It doesnt seem to recognize -d option. Below is the output.
Please Let me know what information you are looking for. I'll add that if that helps.
It is a SOLARIS OS with bash running on it.
I want to change a date from format dd-mmm-yyyy to mm/dd/yyyy. Is there a way to do this with sed or do you have to write a case statement to convert JAN to 01? Thanks (9 Replies)
How to convert the date field from dd/mm/yyyy to yyyy/mm/dd in unix
my script will generate text file which have two fields
one is date and another is name of the server for example this is sample date which I have to sort based on older to newer date the problem is when I found out sort will... (4 Replies)
(Attention: Green PHP newbie !)
I have an online inquiry form, delivering a date in the form yyyy/mm/dd to my feedback form. If the content passes several checks, the form sends an e-mail to me. All works fine. I just would like to receive the date in the form dd/mm/yyyy. I tried with some code,... (6 Replies)
I am trying to get the string containing date - in a specfic format actually, although I think that part is inconsequencial - 1110226^1110226^1110226^1110226^1110226 - through echo or printf or in some other way - created within a cront job and passed as a parameter to a perl script.
Now, I know... (3 Replies)
Hello,
I am writing a script that parses different logs and produces one. In the source files, the date is in DD MM YYYY HH24:MI:SS format. In the output, it should be in DD MON YYY HH24:MI:SS (ie 25 Jan 2010 16:10:10)
To extract the dates, I am using shell substrings, i.e.:
read line
... (4 Replies)
Hi,
How can I convert day of year value in format(yy,doy) to normal formatted (dd.mm.yyyy) string also all of them with awk or awk system function?
in_file.txt
---------
12,043
12,044
12,045
12,046
out_file.txt
----------
12.02.2012
13.02.2012
14.02.2012
15.02.2012
imagine... (5 Replies)
Hi I have a problem with Date format in my code.
1st I am trying to convert today's date to yesterday's using
YESTERDAY3=`perl -e '@y=localtime(time()-86400); printf "%04d/%02d/%02d",$y+1900,$y+1,$y;$y;'`
And once it is done I am trying to using the yesterday date in a grep command to... (3 Replies)
I've seen a lot of posts on this and have tried the following:
echo 1257000000| perl -e '($d,$m,$y)=(localtime(time-86400));$m+=1;$y+=1900;printf "$y/$m/$d\n";'
But I am unable to convert a past Epoch date into a format such as YYYY/MM/DD or MM/DD/YYYY.
I am using bash and don't know... (4 Replies)
I am getting output of YYYY-MM-DD and want to change this to DD/MM/YYYY.
When am running the query in 'Todd' to_date(column_name,'DD/MM/YYYY') am getting the required o/p of DD/MM/YYYY, But when am executing the same query(Netezza) in linux server(bash) am getting the output of YYYY-MM-DD
file... (3 Replies)
Discussion started by: Roozo
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)