Hi all,
I've used various scripts in the past to work out the date last week from the current date, however I now have a need to work out the date 1 week from a given date.
So for example, if I have a date of the 23rd July 2010, I would like a script that can work out that one week back was... (4 Replies)
Hello gurus,
I am hoping someone can help me with the required code/script to make this work. I have the following file with records starting at line 4:
NETW~US60~000000000013220694~002~~IT~USD~2.24~20110201~99991231~01~01~20101104~... (4 Replies)
Give your an integer (e.g. 0x0076f676) representing the number of minutes elapsed since January 1, 1996.
How to calculate the current date which format should be "year-month-day-hour-minutes" ? (3 Replies)
$beginDate = substr(DateCalc("today", "-7Days"),0,8);
This fetches the date 7 days back
Can I fetch the date before 7 years from todays date in Perl using same syntax
Use code tags, see PM. (3 Replies)
Hi,
I want to add some hours and minutes to the current date. For example, if the current date is "July 16, 2012 15:20", i want to add 5 hours 30 minutes to "July 16, 2012 00:00" not to "July 16, 2012 15:20". Please help.
Thanks! (4 Replies)
current date command runs well
awk -v t="$(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)
I am trying to work on a script where it is a *(star) delimited file has a multiple lines starts with RTG and 3rd column=TD8 I want to substring the date part and
I want to replace with currentdate minus 15 days. Here is an example. iam using AIX server
$ cat temp.txt
RTG*888*TD8*20180201~... (1 Reply)
We want to call a parameter file (.txt) where my application read dynamic values when the job is triggered, one of such values are below:
abc.txt
------------------
Code:
line1
line2
line3
$$EDWS_DATE_INSERT=08-27-2019
line4
$$EDWS_PREV_DATE_INSERT=08-26-2019
I am trying to write a... (3 Replies)
Hello All,
we what we call a parameter file (.txt) where my application read dynamic values when the job is triggered, one of such values are below:
abc.txt
------------------
line1
line2
line3
$$EDWS_DATE_INSERT=08-27-2019
line4
$$EDWS_PREV_DATE_INSERT=08-26-2019
I am trying to... (1 Reply)
Discussion started by: pradeepp
1 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)