06-03-2014
Some users have services (there are several different costs) and for example Thomas have bought "D3" service for 6 months. After some time he want to improve it to "D4". So now we're calculating his service virtual value (we know when he bought it and when it's ending), we have that there are left 0,5462 value of his paid virtual money. For example at beginning he paid 100 Eur, so there is 54.62 Eur left. "D4" for one month costs 25 Eur. We get: 54,62 / 25 = 2,1848 months. So now we need to add this time 2,1848 months (in minutes accuracy) to current date/time, (but it must consider, that each month has different number of days) and then we will get exact date that "D4" service will expire. Best way to get how many minutes it need to add to current time to get that expiration date. It can be done with php, bash or smth taht would work in linux.
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LEARN ABOUT PHP
datetime.setdate
DATETIME.SETDATE(3) 1 DATETIME.SETDATE(3)
DateTime::setDate - Sets the date
Object oriented style
SYNOPSIS
public DateTime DateTime::setDate (int $year, int $month, int $day)
DESCRIPTION
Procedural style
DateTime date_date_set (DateTime $object, int $year, int $month, int $day)
Resets the current date of the DateTime object to a different date.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $year
- Year of the date.
o $month
- Month of the date.
o $day
- Day of the date.
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
CHANGELOG
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 5.3.0 | |
| | |
| | Changed the return value on success from NULL to |
| | DateTime. |
| | |
+--------+---------------------------------------------------+
EXAMPLES
Example #1
DateTime.setDate(3) example
Object oriented style
<?php
$date = new DateTime();
$date->setDate(2001, 2, 3);
echo $date->format('Y-m-d');
?>
Procedural style
<?php
$date = date_create();
date_date_set($date, 2001, 2, 3);
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2001-02-03
Example #2
Values exceeding ranges are added to their parent values
<?php
$date = new DateTime();
$date->setDate(2001, 2, 28);
echo $date->format('Y-m-d') . "
";
$date->setDate(2001, 2, 29);
echo $date->format('Y-m-d') . "
";
$date->setDate(2001, 14, 3);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-02-28
2001-03-01
2002-02-03
SEE ALSO
DateTime.setISODate(3), DateTime.setTime(3).
PHP Documentation Group DATETIME.SETDATE(3)