Thanks for the reply. I have tried the below code, it's working fine for most of the dates but with few discrepancies.
The date passed as input parameter will always be the end of the month date and I need the output to be 2 months prior to the input date but want the O/P to be the end of the month date.
The above code is giving me the following output :
Hi i am trying to subtract days from current date. For example todays date is 10/03/2006. If i subtract 2 days it should give 8/03/2006. I am also trying to find the access date of a file in dd/mm/yyyy format. Can any one please help in how to do this.
Ramesh (1 Reply)
I have looked through the forums and found many date / time manipulation tools, but cannot seem to find something that fits my needs for the following.
I have a log file with date time stamps like this:
Jun 21 17:21:52
Jun 21 17:24:56
Jun 21 17:27:59
Jun 21 17:31:03
Jun 21 17:34:07
Jun... (0 Replies)
how can we add or subtract days from the output of date command in unix...
like if i want to subtract a day from the result of date command like this..
v_date=`date +%Y%m%d`
this wud give me 20080519
now i want to subtract one day from this.. so tht it wud give me 20080518..
how do i do... (1 Reply)
Hi All,
I have a requirement in my project where a batch runs on any day of a week. The input files will land to the server every Sunday. I need to read these files based on the current BUS_DAY (which falls any day of the week).
For e.g : if the BUS_DAY is 20120308, we need to derive the... (3 Replies)
Hi Gurus!
I have a static date in a YYYYMMDD format; and I want get the date 2 years in the past and 2 years in the future.
static_date=20010203
old_date=$static_date - 3 years
future_date=$static_date + 2 years
I was only able to research on dates that are current and not on static... (3 Replies)
I got a statement like below to subtract 1 from given date using teradata. I am looking for a one line unix command to perform the same.
select 'parse_this_record', (DATE '${FILE_DATE}' - 1) (FORMAT 'YYYY-MM-DD');
Input: 2012-02-21
Expected Output: 2012-02-20
PS: One liner because I am... (2 Replies)
Hi,
Is there a way to subtract a month or at least 30 days from an arbitrary or user inputted date without the GNU date?
Example:
please Input date: 2011-05-11
then the answer will be 2011-04-11
This code doesn't work:
$date -d "2011-05-11 - 1 month" "+%Y%m%d"
Thanks (7 Replies)
I am trying to achieve to get only the month and the day. Example Feb 5 (as you can see if it is feb 1-9) the space is 2. If it is feb 10-28, the space is only 1. I am trying to right a script that will list a directory and shoot an email if there is an activity in last 7 days. I dont really trust... (5 Replies)
Hi All,
I have a CSV file which is as below. Basically I need to take the year column in it and find if the year is >= 20152 . If that is then I should subtract all values by 6. In the below example in description I am having number mentioned as YYWW so I need to subtract those by -5. Whereever... (8 Replies)
Discussion started by: arunkumar_mca
8 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)