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.
Code:
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 dates wherein I can have it set and just changing it when I want.
Please help...
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,
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)
$beginDate = substr(DateCalc("today", "-7Days"),0,8);
This fetches the date 7 days back
Can I fetch the date before 7 years from todays date in Perl using same syntax
Use code tags, see PM. (3 Replies)
Hi All,
I am getting a date from environment variable and want to do some processing by subtracting 2 months from the date passed through the environment variable.
I am trying the following syntax :
date_var=2014-08-31
date_2M_ago='$date_var+"%d%m%y" --$date_var="2 months ago" '... (3 Replies)
Hi Folks,
I have a file with 2 columns TAB delimited and I want to add '1' to the first column and subtract '-1' from the second column.
What I have tried so far is;
awk -F"\t" '{ $1-=1;$2+=1}1' OFS='\t' file
File
0623 0623
0624 0624
0643 0643
1059 1037
1037 1037
1038 1038... (2 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.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)