@Bakunin: that looks like a neat trick. It does not seem to work on HP-UX, however (max 24 hours)...
It appears to only work on AIX and IRIX to the full extent.
I found the following maximum time shifts using the date utility with the TZ variable...
Of course on Linux one would not need the trick since it has GNU date...
Last edited by Scrutinizer; 12-24-2018 at 12:29 PM..
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 $2}' will this type of command work.
actually my need is
if today is... (17 Replies)
Hi All,
Is it possible to run date -d option in Solaris?
Do we have a work around so that -d option will be recognized
by solaris as it is recognized by linux.
I need this since i am using this in scripting and it works in Linux box. my problem is
it doesn't work in solaris box.
... (6 Replies)
Hi,
Anybody knows how to get what date was 28 days ago of the current system date through UNIX script.
Ex : - If today is 28th Mar 2010 then I have to delete the files which arrived on 1st Mar 2010, (15 Replies)
Hi all,
I am trying to execute the following command in a sun solaris machine and getting the error as below.
bash-2.03$ date -d "1 day ago" +%Y%m%d
date: illegal option -- d
bash-2.03$ uname -a
SunOS gtrd02 5.8 Generic_117350-55 sun4u sparc SUNW,Sun-Fire-V440
Can anybody help me to... (1 Reply)
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)
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)
SunOS -s 5.10 Generic_147440-04 sun4u sparc SUNW,SPARC-Enterprise
Hi,
In a folder, there are files. I have a script which reads the current date and subtract the modification date of each file.
How do I achieve this?
Regards,
Joe (2 Replies)
Discussion started by: roshanbi
2 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)