Can I determine when the particular file was created, in korn-shell. Can please someone help me. If possible please mail the solution to me.
my mail id: bharat.surana@gmail.com (1 Reply)
Hiee .... i want an immediate solution for this....
Let replace.cfg file contains a list of files to be replaced the old dates with new dates......
The script must take 3 parameters....
1) old date
2) new date
3) .config file
Can u get me the solution......
I tried with sed but i... (2 Replies)
Hi,
I am using shell (#!/bin/bash), i am trying to set my date in a script date.sh, i want this dates to be used by another scripts.
I have tried to make:
#!/bin/bash
begin_date =`date +%Y_%m_%d_%H_%M_%S`,
i also tried:
#!/bin/bash
begin_date =20080709 12:14:11, then i write in... (11 Replies)
Hi all,
i need a script that can check if users did an operation within 3 days,if not delete all the logs of users who did not perform any activity after 3 days.therefore script should be able to use current date and verify last date user performed activity and see if it is greater or less than 3... (3 Replies)
Hello gurus,
I have three korn shell script 3.1, 3.2, 3.3. I would like to call three shell script in one shell script.
i m looking for something like this
call 3.1;
If 3.1 = "complete" then
call 3.2;
if 3.2 = ''COMPlete" then
call 3.3;
else
exit
The... (1 Reply)
Hello,
I am trying to show today's date and time in a better format than ‘date' (Using positional parameters). I found a command mktime and am wondering if this is the best command to use or will this also show me the time elapse since 1/30/70? Any help would be greatly appreciated, Thanks... (3 Replies)
I am looking to do something where if I created a file named backup,or whatever it would print a name like “backup_Apr_11_2011”. Thanks citizencro (1 Reply)
Hello,
I need to create a shell script that appends a filename to create a name with the date and time appended that is guaranteed to not exist. That is, the script insures you will not overwrite a file with the same name. I am lost with this one. I know I need to use date but after that I am... (3 Replies)
Hi Team,
I have the below 4 scripts which I will be running in sequential order.
This run will start for today's business date.
If all the 4 scripts are success for today that means script has ran succesfully.
Howver if any one of these 4 scripts failed then it has to take the next... (1 Reply)
Discussion started by: Deena1984
1 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)