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)
:cool:
Hi all,
I have a pecular issue in sorting these files in Solaris environment.
All the below files are modified on November 4th, but I want to sort these files as per date column (eg: 01May07_1623 = ddmmmyy_hhmm)
Nov 4 18:27 SONYELEC00.GI22973.01May07_1623.gpg
Nov 4 18:27... (4 Replies)
Hi Friends,
I've a special requirement, even though I know how to implement this using shell scripting, current requirement is PERL, in which I'm not much familiar !!!.
I've a record, which has around 200 fields, out of which I need to extract only one date value from the 97th field (this... (1 Reply)
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)
Hi,
I have a set of columns in a csv file, my first row being an integer and 2nd being a date. I want to first sort it using the first column and then by the second.
for e.g. i have ,
1234,09/05/2009,hi
5678,01/01/2008,hi
1234,11/03/2006,hello
5678,28/07/2010,hello
i tried this... (5 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)
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)
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 wanted to sort the below data on 4th field(comma seperator) based on month and date and time on AIX OS.
Input data:
3,AJ,30 Jul 06:30,30 Jul 06:30
5,AJ,30 Jul 06:30,30 Jul 06:49
10,AJ,30 Jul 06:30,02 Jan 05:41
4,AJ,30 Jul 06:30,30 Jul 06:36
2,AJ,30 Jul 06:30,28 Jul 06:45
9,AJ,30 Jul... (2 Replies)
Discussion started by: Amit Joshi
2 Replies
LEARN ABOUT DEBIAN
html::formfu::deflator::compounddatetime
HTML::FormFu::Deflator::CompoundDateTime(3pm) User Contributed Perl Documentation HTML::FormFu::Deflator::CompoundDateTime(3pm)NAME
HTML::FormFu::Deflator::CompoundDateTime - CompoundDateTime deflator
SYNOPSIS ---
element:
- type: Multi
name: date
elements:
- name: day
- name: month
- name: year
deflator:
- type: CompoundDateTime
# set the default
$form->get_field('date')->default( $datetime );
DESCRIPTION
For use with a HTML::FormFu::Element::Multi group of fields.
Sets the default values of several fields from a single DateTime value.
By default, expects the field names to be any of the following:
year
month
day
hour
minute
second
nanosecond
time_zone
METHODS
field_order
Arguments: @order
If your field names don't follow the convention listed above, you must provide an arrayref containing the above names, in the order they
correspond with your own fields.
---
element:
- type: Multi
name: date
elements:
- name: m
- name: d
- name: y
deflator:
- type: CompoundDateTime
field_order:
- month
- day
- year
AUTHOR
Carl Franks
LICENSE
This library is free software, you can redistribute it and/or modify it under the same terms as Perl itself.
perl v5.14.2 2012-01-23 HTML::FormFu::Deflator::CompoundDateTime(3pm)