$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
Related
I have a string: 30/06/18 (30th June 2018)
I am converting to a date:
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18')->format('d-m-Y');
echo $calcFieldDate;
Result: 18-06-2018
Now I want to add 20 days to the date:
$expiryDate = date("d-m-Y", strtotime("+20 days", $calcFieldDate));
echo $expiryDate;
Expected Result: 08-07-2018
Actual Result: 31-01-1970
I am obviously creating a date format which is then subsequently being treated as a string...
Every time I try a conversion, I just hit another road block - is there anyway to create a date that is then treated like a date?
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18')->format('d-m-Y');
echo $calcFieldDate;
Result:30-06-2018
$expiryDate = date("d-m-Y", strtotime("+20 days", strtotime($calcFieldDate)));
echo $expiryDate;
Result:20-07-2018
Strtotime() The second parameter is the timestamp
You actually don't need to revert using strtotime and date functions, you can actually use DateTime to simply add dates into it:
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18');
echo $calcFieldDate->format('d-m-Y'); // get inputted date
$expiryDate = clone $calcFieldDate; // clone the original date object
$expiryDate->modify('+20 days'); // adjust the cloned date
echo $expiryDate->format('d-m-Y'); // show the adjusted date
This will sort your problem.
$str="30/06/18 (30th June 2018)";
$arr_temp=explode(" ",$str);
$str_date=str_replace("/","-",$arr_temp[0]);
$dt = DateTime::createFromFormat('d-m-y',$str_date);
$date=$dt->format('d-m-Y');
$new_date=date('d-m-Y',strtotime("+20 days",strtotime($date)));
echo $new_date;
This question already has answers here:
How to get time difference in minutes in PHP
(21 answers)
Closed 5 years ago.
I have this php code
$today_date = date('d/m/Y H:i:s');
$Expierdate = '09/06/2017 21:45:03';
$remaindate = date_diff($today_date,$Expierdate);
echo $remaindate;
and i need result from difference between two date.
date_diff() needs a DateTimeInterface as an argument. In other words, you need to create a DateTime object first, using new DateTime() as shown below.
$today_date = new DateTime();
$Expierdate = new DateTime('09/06/2017 21:45:03');
$remaindate = $today_date->diff($Expierdate);
echo $remaindate->format('%a days');
Live demo
The above would output
90 days
Because today is June 8th, and the format 09/06/2017 is September 6th - because you're using American format (MM/DD/YYYY).
If you ment June 9th (tomorrow), you need to use European format (MM-DD-YYYY, note the dash instead of slash). You can alternatively use DateTime::createFromFormat() to create from a set format, so your current format, 09/06/2017, is interpreted as June 9th. The code would then be
$today_date = new DateTime();
$Expierdate = DateTime::createFromFormat('d/m/Y H:i:s', '09/06/2017 21:45:03');
$remaindate = $today_date->diff($Expierdate);
echo $remaindate->format('%a days');
Output (live demo)
1 days
In any case, $remaindate holds some properties which can be used (see the manual), or you can format it to your liking by supplying the desired formation into the format() method.
new DateTime()
DateTime::diff()
DateTime::format()
DateTime::create_from_format()
I have a MySQL DB table with a column named "timestamp", a type of timestamp, and attribute of on update CURRENT_TIMESTAMP and a default of CURRENT_TIMESTAMP.
If I add a record to the table, specifying other values, but not the timestamp, then the timestamp is automatically added like 2016-12-28 17:02:26.
In PHP I query the table using the following query
SELECT * FROM history WHERE user_id = 9 ORDER BY timestamp ASC
The result of the query is saved into $rows and I use a foreach to create an array with some of the other values formatted. I am attempting to format the time stamp to UK type 24-hour date time: dd/mm/yy, HH:MM:SS.
