hello !
can so help me ? i'm writting a shell srcipt ( UNIX AIX) and I'd like to convert a date as 20041108 in lilian format.
thanks so much (2 Replies)
Hi ,
I have a csv file and column 1 is date in the cursed YYYY-M-D format. Is there any way to convert 1st column to an absolute date say using 1977-1-1 as day 1?
Thanks a lot. (2 Replies)
Hi Guys,
I was working some time ago n was in need to calculate date 30/31 days from today including Feb (Leap yr stuff). Today date is variable depending on day of execution of script. I tried searching but was not able to get exactly what I needed....So at that I time I implemented by my own... (3 Replies)
Hi,
I'm writing an batch file to create report
In the batch file iam passing two arguments:startdate and finishdate
Ex: startdate=07-sep-2009 finishdate=07-sep-2011
I need to have script that takes command line argument as input and gives me out currentdate last year and current date... (2 Replies)
Hi i am writing a cron job.
so for it i need the 60 days old date form current date in variable.
Like today date is 27 jan 2011 then output value will be stote in variable in formet Nov 27.
i am using EST date, and tried lot of solution and see lot of post but it did not helpful for me. so... (3 Replies)
Hi,
I want to subtract 'n' days from the current timestamp in a k shell script. Is there any inbuilt function to do it
or any workaround solution to get the date. And I want the output to be in YYYY:MM:DD HH:MM:SS format. Please help.
Thanks in advance. (4 Replies)
hi friends,
I m trying to write a script which compares to dates.
for this i am converting dates into no using synatx
as below
v2=`date | awk '{print $2,$3,$4}'`
v3=`date +%s -d "$v2"`
this syntax is working in bash shell ,but fails in ksh shell.
please suggest on this. (12 Replies)
Hi,
I have a requirement where I am getting date in string format (20161130). I need to add 20 days(not no. 20) to the above string. The o/p should 20161220.
In case of 20170228, it should show 20170320.
Could you please help me with the command to achieve this.
Note: I am using AIX 7.1... (5 Replies)
Discussion started by: satyaatcgi
5 Replies
LEARN ABOUT PHP
datetime.add
DATETIME.ADD(3) 1 DATETIME.ADD(3)DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::add (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_add (DateTime $object, DateInterval $interval)
Adds the specified DateInterval object to 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.add(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-11
Example #2
Further DateTime.add(3) examples
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-01 10:00:30
2007-06-05 04:03:02
Example #3
Beware when adding months
<?php
$date = new DateTime('2000-12-31');
$interval = new DateInterval('P1M');
$date->add($interval);
echo $date->format('Y-m-d') . "
";
$date->add($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-01-31
2001-03-03
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.sub(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.ADD(3)