);
$date_lower = DateTime::createFromFormat('m-d-Y H:i:s', $my_date);
$date_upper = DateTime::createFromFormat('m-d-Y H:i:s', $my_date);
$date_lower->sub(new DateInterval('PT2H'));
$date_upper->add(new DateInterval('PT2H'));
var_dump($date >= $date_lower && $date <= $date_upper); // bool(true)
I find it more readable.
You can also use another timezone if necessary, check the third argument of createFromFormat.
I suggest you to never pass times and dates with format string. Just convert it later. You just pass the timestamp as a get variable and then you format it in the script.
It's the best solution and also the cleanest.
Then use the following code:
$flexibility = X seconds;
if ($date_time < time() - $flexibility or $date_time > time() + $flexibility)
{ /*Error*/ }

Want to display the current date/time

Hi I'm using php and sql through odbc to write a program and i hav got abit stuck in a part where i want to display the current date/time in the format date('Y-m-d H:i:s) but it only displays the gmt time. I want to add 8hours to it.Can any of you b able to help me.Thank you so much
Check out date_default_timezone_set. You can do something like:
date_default_timezone_set('America/Los_Angeles');
print 'Current datetime is: ' . date('Y-m-d H:i:s');
You could use that to set the timezone to whatever timezone you need time to be at, and then use date normally. Alternatively, you can do this, using strtotime:
print 'Current datetime is: ' date('Y-m-d H:i:s', strtotime('+8 hours'));
If you're looking for a way to display a timestamp in a user's local time, you can use JavaScript:
function showtime(t)
{
if (t == 0)
{
document.write("never");
return;
}
var currentTime = new Date(t);
var month = currentTime.getMonth() + 1;
var day = currentTime.getDate();
var year = currentTime.getFullYear();
var hours = currentTime.getHours();
var minutes = currentTime.getMinutes();
var seconds = currentTime.getSeconds();
document.write();
if (minutes < 10){
minutes = "0" + minutes;
}
if (seconds < 10){
seconds = "0" + seconds;
}
document.write(month + "/" + day + "/" + year + " " +
hours + ":" + minutes + ":" + seconds + " ");
if(hours > 11){
document.write("PM");
} else {
document.write("AM");
}
}
Then if you need to display a time, just make a call to it in the HTML and splice in the value from PHP:
<script type="text/javascript">showtime(<?=$time."000"?>)</script>
I would steer clear of the timezone method.
If i understood correctly, you want to add time, thus change it. An example could be, A task has been created NOW, and must be complete in 8 hours. The timezone method would only change the display of the date and time. Only change the timezone setting if you know your visitor's timezone, and datetime's must be shown relative to them.
Now: 1234418228 is 2009/02/12 00:57:08 in Montreal or 2009/02/11 09:57:08 in San Francisco. It's the exact same moment.
Appending to the first answer, date() and strtotime() are your friends.
strtotime( "+8 hours", $now )
$now being a timestamp of when it's supposed to relate to. So if your start time isn't time(), you can still use that. eg
strtotime( "+8 hours", strtotime( "2009/03/01 00:00:00" ); (8AM on 2009/03/01)
However, when dealing with intervals counted in weeks, or less, i prefer doing it 'mathematically'
$StartTime = strtotime( "2009/03/01 13:00:00" );
$EndTime = $StartTime + ( 8 * 60 * 60 );
date( "Y/m/d H:i:s", $EndTime ) ==> "2009/03/01 21:00:00"
3600 seconds in an hour, 86400 in a day.
You can't use this method for months, quarters or years because the number of seconds they last varies from one to the next.
If you want to use time for a certain timezone, then using date_default_timezone_set() is preferred. anyway you can provide the date() function another parmater: int timestamp. an integer representing the timestamp you would like date() to return the information about.
so if you would like to show date('Y-m-d H:i:s') for now you can use this:
$now = date('Y-m-d H:i:s', time() ); // time() returns current timestamp.
// if you omit the second parameter of date(), it will use current timestamp
// by default.
$_8hoursLater = date('Y-m-d H:i:s', time()+60*60*8 );
$_8hoursBefore = date('Y-m-d H:i:s', time()-60*60*8 );

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