I have tried both the date() and strftime() functions as follows:
$formatted_datetime = strftime("%d %m %y %H %M %S", $row["timestamp"]);
$formatted_datetime = date("d/m/y, H:i:s", $row["timestamp"]);
Both of these result in the following notice and the date time being output incorrectly like 01 01 70 00 33 36:
Notice: A non well formed numeric value encountered in /home/ubuntu/workspace/pset7/public/history.php on line 20
I am new to PHP and MySQL and so far none of the other questions or documentation I have seen have successfully addressed performing this conversion.I do not understand why strftime() does not work, nor how to do this properly?
To do this the OO (and most flexible) way use DateTime class and use the static createFromFormat method to instantiate a new DateTime object:
$new_datetime = DateTime::createFromFormat ( "Y-m-d H:i:s", $row["timestamp"] );
Now you can use the $new_datetime object to generate any string representation you'd like by calling the object's format method:
echo $new_datetime->format('d/m/y, H:i:s');
To boot, you since you've a DateTime object you can now also to any manner of transformation (like shifting timezones or adding days), comparison (greater or less than another DateTime), and various time calculations (how many days/months/etc... between this and another DateTime).
DateTime:http://php.net/manual/en/class.datetime.php
Learn it. Love it. Live it.
Normally in MySQL date timestamp save time as YYYY-MM-DD HH:MM:SS (2016-12-20 18:36:14) formate you can easily convert them as your wish using date formate but have to convert your input to time first. Following will do the trick
$formatted_datetime = date("d/m/y, H:i:s", strtotime($row["timestamp"]));
Why not make MySQL do the work? And do you really want mm/dd/yy, not dd/mm/yy?
SELECT *, DATE_FORMAT(timestamp, '%m/%d/%y, %T') formatted_date FROM ....
Then you can extract the formatted date as $row['formatted_date'].
See https://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_date-format
The date is of timestamp type which has the following format: ‘YYYY-MM-DD HH:MM:SS’ or ‘2008-10-05 21:34:02.’
$res = mysql_query("SELECT date FROM times;");
while ( $row = mysql_fetch_array($res) ) {
echo $row['date'] . "
";
}
The PHP strtotime function parses the MySQL timestamp into a Unix timestamp which can be utilized for further parsing or formatting in the PHP date function.
Here are some other sample date output formats that may be of practical use:
echo date("F j, Y g:i a", strtotime($row["date"])); // October 5, 2008 9:34 pm
echo date("m.d.y", strtotime($row["date"])); // 10.05.08
echo date("j, n, Y", strtotime($row["date"])); // 5, 10, 2008
echo date("Ymd", strtotime($row["date"])); // 20081005
echo date('\i\t \i\s \t\h\e jS \d\a\y.', strtotime($row["date"])); // It is the 5th day.
echo date("D M j G:i:s T Y", strtotime($row["date"])); // Sun Oct 5 21:34:02 PST 2008
I would use the Carbon class and its String Formatting
$carbon = Carbon::instance('2016-12-28 17:02:26');
Then you can format it the way you want.
What seemed to be a fairly standard date conversion is giving me unusual results. I want to get $pay_date into the 'j/d/Y' format, but can only get the correct date if the format is 'Y-m-d'.
Here's my code:
$payment = '2014-09-09';
$pay_date = strtotime("+1 month", strtotime($payment) );
$converter = date('j/d/Y', $pay_date );
echo $converter;
result: 9/09/2014 (should be 10/09/2014)
If I keep these variables, but change $converter to this format:
$converter = date('Y-m-d', $pay_date );
the result is correct - 2014-10-09
I also tried this:
$convert2 = DateTime::createFromFormat('Y-m-d', $pay_date)->format('j/d/Y');
echo $convert2;
result: 9/09/2014
but:
$convert2 = DateTime::createFromFormat('Y-m-d', $pay_date)->format('Y-m-d');
echo $convert2;
gives me the correct result: 2014-10-09
j and d are the same formatting options, they both represent DAYs; from the manual for PHP's date function:
d Day of the month, 2 digits with leading zeros 01 to 31
j Day of the month without leading zeros 1 to 31
Use the format 'm/d/Y'.
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"