php - Year Changed To The Current Year After Date Being Substituted - php

I am trying to set publish date based on user choice and give it interval in a loop. But after it is substituted with the intervals, the year changed to the current year.
Here is the sample of my code:
$datestart = "2012-03-06";
$datenow = date("$datestart H:i:s", current_time( 'timestamp' ));
$newpostdate1 = $datenow + strtotime("0 years 0 months 1 days 0 hours 0 minutes 0 seconds");
$newpostdate = date("Y-m-d H:i:s", $newpostdate1);
echo $datenow . " " . $newpostdate;
$datenow Will return 2012-03-06 16:19:33 while $newpostdate return the current date plus 1 day i.e: 2014-03-15 17:02:23.
Why $newpostdate returning the current date plus next 1 day instead of 2012-04-06 16:19:33 ?

..because what you're doing doesn't do what you think it does.
First, you set $datenow to a string (not a date object), with value "2012-03-06 " + the current time (assuming that's what current_time returns).
Then you call strtotime with the value "1 days" (well, your string has a bunch of other zero-valued fields, but they don't change the result), which returns the current time + 24 hours as a number (the number of seconds since 1970).
Then you take that value and add it with + to the above string. This causes the string to be interpreted as a number, so it turns into 2012 (and the rest of the string is ignored). So the result is a timestamp representing the current time + one day + 2,012 seconds - or one day, 33 minutes and 32 seconds from the time the code is run. Which you then format as a string.
You could use John Conde's solution to get a more meaningful result. (I assume your real problem is different, else why not just start out by setting the string to '2012-03-07' in the first place?)

The first parameter of date() is the format you want the timestamp passed as the second parameter to be displayed as. So basically you are using date() incorrectly.
I think this is what you are looking for:
$date = new DateTime($datestart);
$date->modify('+1 day');
echo $date->format(Y-m-d H:i:s);

Related

Add 30 seconds interval in a given datetime

Need some help.
I am using PHP.
So I have coordinates data.
Specifically Longtitude and Latitude.
So let's say I have 15 data of Long and Lat to be inserted on a table.
However, the api gives me only a single datetime because these coordinates are in array.
For example:
[14.4364372;121.0125753, 14.4364375;121.0125755, 14.4364377;121.0125758, 14.436436;121.012574, 14.4364342;121.0125721, 14.4364326;121.0125704, 14.436433;121.0125707, 14.4364334;121.0125711, 14.4364338;121.0125716, 14.4364342;121.012572, 14.4364345;121.0125724, 14.4364348;121.0125728, 14.4364351;121.0125731, 14.4364353;121.0125733, 14.4364356;121.0125735]
So first you will explode it to get rid of the delimeter ','
And loop it to count how many are data,then explode again the delimiter ';' to count the long and lat given.
But it only have a single datetime.
What i want here is to insert these data including datetime but with interval of 30 seconds per insert.
How can i do that?
Expected output would be like this:
INSERT INTO table_gps(ticket,datetime,long,lat) VALUES(0,09/16/2016 03:30:26 pm, 14.4364363,121.0125745)
INSERT INTO table_gps(ticket,datetime,long,lat) VALUES(0,09/16/2016 03:30:56 pm, 14.4364364,121.0125746)
Here is my code:
$msg= 'YC GPS2~09/16/2016 13:29:46~-~[14.4364362;121.0125744, 14.4364363;121.0125745]';
if(strpos($msg,'YC GPS2')!==false){
//explode data
$explode=explode('~',$msg);
$ky=$explode[0];
$datetime=$explode[1];
$ticket=$explode[2];
$coords=$explode[3];
$coords=explode(',',$coords);
for($i=0;$i<count($coords);$i++){
$xpVal=$coords[$i];
if($xpVal){
$xp3=explode(';',$xpVal);
$lng=$xp3[0];
$lat=$xp3[1];
$lng=str_replace('[','',$lng);
$lat=str_replace(']','',$lat);
$time = date("m/d/Y h:i:s a", time() + 30);// where 30 is the seconds
echo "INSERT INTO GPS2(ticket,datetime,long,lat) VALUES(".$ticket.",".$time.",".$lng.",".$lat.") ";
}
}
}else{
echo '0';
}
Thanks.
In order to add how many seconds you want to particular date in PHP you can use the following.
$time = date("m/d/Y h:i:s a", time() + 30);// where 30 is the seconds
The following is really easy way to add days, minutes, hours and seconds to a time using PHP. Using the date function to set the format of the date to be returned then using strtotime to add the increase or decrease of time then after a comma use another strtotime passing in the start date and time.
//set timezone
date_default_timezone_set('GMT');
//set an date and time to work with
$start = '2014-06-01 14:00:00';
//display the converted time
echo date('Y-m-d H:i',strtotime('+1 hour +20 minutes',strtotime($start)));
Times can be entered in a readable way:
+1 day = adds 1 day
+1 hour = adds 1 hour
+10 minutes = adds 10 minutes
+10 seconds = adds 10 seconds

