I'm trying to apply the straight line deprecation for this, I've two date starting date 2018-09-17 (YYYY-MM-DD) ending date 2018-12-30. I need answer in months which is 4 not 3 because, it give me 3.
I've tried starting date 2018-09-01 and ending date 2019-01-01. It give me correct answer 4 what I want.
It give me answer 4
$d1 = new DateTime("2018-09-01");
$d2 = new DateTime("2019-01-01");
echo $d1->diff($d2)->m . " Months";
I need answer 4 for these two date
2018-17-09 to 2018-30-12
Is there any way to get the answer 4 in months including the starting date. I am getting the answer 4 from these date (2018-09-01) to (2019-01-01). In short, I want to include the current date month.
You can use Carbon:
use \Carbon\Carbon;
$from = Carbon::createFromFormat('Y-d-m', '2018-17-09');
$to = Carbon::createFromFormat('Y-d-m', '2018-30-12')->addMonth();
return $to->diffInMonths($from);
Live demo here
You can easily use format method and pass "n" to it this will get the month as a numeric value then you can define your logic.
An example:
$date1 = (new Datetime("2018-09-17"))->format("n");
$date2 = (new Datetime("2018-12-30"))->format("n");
echo $date2 - $date1 + 1;
Related
I am trying to compare to date to figure out how much time is between them, which I know how to do, date_diff(), but I want to then compare the time between the dates and if it is greater than 7 days do something and if not do something else. I think it sounds easy and I know there are probably fairly simple solutions to do so but I am just not a fan of dates and comparisons. Here is a snippet of what I got so far as it is just one case of a switch statement so the rest are basically identical.
$array = array();
$today = date("Y-m-d"); // get today's date
foreach($arrayOfObjs as $obj){
if ($obj->get("renewalDate") >= $today){
array_push($array, $obj->get("renewalDate"));
}else{
switch($obj->get("recurrencePeriod")){
case 1:
/*
* All cases follow same structure
* Build the date in format Y-m-d from renewalDate out of the obj.
* Loop through the date while it's less than today.
* After date is greater than today return date add to array
*/
$date = DateTime::createFromFormat("Y-m-d", $obj->get('renewalDate'));
while($date <= $today){
$date->add(new DateInterval('P7D'));
}
$diff = date_diff($today, $date);
if($diff->format('%a') <= 7){
$obj->renewalDate($date);
array_push($array, $obj);
}
break;
Basically, my database stores dates and those dates could be passed but it could be a reoccurring event. To calculate the next time that event would happen I check if the data in the database is before today's date and if it is then I continue to add the incremental amount (in this case 7 for a weekly reoccurring event) and compare the dates again. After the date that is incremented passes today's date I want to find out if it is within 7 days and if so add it to an array to get returned. I know... since I'm adding 7and it's within 7 days the first reoccurring event will always be within 7 days but that is not the case for monthly events or anything greater.
All cases are broken so I only included this one for simplicity. I can get date_Diff to return something like 7 or 12 or whatever the number may be but how can I check if that number is within the 7 days I want?
Thanks, I will include more information if needed to clarify any misunderstandings.
I'm not entirely sure what you are trying to do, but how about the following if you are just projecting dates forward and backwards and want to know if they are 7 days or more either way:
$today = date("Y-m-d");
$todaytime = strtotime($today);
$testdate = "2017-06-31";
$testtime = strtotime($testdate);
$back7days = strtotime("-7 days",$todaytime);
if($testtime < $back7days)
echo "X"; // do somthing if testdate was more than 7 days ago
$fwd7days = strtotime("+7 days", $todaytime);
if($testtime > $fwd7days)
echo "Y"; // do somthing if testdate is more than 7 days in future
Just make sure that you use less-than or less-than-and-equals comparators etc to handle the boundary conditions you need.
Suppose I have a date specific as 2016-11-14 .
Also I have another date (starting) 02-14-2017 and ending 02-30-2017
I want to count the months that has passed is the date 2016-11-14 until the set of Range date.
I tried searching but I cant find a link on how to find months that has passed from a date to set of range date
DateTime() makes doing date math easy:
$datetime1 = new DateTime('2016-12-01');
$datetime2 = new DateTime('2017-02-01');
$interval = $datetime2->diff($datetime1);
echo (($interval->format('%y') * 12) + $interval->format('%m'));
Demo
$date_raw = '05/05/1995';
$newDate = (date('j F Y', strtotime('-192years -14months -2days', strtotime($date_raw))));
print "New Date: $newDate <br>";
I'm trying to subtract 100+ years from a given date. but the value i get is real till 92 years only. after that i don't get correct subtraction. whats the reason?
