10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
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
2. Shell Programming and Scripting
I need the date format in YYYY.MM format and I am able to get current month date as well as previous month date with below command
PM=`date +'%Y.%m' -d 'last month'`
CM=`date +'%Y.%m' -d 'now'`
but I need to get YYYY.MM date format for previous 12 months so could you please help me how I get... (2 Replies)
Discussion started by: Amit Joshi
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3. Shell Programming and Scripting
I am changing epoch times to dates. I was able to do the following:
echo "$varx" | gawk '{print strftime("%c", $0)}'
Mon Dec 31 16:26:40 2012
This changes the epoch date (which is what varx is) into localtime.
However, my problem is that I only want 12/31/2012 and not the Mon Dec 31... (2 Replies)
Discussion started by: newbie2010
2 Replies
4. Shell Programming and Scripting
How can I convert any user inputted date into yyyy/mm/dd ?
For example user can input date one of the following 20120121 , 2012-01-21 ,01/21/2012,01/21/2012 etc
But I need to convert any of the date entered by user into yyyy/mm/dd (2012/01/2012). Any suggestion. Thanks in advance
this is... (1 Reply)
Discussion started by: ZeroHedge
1 Replies
5. Shell Programming and Scripting
Hi,
I am having one log files. contains som data according to date. And it is going to append .Eg:abc.log contains below data
2011-10-19 abjhgj
2011-10-19 gjhgjgj
2011-10-20 hhhjh
2011-10-20 hhhhjj
2011-10-21 gg
.
.
.
2011-11-24 yyy
from log files i want catch only... (2 Replies)
Discussion started by: aish11
2 Replies
6. Shell Programming and Scripting
could you please help be on the below code ..
Requirement is when i pass the parameter(for below 2) i should get current time -2 hours in the format :wall:..
cur_dt=`$ICEBIN/sqsh -S$DSQUERY -U $BATCHID -P $PASSWD -h -C"select getdate()" | sed '2d'`
pr_dt="`$ICEBIN/sqsh -S$DSQUERY -U $BATCHID... (2 Replies)
Discussion started by: karthicss
2 Replies
7. Shell Programming and Scripting
(Attention: Green PHP newbie !)
I have an online inquiry form, delivering a date in the form yyyy/mm/dd to my feedback form. If the content passes several checks, the form sends an e-mail to me. All works fine. I just would like to receive the date in the form dd/mm/yyyy. I tried with some code,... (6 Replies)
Discussion started by: keyboarder
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8. Shell Programming and Scripting
Frens,
I have a logfile which has got manier things in it and somewer in that file i have a date in mm/dd/yyyy hh:mm format....
I just need to get the date and time nothing else....
i have searched in this forum for it but got only date in mm-dd-yyyy format..
but i need to search for... (10 Replies)
Discussion started by: smarty86
10 Replies
9. UNIX for Dummies Questions & Answers
I have a file with some date columns in MM/DD/YYYY format:
SMPBR|DUP-DO NOT USE|NEW YORK||16105|BA5270715|6/6/2007 |MWERNER|109||||JOHN||SMITH|MD|72211118||||||74559|21 WILMINGTON RD||D|11/6/2003|SL# MD CONTACT-LIZ RICHARDS|||0|Y|N||1411458|
And I want to convert the date format to:
... (5 Replies)
Discussion started by: ChicagoBlues
5 Replies
10. Shell Programming and Scripting
Hi
I have a sed command
sed -e "s/sub_date/=$(date +"%d/%m/%Y")/" sub_create_tmp
I want to substitute with the current date in dd/mm/yyyy format .
But the result is an error " cannot parse " .
Pls help .
sars (2 Replies)
Discussion started by: sars
2 Replies
DateFormat(3) User Contributed Perl Documentation DateFormat(3)
NAME
Log::Log4perl::DateFormat - Log4perl advanced date formatter helper class
SYNOPSIS
use Log::Log4perl::DateFormat;
my $format = Log::Log4perl::DateFormat->new("HH:mm:ss,SSS");
# Simple time, resolution in seconds
my $time = time();
print $format->format($time), "
";
# => "17:02:39,000"
# Advanced time, resultion in milliseconds
use Time::HiRes;
my ($secs, $msecs) = Time::HiRes::gettimeofday();
print $format->format($secs, $msecs), "
";
# => "17:02:39,959"
DESCRIPTION
"Log::Log4perl::DateFormat" is a low-level helper class for the advanced date formatting functions in
"Log::Log4perl::Layout::PatternLayout".
Unless you're writing your own Layout class like Log::Log4perl::Layout::PatternLayout, there's probably not much use for you to read this.
"Log::Log4perl::DateFormat" is a formatter which allows dates to be formatted according to the log4j spec on
http://java.sun.com/j2se/1.5.0/docs/api/java/text/SimpleDateFormat.html
which allows the following placeholders to be recognized and processed:
Symbol Meaning Presentation Example
------ ------- ------------ -------
G era designator (Text) AD
y year (Number) 1996
M month in year (Text & Number) July & 07
d day in month (Number) 10
h hour in am/pm (1~12) (Number) 12
H hour in day (0~23) (Number) 0
m minute in hour (Number) 30
s second in minute (Number) 55
S millisecond (Number) 978
E day in week (Text) Tuesday
D day in year (Number) 189
F day of week in month (Number) 2 (2nd Wed in July)
w week in year (Number) 27
W week in month (Number) 2
a am/pm marker (Text) PM
k hour in day (1~24) (Number) 24
K hour in am/pm (0~11) (Number) 0
z time zone (Text) Pacific Standard Time
Z RFC 822 time zone (Text) -0800
' escape for text (Delimiter)
'' single quote (Literal) '
For example, if you want to format the current Unix time in "MM/dd HH:mm" format, all you have to do is this:
use Log::Log4perl::DateFormat;
my $format = Log::Log4perl::DateFormat->new("MM/dd HH:mm");
my $time = time();
print $format->format($time), "
";
While the "new()" method is expensive, because it parses the format strings and sets up all kinds of structures behind the scenes, followup
calls to "format()" are fast, because "DateFormat" will just call "localtime()" and "sprintf()" once to return the formatted date/time
string.
So, typically, you would initialize the formatter once and then reuse it over and over again to display all kinds of time values.
Also, for your convenience, the following predefined formats are available, just as outlined in the log4j spec:
Format Equivalent Example
ABSOLUTE "HH:mm:ss,SSS" "15:49:37,459"
DATE "dd MMM yyyy HH:mm:ss,SSS" "06 Nov 1994 15:49:37,459"
ISO8601 "yyyy-MM-dd HH:mm:ss,SSS" "1999-11-27 15:49:37,459"
APACHE "[EEE MMM dd HH:mm:ss yyyy]" "[Wed Mar 16 15:49:37 2005]"
So, instead of passing
Log::Log4perl::DateFormat->new("HH:mm:ss,SSS");
you could just as well say
Log::Log4perl::DateFormat->new("ABSOLUTE");
and get the same result later on.
Known Shortcomings
The following placeholders are currently not recognized, unless someone (and that could be you :) implements them:
F day of week in month
w week in year
W week in month
k hour in day
K hour in am/pm
z timezone (but we got 'Z' for the numeric time zone value)
Also, "Log::Log4perl::DateFormat" just knows about English week and month names, internationalization support has to be added.
COPYRIGHT AND LICENSE
Copyright 2002-2009 by Mike Schilli <m@perlmeister.com> and Kevin Goess <cpan@goess.org>.
This library is free software; you can redistribute it and/or modify it under the same terms as Perl itself.
perl v5.12.1 2010-02-07 DateFormat(3)