PHP Time Since Function Bug

i am writing a time since function to return the time since a given mysql datetime. When taking the $oldtime from current time() it is returning a negative int when i need a positive int. I have written similar functions before in other languages but i have become blind to this problem, so any help would be much appreciated.
function timeSince($time){
$today = date("Y");
$oldtime = strtotime($time);
$time = time() - $oldtime;
$tokens = array (
3600 => 'h',
60 => 'm',
1 => 's'
);
if($time >= 86400){
}
}
echo timeSince('2016-02-25 14:35:00');
it could be much more convenient if you use PHP's DateTime and DateInterval classes and their methods:
function timeSince($datetime) {
$now = strtotime("now");
$then = strtotime($datetime);
$dt_now = new DateTime("#" . $now);
$dt_then = new DateTime("#" . $then);
//DateTime's diff method returns a DateInterval object which got a format method:
return $dt_now->diff($dt_then)->format('%a days, %h hours, %i minutes and %s seconds');
}
some test cases:
//my local date & time is around "2016-02-25 19:49:00" when testing
echo '<pre>';
echo timeSince('2016-02-25 19:30:00');
//0 days, 0 hours, 19 minutes and 11 seconds
echo PHP_EOL;
echo timeSince('2013-11-02 15:43:12');
//845 days, 4 hours, 4 minutes and 3 seconds
echo PHP_EOL;
echo timeSince('2017-01-31 00:22:45');
//340 days, 4 hours, 35 minutes and 30 seconds
echo PHP_EOL;
echo timeSince('1950-05-14 07:10:05');
//24028 days, 12 hours, 37 minutes and 10 seconds
echo PHP_EOL;
code partially based on this answer: https://stackoverflow.com/a/19680778/3391783
strtotime uses timezone in your PHP settings. Depending on timezone set, it might convert to the time that is yet to happen. For example, on my ukrainian server, strtotime('2016-02-25 14:35:00') converts to 1456403700, on a server in another timezone (US/Pacific) it converts to 1456439700.
Quote from PHP docs:
The function expects to be given a string containing an English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.
Each parameter of this function uses the default time zone unless a time zone is specified in that parameter. Be careful not to use different time zones in each parameter unless that is intended. See date_default_timezone_get() on the various ways to define the default time zone.
You can add UTC/GMT offset to your datetime (1st param), for example strtotime('2016-02-25 14:35:00 +0800') or ('2016-02-25 14:35:00 GMT+08:00') will convert to 1456382100
In your example, $oldtime must be smaller value than current time().
So, if you want to count time between larger value, simply reverse your equation:
This line:
$time = time() - $oldtime;
Becomes:
$time = $oldtime - time();

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

PHP Adding 15 minutes to Time value

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));

How can I get how old a user is/ how long it has been by comparing dates with PHP/javascript?

I'm looking for a way to compare dates and find the difference with PHP or jQuery (JavaScript). For instance if a user's birthday is 1/16/95 or January 16, 1995 and today is December 24, 2011 how could I get "x Years and x Days"? Or if some one were to save a file and the date of creation is 1/16/95 or January 16, 1995 and 25 seconds passed since creation how could I get that time?
Take a look at DateTime::diff:
PHP Manual
Your code could look like this:
$datetime1 = new DateTime('1994-12-16');
$today = new DateTime();
$interval = $datetime1->diff($datetime2);
echo $interval->format('%y years %d days');
If I where you I would use the PHP time() function.
This value is always unique and is always interpreted the right way.
You can always use strtotime php function to convert a string to a timestamp.
Its easy calculating with those timestamps because difference A-B gives you time difference in seconds.
Hope it helps
You just need to convert the date to UNIX time (seconds since Jan 1, 1970), then subtract from "now". That will give you the difference in seconds, which you can then convert to a time string.
PHP has, as expected, some built-in stuff to do it, so it's very simple:
$birth = new DateTime('12/16/94');
$age = $birth->diff(new DateTime("now"));
echo $age->format('%y years and %d days');
(see valid time strings)
In Javascript, you need to do it yourself (but I did it for you):
function timeSince(_date){
// offset from current time
var s = Math.floor( (+new Date - Date.parse(_date)) / 1000 )
, day = 60*60*24
, year = day*365;
var years = Math.floor(s/year)
, days = Math.floor(s/day) - (years*365)
// handle plurals
, ys = ' year' + (years>1 ? 's' : '')
, ds = ' day' + (days>1 ? 's' : '');
return years + ys + ' and ' + days + ds;
};
timeSince(new Date('06/25/1985')); // => '25 years and 278 days'
This is getting the difference in seconds from the date to now, then dividing the value by an year/day's length in seconds. The allowed time strings are much more limited in js though.

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