If you need to work with dates that fall outside the range of a 32-bit signed unix timestamp (1901-12-13 to 2038-01-19), then start using DateTime objects
$date_raw = '05/05/1995';
$newDate = (new DateTime($date_raw))
->sub(new DateInterval('P192Y14M2D'))
->format('j F Y');
echo $newDate, PHP_EOL;
gives
3 March 1802
or you can do this in the following way
// set your date here
$mydate = "2018-06-27";
/* strtotime accepts two parameters.
The first parameter tells what it should compute.
The second parameter defines what source date it should use. */
$lastHundredyear = strtotime("-100 year", strtotime($mydate));
// format and display the computed date
echo date("Y-m-d", $lastHundredyear);
this will give you following output
1918-06-27
I have a PHP date in a database, for example 8th August 2011. I have this date in a strtotime() format so I can display it as I please.
I need to adjust this date to make it 8th August 2013 (current year). What is the best way of doing this? So far, I've been racking my brains but to no avail.
Some of the answers you have so far have missed the point that you want to update any given date to the current year and have concentrated on turning 2011 into 2013, excluding the accepted answer. However, I feel that examples using the DateTime classes are always of use in these cases.
The accepted answer will result in a Notice:-
Notice: A non well formed numeric value encountered......
if your supplied date is the 29th February on a Leapyear, although it should still give the correct result.
Here is a generic function that will take any valid date and return the same date in the current year:-
/**
* #param String $dateString
* #return DateTime
*/
function updateDate($dateString){
$suppliedDate = new \DateTime($dateString);
$currentYear = (int)(new \DateTime())->format('Y');
return (new \DateTime())->setDate($currentYear, (int)$suppliedDate->format('m'), (int)$suppliedDate->format('d'));
}
For example:-
var_dump(updateDate('8th August 2011'));
See it working here and see the PHP manual for more information on the DateTime classes.
You don't say how you want to use the updated date, but DateTime is flexible enough to allow you to do with it as you wish. I would draw your attention to the DateTime::format() method as being particularly useful.
strtotime( date( 'd M ', $originaleDate ) . date( 'Y' ) );
This takes the day and month of the original time, adds the current year, and converts it to the new date.
You can also add the amount of seconds you want to add to the original timestamp. For 2 years this would be 63 113 852 seconds.
You could retrieve the timestamp of the same date two years later with strtotime() first parameter and then convert it in the format you want to display.
<?php
$date = "11/08/2011";
$time = strtotime($date);
$time_future = strtotime("+2 years", $time);
$future = date("d/m/Y", $time_future);
echo "NEW DATE : " . $future;
?>
You can for instance output it like this:
date('2013-m-d', strtotime($myTime))
Just like that... or use
$year = date('Y');
$myMonthDay = date('m-d', strtotime($myTime));
echo $year . '-' . $myMonthDay;
Use the date modify function Like this
$date = new DateTime('2011-08-08');
$date->modify('+2 years');
echo $date->format('Y-m-d') . "\n";
//will give "2013-08-08"
I was looking at this post, and it is close to what I need:
PHP - How to count 60 days from the add date
However, in that post, the calculation is performed by adding 60 days to the current date. What I need to do is calculate the date based on a variable date (and not the current date).
Something like this:
$my_date = $some_row_from_a_database;
$date_plus_10_days = ???;
Anyone know how to do that?
Thanks
You can put something before the "+10 days" part:
strtotime("2010-01-01 +10 days");
Use date_add
http://www.php.net/manual/en/datetime.add.php
$my_date = new DateTime($some_row_from_a_database);
$date_plus_10_days = date_add($my_date, new DateInterval('P10D'));
You will have to look into strtotime(). I'd imagine your final code would look something like this:
$dateVariable = strtotime('2017-01-29');//your date variable goes here
$date_plus_60_days = date('Y-m-d', strtotime('+ 60 days', $dateVariable));
echo $date_plus_60_days;
If you are using PHP >= 5.2 I strongly suggest you use the new DateTime object. For example like below:
$date_plus_60_days = new DateTime("2006-12-12");
$date_plus_60_days->modify("+60 days");
echo $date_plus_60_days->format("Y-m-d");
I see you are retriving data from a database.
If you are using mysql you can do it on the select:
Example: you need the last date of the table and this date-7 days
select max(datefield) as ultimaf, DATE_SUB(max(datefield),INTERVAL 7 DAY) as last7
from table
It´s easy use curdate() if you want todays date.
If you need a dynamic between that selects the count of last 7 days:
select count(*) from table
where DATE_SUB(CURDATE(),INTERVAL 7 DAY)<=datefield"
date('Y-m-d H:i:s', strtotime("2014-11-24 06:33:39" +35 days"))
this will get the calculated date in defined format.
Suppose today's date is
date_default_timezone_set('Asia/Calcutta');
$today=date("Y-m-d");
And i can add 10 days in current date as follows :
$date_afte_10_days = date('Y-m-d', strtotime("$today +10 days"));