I don't think that the date command is intended for that. I can sort-of get it to work by abusing the timezone concept. Note that I had to switch signs.
Hi Friends :)
I have a long file having fields in the form :
Field1 yy/mm/dd hh:mm:ss Duration(Sec)
line 1) 123123 05/11/30 12:12:56 145
line 2) 145235 05/11/30 12:15:15 30
line 3) 145264 05/11/30 13:14:56 178
.
.
I want to subtract yy/dd/dd hh:mm:ss in line (2) from yy/mm/dd hh:mm:ss in... (1 Reply)
Hello,
Im writing a script using the ksh shell. I have 2 variables in the script:
CURRTIME
PREVTIME
Example, if CURRTIME=13:00, I want to somehow calculate what the time was an hour ago so that PREVTIME=12:00
Right now I have the following:
CURRTIME=`date +%H:%M`
How can I... (4 Replies)
i have the time 20100421043335 in format (date +%Y%m%d%H%M%S),and i want to be able to get the previous time 2 minutes ago,which is
20100421043135 (9 Replies)
need some help on the below requirement:
File1:
SV,22,20100501140000,JFK,RUH
SV,29,20100501073000,BOM,RUH
SV,29,20100501073000,SIN,RUH
third filed is datetime which is of the format (yyyymmddhh24miss)
File2
JFK,+,0500
BLR,-,0530
SIN,-,0800
for every line of file 1, take 4... (9 Replies)
Hi,
Need to subtract 5 seconds after syncing my Linux server from NTP like;
#ntpdate time.myorg.int.
This script will only run once in each morning at 9 AM.
Please help me. (4 Replies)
Hello All ,
Please support for below request
how to change format and subtract time and date and get average.
xxx 13-OCT-15 11.32.18.241000 AM 13-OCT-15 11.35.49.089080 AM
xxx 13-OCT-15 11.32.24.000000 AM 13-OCT-15 11.45.17.810904 AM
xxx 13-OCT-15 11.32.25.232000 AM ... (1 Reply)
INPUT:
16:45:51 10051 77845
16:45:51 10051 77845
16:46:52 10051 77846
16:46:53 10051 77846
Match the last PID then subtract second line time with first line.
Please help me with any command or script. (3 Replies)
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)
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.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)