It seems you're storing the date in a variable somewhere in your script which is fine, however, if you need a start and end date then you will need to store the date command again.
This User Gave Thanks to in2nix4life For This Post:
Hi there,
I am trying to do the following in Unix (Solaris 2.7):
1...Find the actual date 7 days ago in format (dd-mmm-yyyy) and with the weekday prefably,
2...Find the date 5 days after the above in the same format.
How can I do this in Unix instead of accessing our Oracle database and... (5 Replies)
Good day folks,
This is my first post on this board and I thank you in advance for helping me with this issue.
Any idea how I can synchronize server time with another timeserver but have my server lag behind by 2 seconds?
Meaning...I need a simple unix script that I can run as crone that... (2 Replies)
Hi
Do you have any pointers how to validate numbers (not to contain alphabets and special characters) and date(MM/DD/YYYY) format.
I used following regular expression to validate integer, which is not working in the default shell:
nodigits="$(echo $testvalue | sed 's/]//g')"
... (4 Replies)
Hi,
I wrote a simple shell script which accepts the input value yearmonth in the format YYYYMM and displays the date as YYYY-MM-DD.Day will be 01 always.Please find the code below
#!/bin/ksh
export yearmonth_date=$1
print_usage() {
echo "usage: ${0##*/} <yearmonth_date> \n" \
... (1 Reply)
When I write
Select date_field from TableA fetch first row only
I am getting the output as 09/25/2009.
I want to get the output in the below format
2009-09-25 i.e., MM-DD-YYYY. Please help (7 Replies)
I need to generate a report on the first of each month that will show me how many points are going to expire on a particular scheme for the next three months. The points expiry is not an issue what I have an issue with is finding the command to give me the last day of each month for the current and... (2 Replies)
Hi all,
Need an urgent help on the below scenario.
script:
awk -F","
'BEGIN { #some variable assignment}
{ #some calculation and put values in array}
END {
year=#getting it from array and assume this will be 2014
month=#getting it from array and this will be 05
date=#... (7 Replies)
HI,
Can anyone tell me how to pull the date and file name separated by a space using the find command or any other command. I want to look through several directories and based on a date timeframe (find -mtime -7), output the file name (without the path) and the date(in format mmddyyyy) to a... (2 Replies)
current date command runs well
awk -v t="$(date +%Y-%m-%d)" -F "'" '$1 < t' myname.dat
subtract 30 days fails
awk -v t="$(date --date="-30days" +%Y-%m-%d)" -F "'" '$1 < t' myname.dat
awk command in hp unix subtract 30 days automatically from current date without date illegal option error... (20 Replies)
Discussion started by: kmarcus
20 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)