Hi All,
I need to find the previous month last day minus one day, using shell script. Can you guys help me to do this.
My Requirment is as below:
Input for me will be 2000909(YYYYMM)
I need the previous months last day minus 1 day timestamp. That is i need 2000908 months last day minus ... (3 Replies)
Hi
I have a table with name, date in format DD.MM.YYYY.
I need to something like this (I try to explain in pseudo code)
if SYSDATE (current date) minus 6 months > $expiry date
print OK
else print NOK with $name and $expiry date
I know this is possible with Oracle. How to do this... (0 Replies)
Hi guys,
I had a scenario...
1. I had to get the previous days date in yyyymmdd format
2. i had to create a file with Date inthe format yyyymmdd.txt format
both are different
thanks guys in advance.. (4 Replies)
I need to get the next day's date of the user entered date
for example:
Enter date (yyyy/mm/yy):
2013/10/08I need to get the next day's date of the user entered date
Desired Output:
2013/10/09Though there are ways to achieve this is Linux or Unix environment (date command) ,I need to... (1 Reply)
Hi there,
is it possible to get the actual date minux six months with just a simple command?
It's easy with Linux but on HP Unix (for me) impossible ;)
Best wishes (3 Replies)
I am running a script in ksh to get the 2 months back date from system date.The below code is giving correct date output from putty command prompt.But while running the script is .ksh file it is giving the error below.Please suggest.
; d=a; y=a
m-=num
while(m < 1) {m+=12; y--}... (1 Reply)
Hello Folks,
I have a variable output holding date as below -
output = "20141220"
I need to extract a day out of it and store it in another variable i.e. something similar to below -
output1=20141219"
and if the month is changing i.e. date in on 31st or 1st it should be taken care of
"date... (5 Replies)
I Have text like
XXX_20190908.csv.gz need to replace Only date in this format with current date every day
Thanks! (1 Reply)
Discussion started by: yamasani1991
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)