Hi,
I have a date input in MMDDYYYY format..
I have to give the day (whether that DD is sunday/monday...)
Is there any command for it...
Or do I have to write a script for that...
Thanks in Advance
Yeheya (1 Reply)
Hi,
I have a date field which is a variable field being passed to the script from outside.
I need to know how i can add 1 day to it.
I have seen example of date subtraction but while adding each time date reaches 31 or 30 i have to put if else condition.
Or for leap year also.
Kindly let... (4 Replies)
Hi, I have been trying just about every unix command to come up with yesterday's date (today's date - 1). I have seen all of the help on this forum, and none of it seems to work for me here. We are using Sun Solaris 9 Unix. I am using this script to create a .txt file with ftp commands that I will... (2 Replies)
i would need some help in setting the value of a variable (TIME_NOW) depending on the time of the day ...e.g.
if today's date is 12th April 2009 and if the current time is between midnight and 16:59:59 hrs then the TIME_NOW should be yesterday's date i.e. TIME_NOW=11
else if the current time... (3 Replies)
It's easy as pie to get the date minus one day on opensolaris:
date -d "-1 day" +"%Y%m%d"run this command on our crappy Solaris 10 machines however (which I'm guessing doesn't have GNU date running on it) and you get:
date: illegal option -- d
date: illegal option -- 1
date: illegal option --... (5 Replies)
Hi guys,
I had a scenario...
1. I had to get the previous days date in yyyymmdd format
2. i had to create a file with Date inthe format yyyymmdd.txt format
both are different
thanks guys in advance.. (4 Replies)
I need to get the next day's date of the user entered date
for example:
Enter date (yyyy/mm/yy):
2013/10/08I need to get the next day's date of the user entered date
Desired Output:
2013/10/09Though there are ways to achieve this is Linux or Unix environment (date command) ,I need to... (1 Reply)
I Have text like
XXX_20190908.csv.gz need to replace Only date in this format with current date every day
Thanks! (1 Reply)
Discussion started by: yamasani1991
1 Replies
LEARN ABOUT PHP
datetime.sub
DATETIME.SUB(3) 1 DATETIME.SUB(3)DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_sub (DateTime $object, DateInterval $interval)
Subtracts the specified DateInterval object from the specified DateTime object.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $interval
- A DateInterval object
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
EXAMPLES
Example #1
DateTime.sub(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-20');
date_sub($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-10
Example #2
Further DateTime.sub(3) examples
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-19 13:59:30
1992-08-15 19:56:58
Example #3
Beware when subtracting months
<?php
$date = new DateTime('2001-04-30');
$interval = new DateInterval('P1M');
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-03-30
2001-03-02
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.add(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.SUB(3)
04-13-2018
