hii all.
I have to get the date of the 7th day past from the current date.
if i give the current date as sep 3 then i must get the date as 27th of august.
can we get the values from the "cal" command.
cal | awk '{print $2}' will this type of command work.
actually my need is
if today is... (17 Replies)
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)
Hi i am writing a cron job.
so for it i need the 60 days old date form current date in variable.
Like today date is 27 jan 2011 then output value will be stote in variable in formet Nov 27.
i am using EST date, and tried lot of solution and see lot of post but it did not helpful for me. so... (3 Replies)
I am trying to find out the number of days between the current date and user defined date.
I took reference from here for the date2jd() function.
Modified the function according to my requirement. But its not working properly.
Original code from here is working fine.
#!/bin/sh... (1 Reply)
Well guys,
I know the right syntax for displaying the current date is $(date). However, I am planning to send emails to some customers which displays their subscription date, and then the expiry. The expiry being 30 days from the current date.
What would the right syntax be? (6 Replies)
hi all..
i want 2 know how 2 find 7days past date from current date..
when i used set datetime = `date '+%m%d%y'` i got 060613..
i just want to know hw to get 053013..
i tried using date functions but couldnt get it :( i use c shell and there is no chance that i can change that ..... (3 Replies)
I have to display only those subscribers which are in "unconnected state" and the date is 90 days older than today's date.
Below command is used for this purpose:
cat vfsubscriber_20170817.csv | sed -e 's/^"//' -e '1d' | nawk -F '",' '{if ( (substr($11,2,4) == 2017) && ( substr($11,2,8) -lt... (1 Reply)
Hi all,
I have been researching to obtain SSL certification expiry for most of our webistes. For some cases, some hosts where not directly accessible so i finally got a solution working with curl using my proxy. This lists the expiry date which i'm finally looking for.
# curl --proxy... (4 Replies)
current date command runs well
awk -v t="$(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)
Discussion started by: kmarcus
20 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)