hi all,
I'm trying to do a cp only on files I created on a given day or within a certain date range.
What's the best way to do this?
Cheers,
KL (1 Reply)
Hi Experts,
I have files name
report_20090416
report_20090417
report_20090418
report_20090420
report_20090421
I have 2 input from user
From Date: 20090417
To Date: 20090420
and I need to grep only those line in between. Output should be
report_20090417
report_20090418... (3 Replies)
Hi ,
I need a function that verfies the given date is between start date and end date .
I have written this but this not working if start date is 1900/01/01
Below is my code
validateDate()
{
RC=$#
if
then
return 0
else
... (2 Replies)
I have a number of instances wher I need to run reports for the previous month and need to include the last months date range in the sql.
I want to create a string which consists of the first and last dates of last month separated with an ' and ' ie for this month (Feb) I want it to say
'01/01/10... (3 Replies)
Hi All,
Can anybody help me out a Shell script which pulls the files based on date range
Example
./test.sh start_date End_date (20110901 20110930)
or
./test.sh ( if we don't provide any input)
it should take sysdate-1 ( yesterdays date)
it should have both conditions
Plzz help me... (1 Reply)
Hi Everyone How all are doing today,
Want some help from All in Unix, What I am trying to do is , A shell file should be a called with two date parameters suppose the shell file name is run.sh
run.sh <start_date> <end_date>
If end date is not given it will pick today's date. The date should... (7 Replies)
Solaris 10
ksh88
Sorry for re-hashing some of this, but I can't find a proper solution in the forums.
Starting with /a/archive containing (on and on date formatted directories)
20060313 20080518 20100725 20121015
20060314 20080519 ... (1 Reply)
Hi
i am try to run a script by using a dates here is what i am doing
EXPORTDATE=`date --date "2 days ago" +%Y-%m-%d`
sh /path/to/the/files.sh ${EXPORTDATE}
the above code runs the job for one day,if i want to run the job for all the past 4 days how can i pass the date as a... (1 Reply)
My unix version is IBM AIX Version 6.1
I tried google my requirement and found the below answer,
find . -newermt “2012-06-15 08:13" ! -newermt “2012-06-15 18:20"
But newer command is not working in AIX version 6.1 unix
I have given my requirement below:
Input:
atr files:
... (1 Reply)
Dear all,
how can I select in the file below only the files created between Aug 14 2014 and Feb 03 2015?
EZA2284I -rw-r--r-- 1 30 8 356954 Aug 15 2014 file1
EZA2284I -rw-rw-r-- 1 30 8 251396 Feb 05 12:53 file2
EZA2284I -rw-rw-r-- 1 30 8 ... (3 Replies)
Discussion started by: simomuc
3 Replies
LEARN ABOUT PHP
datetime.add
DATETIME.ADD(3) 1 DATETIME.ADD(3)DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::add (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_add (DateTime $object, DateInterval $interval)
Adds the specified DateInterval object to 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.add(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-11
Example #2
Further DateTime.add(3) examples
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-01 10:00:30
2007-06-05 04:03:02
Example #3
Beware when adding months
<?php
$date = new DateTime('2000-12-31');
$interval = new DateInterval('P1M');
$date->add($interval);
echo $date->format('Y-m-d') . "
";
$date->add($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-01-31
2001-03-03
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.sub(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.ADD(3)