I have a date value stored in a variable. I need to extract the time part of the value in to a separate variable and then add/subtract time from it.
The date variable is set with date('YmdHis'), giving (for example) 20110805124000 for August 5th 2011, 12:40:00
From the value 20110805124000 (which is stored in the variable $fulltime), I need to store the time only in the format 12:40 (so ignoring the year, month, day and seconds and adding the colon between the hour and minute) in a variable called $shorttime. I then need to add a number of hours to that time (so for example +3 hours would change the value in the $shorttime variable to 15:40). The number of hours I need to add is stored in a variable called $addtime, and this value could be a negative number.
Is this easily doable? Could anyone help?
Thanks :)
$time = '2013-01-22 10:45:45';
echo $time = date("H:i:s",strtotime($time));
It will give the time 10:45:45 from datetime.
<?PHP
$addhours = 3;
$date = DateTime::createFromFormat('YmdHis', '20110805124000');
$shorttime = $date->format("H:i");
$newdate = $date->add(DateInterval::createFromDateString($addhours . "hours"));
$newtime = $newdate->format("H:i");
echo $shorttime . "<br />";
echo $newtime . "<br />";
?>
for your reference:
http://www.php.net/manual/en/datetime.createfromformat.php
http://www.php.net/manual/en/dateinterval.createfromdatestring.php
hello the way I usually do it is like this, maybe it's a bit long but it works for me...
$DateIn = new DateTime($In);
$DateOut = new DateTime($Out);
$HourOut = new DateTime($H_Out);
$FechaEntrada = $DateIn->format('d-m-Y');
$FechaSalida = $DateOut->format('d-m-Y');
$HoraSalida = $HourOut->format('H:i a');
the guide I used was the PHP one DateTime::format()
Related
I have been trying to figure this out for a week now. My wife has started a new taxi-company and she asked me to code a simple webpage for here where she could press a button to save a timestamp, then the press is again when she gets off work, it then creates a second timestamp
I have an MYSQL database with rows containing the start time and stop time. I have managed to use the diff function to see how much time it is between the two timestamps but now comes the tricky part.
Since it's different payments at different times of the day I need to divide the time at a shortened time.
Up to 19:00 she works "daytime" and after that, she works "nighttime" until 06:00 the other day, then there is "weekend daytime" and "weekend nighttime" as well.
So if she creates a timestamp whit the date and time: 2018-08-08 06:30 and then another timestamp at 2018-08-08 21:00, then I need a script that puts these data in ex "$daytimehours = 12" "$daytimeminutes = 30" and "$nighttimehours = 3" "$nighttimeminutes = 0"
I have managed to create a script that almost works, but it is several pages long, and it contains one if-statement for each different scenario daytime-nighttime, nighttime-daytime etc.
So do anyone has a good idea on how to solve this? or maybe just point me in the right direction. I would be happy to pay some money to get this to work.
My solution is
<?php
date_default_timezone_set('Asia/Almaty');
$endDate = '2018-08-08 21:00';
$startDate = '2018-08-08 06:30';
$nightmare = date('Y-m-d 19:00');
$startDay = date('Y-m-d 06:00');
$diffMorning = strtotime($nightmare) - strtotime($startDate);
$diffNight = strtotime($endDate) - strtotime($nightmare);
echo gmdate('H:i', $diffMorning) . "\n"; // this is the difference from start day till 19:00
echo gmdate('H:i', $diffNight); // this is the difference on nightmare
$total = $diffMorning + $diffNight;
echo intval($total/3600) . " hours \n";
echo $total%3600/60 . " minutes \n";
echo $total%3600%60 . ' seconds';
You can check via online compiler
given two dates stated as:
$endDate = '2018-08-08 21:00';
$startDate = '2018-08-08 06:30';
you can use the PHP Date extension to achieve the difference like this:
$start = date_create($startDate);
$end = date_create($endDate);
$boundnight = clone($end);
$boundnight->setTime(19,00);
$total_duration = date_diff($end,$start);//total duration from start to end
$day_duration = date_diff($boundnight,$start);//daytime duration
$night_duration = date_diff($end,$boundnight);// nighttime duration
you can use the format method to print a human readable string this way:
$total_duration=$total_duration->format('%H:%I');
$day_duration=$day_duration->format('%H:%I');
$night_duration=$night_duration->format('%H:%I');
At this step there is nothing left but you say you want to convert each duration in minutes.So let's build a function :
function toMinute($duration){
return (count($x=explode(':',$duration))==2?($x[0]*60+$x[1]):false);
}
Then you can use it this way:
$total_duration = toMinute($total_duration);
$day_duration = toMinute($day_duration);
$night_duration = toMinute($night_duration);
The output of
var_dump($total_duration,$day_duration,$night_duration) at this step is:
int(870)
int(750)
int(120)
I ran time() at 6:38:47 and it returned a different value than strtotime of the same time. Why is this?
It should absolutely return the same value. Run this code:
<?php
$time = time();
$string = date('Y-m-d H:i:s', $time);
$strtotime = strtotime($string);
print "time = $time\n";
print "string = $string\n";
print "strtotime = $strtotime\n";
print "difference = ".($time-$strtotime)."\n";
?>
My output right now:
time = 1377686839
string = 2013-08-28 12:47:19
strtotime = 1377686839
difference = 0
Are you getting a difference with this? You could also post your test code, maybe there's a mistake in there.
strtotime interprets the time string argument according to the system timezone; time is timezone-independent because it just returns the number of seconds since the start of the Unix epoch.
If your system timezone is anything other than UTC you should expect the values to differ, since the time string argument you passed to strtotime was hardcoded. Your notion of "current wall clock time" and that of the system are different, hence the difference in the timestamps.
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
I have this code:
$x = date("Y-m-d", strtotime($post['commissionEligibilityDate'] . "+ " . $post['billingPeriodExpiration'] . " months"))
$post['commissionEligibilityDate'] = 2011-11-08 <br/>
$post['billingPeriodExpiration'] = 2 <br/>
so $x returns 2012-01-08.
I have another variable $singleDate and it's equal to 1. What I am trying to do replace the 08 with 01. How can I do that?
You don't have to use Y-m-d, you can use Y-m-01 or your variable:
$x = date("Y-m-".$singleDate, strtotime($post['commissionEligibilityDate'] . "+ " . $post['billingPeriodExpiration'] . " months"))
You could use the DateTime class:
$d = new DateTime($x);
$year = $d->format('Y');
$month = $d->format('m');
$d->setDate($year, $month, '01');
echo $d->format('Y-m-d');
str_replace('08','01',$post['commissionEligibilityDate']);
$explode = explode("-",$post['commissionEligibilityDate']);
$explode[2] = $singleDate;
$post['commissionEligibilityDate'] = implode("-",$explode);
$post['commissionEligibilityDate'] will now echo 2011-11-1
Almost anything is possible with PHP. My suggestion seeing as your looking to only get one number from a date that you consider checking out http://php.net/manual/en/function.date.php to see the various ways of handling the date() output. As you could easily output a day then add to it $z = date('d', time())+1; for example.
I am not sure what your doing with your dates specifically but to me it sounds like you might have a misconception of what they are, and how to work with them. Basically the short idea of it is, a date defined in a variable is a string. You can make them anyway you want even without the use of date() then store them, as long as they are in the right format when you go to store them ie yyyy-mm-dd you should be fine.
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 '.