Please write your query clearly, try something like this just an example, remaining digits after decimal you have to convert to days and then you have to add .
I want to split a given month into weeks. For example if I give the date in dd/mm/yy format say 01/02/08 it should give output in the given format :
week1 : start date and end date.
week2 : ""
week3 : ""
week4 : "" (5 Replies)
munt=`date '+%m` will isolate the month in digit form 02 = Feb
Trying to get the same out of perl just cant see it
$stimx = localtime($^T);
print ((split/ /,$stimx)); (4 Replies)
Hello All,
I am trying to come up with a shell script to count a specific word in a logfile on each day of this month, last month and the month before. I need to produce this report and email it to customer.
Any ideas would be appreciated! (5 Replies)
I am writing the code in perl.
I have an array in perl and each variable in the array contains the data in the below format
Now I need to check the below variable w.r.t system month I need to store the date and time(Tue Aug 7 03:54:12 2012) from the below data into file if contains only 'Aug'... (5 Replies)
Hi,
I have a script that accepts an input date from the user in yyyy-mm-dd format.
I need to get the mm-dd part and convert it to month name.
example:
2011-11-15
I want that to become "Nov 15"
I don't have the GNU date, I am using an AIX os.
Thanks. (1 Reply)
Hi,
I need all file names in a folder which has date >= 10th of last month,
Example
: files in folder
AUTO_F1_20140610.TXT
BUTO_F1_20140616.TXT
CUTO_F1_20140603.TXT
FA_AUTO_06012014.TXT
LA_AUTO_06112014.TXT
MA_AUTO_06212014.TXT
ZA_AUTO_06232014.TXT
Output:
AUTO_F1_20140610.TXT... (9 Replies)
Hi Guys,
i am having .sql script which inserts data from one table to another table based on date condition, i need to pass range on based on how many number of months, for e.g
set timing on;
whenever sqlerror exit failure;
spool myscript.log append
accept start_date... (7 Replies)
# Sample input
common-name www.test.com.au
expiration Dec 21 01:00:31 2017 GMT
common-name www.test1.com.au
expiration Jan 19 04:41:03 2018 GMT
# Desired Output
# Field 1: Domain name
# Field 2: Date/time converted to Austraian format DD/MM/YYYY and on the same line as Domain Name.
#... (7 Replies)
Hello,
I need to find the date of next first Friday of the month and set as a variable in a bash script
ie -
FIRSTFRIDAY=$(date -dfirst-friday +%d)
I know date -dfirst-friday doesn't work, but unsure if I can use this / cal + awk or something else to find the right date of the... (7 Replies)
Discussion started by: summerdays
7 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)