hey folks,
been awhile (actaully a long while) since i last touched C. And the 3 books i've read don't really have much about using time.h
Question:
How would i be able to assign a variable the value of the current date minus 2 mths, keeping in mind the yr.
IE. would like to see Nov.31/2001... (1 Reply)
I'm writing a ksh script in which I want to present the user with a choice of choosing any of the last 15 days. i.e., a list like the following:
20040510 (today)
20040509 (yesterday)
20040508
.
.
.
20040426
Is there an easy way to produce the date from x number of days ago other than... (3 Replies)
I need to achieve the following.....I seached the forum but could not find it...
This is I have in a file...
"CH","TIA","10/27/2006",000590
I need the date in the third field to be attached to fileas 20061027_test.txt
How do I do it. (6 Replies)
HI,
I'm comparing my file date with the system date and if both the dates are equal I'm doing some operation. I use two variables for these two dates. I use the following command in my query. if ....
But here the current date $cd shows 01 and filedate $fdate shows 1. The file is created on 1 of ... (6 Replies)
Hi Gurus,
How to minus 15 minuets from current system time.
For example if current time is " Wed Oct 14 12:12:38 BST 2009", i need "Wed Oct 14 11:57:38 BST 2009"
Thanks (2 Replies)
How can i print a future time, so i get current time by date "+%H:M" but how can i say add 20 minutes to the current time and display as I have just done for current time. (1 Reply)
I have a file with a field containing the following:
"7/3/2009 7:07:12 PM","xxxx"
I need to be able to split this field up into two into a different format with the time being converted into 24 hour:
so that i can get the following:
"20090307","19:07:12","xxxx" (8 Replies)
I have a data file. Seperated by "|". The 19 th filed is a date field that occurs like this
11/02/2001
i need to convert into the below format
2001-11-02
for e.g..
i/p
o/p should be
can somebody throw some light (5 Replies)
In my shell script I take date as a input parameter from command line in the format "21 Oct 2011" which would be
date +'%d %b %Y'
Now i need to do two things here.
1) Validate the date entered by user
2) Calculate yesterday's date from the input. So in this case it should be: "20 Oct 2011"... (9 Replies)
Hallo Team
I can perform the task manually but i would like to automate this process. ok here goes. I have a perl script which runs every Wednesday every week and the name of the script is check_19.pl
This is how the script looks like :
#!/usr/bin/perl -w
#use strict;
use DBI;
#... (1 Reply)
Discussion started by: kekanap
1 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)