You should use bash's integer arithmetics. And, btw, using a variable with the same name as a command works, but can become veeery misleading sometime! Try
You can avoid the base 10 indicator if you use date +%e when assigning the variable.
Cound anyone help me on how to compare date in Unix using if function on bash file?
current=date
if ###syntax is wrong, could anyone correct it for me
then
rm -rf /usr/local/src
fi
Thank You... (17 Replies)
hi guys.
in bash is there any other way of limiting the time displayed to HH:MM
appart from
(date +"%H:%M") and (date +"%R")?
i want to input time into a database in the form HH:MM
have tried NOW() but this gives me HH:MM:SS
thanks in advance (1 Reply)
Hi
I have this simple script:
#!/bin/bash
date1=2009:07:15:12:36
date2=2009:07:15:12:16
echo $date1
echo $date2
datediff=
#datediff=date1-date2
echo datediff is$datediff
How do i return the difference in seconds? (6 Replies)
1. The problem statement, all variables and given/known data:
I have standard web server log file. It contains different columns (like IP address, request result code, request type etc) including a date column with the format .
I have developed a log analysis command line utility that displays... (1 Reply)
Hello all,
I'm trying to substract 1 minute from the current date and take the hour and minute (for filename purpose).
1) If I want hour an minute from current time I can use:
timetmp=$(date +"%H:%M")
2) To substract 1 minute from current time I can use:
timetmp=$(date --date "$dte -1... (8 Replies)
I am creating a startup script for an application. This application's startup script is in bash. It will also need to call a perl script (which I will not be able to modify) for the application environment prior to calling the application. The problem is that this perl script creates a new shell... (5 Replies)
date --date='10:30am + 1 hour' +%H:%M
11:30 produces
date --date='10:30pm + 1 hour' +%H:%M produces
23:30
I want to do the following:
TIME="1:30pm"
date --date='$TIME + 1 hour' + %H:%M
to produce
14:30 (1 Reply)
Hi,
I am trying to add few (say 3 days) to sysdate using -
date -d '+ 3 days' +%y%m%d
and it works as expected.
But how to add few (say 3 days) to a literal date value and how bash treats a literal value as a date. Can we say just like in ORACLE TO_DATE that my given literal date value... (2 Replies)
I all
I have written a bash script for compare two date. One of those is a result of query, and another is current date.
I have a problem with the format, because the first is 09/12/19 18:50:30 but for having this result I have to do
d1DB=$(date -d "$valData" +'%m/%d/%y %T')
and the second... (9 Replies)
Discussion started by: rdie77
9 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)