Sponsored Content
Top Forums UNIX for Dummies Questions & Answers Date format conversion function Post 302087226 by Cecile on Wednesday 30th of August 2006 12:21:23 PM
Old 08-30-2006
date format conversion

I can have either :

070829 --> I want 2007-08-29
890829 --> 1989-08-29
 

10 More Discussions You Might Find Interesting

1. UNIX for Dummies Questions & Answers

date format conversion

hi, i have a file in which i get date format as 22/APR/2010... now i want the date format to be in 22-04-2010 if the month changes to may the file should also have 05 as month.... pls help (3 Replies)
Discussion started by: siva_nagarajan
3 Replies

2. Shell Programming and Scripting

Date conversion from Standard/given format to seconds/epoch

I am trying get time difference of two dates in secs. Initially I want to convert a standard date format to epoch for two dates and then subtract the two epoch dates. Example : date -d "2007-09-01 17:30:40" '+%s' But this gives me below error date: illegal option -- d Usage: date OS: AIX... (6 Replies)
Discussion started by: bpaac
6 Replies

3. Shell Programming and Scripting

Date format conversion

Hi All, Can someone please let me know how can i convert the date format in unix as follow: From: 24 Oct 2011 i.e $(date +'%d %b %Y') To: 111024 i.e $(date +%y%m%d) Thanks in advance (3 Replies)
Discussion started by: davidtd
3 Replies

4. Shell Programming and Scripting

Military type format date/time conversion

Hello All, I have a requirement to convert a 12 hour format to 24 hour time format and the sample input /out put is below Input Time format : Nov 2 2011 12:16AM Out Put Format : Nov 2 2011 0:16 Input : Nov 2 2011 4:16PM Out Put: Nov 2 2011 16:16 I have done this using a... (6 Replies)
Discussion started by: jambesh
6 Replies

5. Shell Programming and Scripting

Date conversion from 24 hr format to 12 hr format

hi i want to convert date procured from sone operation which will be in 24hr format to 12 hr format displaying AM and PM # date -d @1362545068 Tue Mar 5 23:44:28 EST 2013 # this Tue Mar 5 23:44:28 EST 2013 i want to convert it so that output is as below Tue... (2 Replies)
Discussion started by: vivek d r
2 Replies

6. UNIX for Dummies Questions & Answers

Rename all Files in a UNIX Directory from one date format to another date format

Hi Unix Gurus, I would like to rename several files in a Unix Directory . The filenames can have more than 1 underscore ( _ ) and the last underscore is always followed by a date in the format mmddyyyy. The Extension of the files can be .txt or .pdf or .xls etc and is case insensitive ie... (1 Reply)
Discussion started by: pchegoor
1 Replies

7. Shell Programming and Scripting

Date conversion and Format

Hello , I have a record in below format Hostname | Query: 0 | Release: 0 | files: 2 | Files_examined: 2 | SET timestamp=1396778638; | select * from test I need output in below format Hostname | 0 | 0 | 2 | 2 | 04/06/2014|03:03:58 | select * from test I was able to get above output... (1 Reply)
Discussion started by: Tomlight
1 Replies

8. Shell Programming and Scripting

Date format conversion

Hi, i have to check the file whether it is created today. here is the ls -l o/p -rw-r----- 20000 50000 130 Dec 12 10:21 file.txt im able to check if file is created today or not if the timestamp is in 2014-12-12 format by comparing $(date +Y-%m-%d) but stuckup as it is of format Dec 12... (6 Replies)
Discussion started by: JSKOBS
6 Replies

9. Shell Programming and Scripting

Bash Scripting with date format conversion

I have a script below and wanted to change the output into three different file format (3 separate script) #!bin/bash #input file format postwrf_d01_20131206_0600_f08400.grb2 #postwrf_d01_YYYYMMDD_ZZZZ_f0HHHH.grb2 #zzzz= 0000,0600,1200,1800 (in UTC) #HHHH=00000,00600,01200,01800 ..ect (in... (1 Reply)
Discussion started by: cumulus_255
1 Replies

10. UNIX for Beginners Questions & Answers

Date format conversion how to change this from using nawk to awk

Hi, I have a file where I need to change the date format on the nth field from DD-MM-YYYY to YYYY-MM-DD so I can accurately sort the record by dates From regex - Use sed or awk to fix date format - Stack Overflow, I found an example using nawk. Test run as below: $: cat xyz.txt A ... (2 Replies)
Discussion started by: newbie_01
2 Replies
HTTP::Date(3)						User Contributed Perl Documentation					     HTTP::Date(3)

NAME
HTTP::Date - date conversion routines SYNOPSIS
use HTTP::Date; $string = time2str($time); # Format as GMT ASCII time $time = str2time($string); # convert ASCII date to machine time DESCRIPTION
This module provides functions that deal the date formats used by the HTTP protocol (and then some more). Only the first two functions, time2str() and str2time(), are exported by default. time2str( [$time] ) The time2str() function converts a machine time (seconds since epoch) to a string. If the function is called without an argument or with an undefined argument, it will use the current time. The string returned is in the format preferred for the HTTP protocol. This is a fixed length subset of the format defined by RFC 1123, represented in Universal Time (GMT). An example of a time stamp in this format is: Sun, 06 Nov 1994 08:49:37 GMT str2time( $str [, $zone] ) The str2time() function converts a string to machine time. It returns "undef" if the format of $str is unrecognized, otherwise whatever the "Time::Local" functions can make out of the parsed time. Dates before the system's epoch may not work on all operating systems. The time formats recognized are the same as for parse_date(). The function also takes an optional second argument that specifies the default time zone to use when converting the date. This parameter is ignored if the zone is found in the date string itself. If this parameter is missing, and the date string format does not contain any zone specification, then the local time zone is assumed. If the zone is not ""GMT"" or numerical (like ""-0800"" or "+0100"), then the "Time::Zone" module must be installed in order to get the date recognized. parse_date( $str ) This function will try to parse a date string, and then return it as a list of numerical values followed by a (possible undefined) time zone specifier; ($year, $month, $day, $hour, $min, $sec, $tz). The $year will be the full 4-digit year, and $month numbers start with 1 (for January). In scalar context the numbers are interpolated in a string of the "YYYY-MM-DD hh:mm:ss TZ"-format and returned. If the date is unrecognized, then the empty list is returned ("undef" in scalar context). The function is able to parse the following formats: "Wed, 09 Feb 1994 22:23:32 GMT" -- HTTP format "Thu Feb 3 17:03:55 GMT 1994" -- ctime(3) format "Thu Feb 3 00:00:00 1994", -- ANSI C asctime() format "Tuesday, 08-Feb-94 14:15:29 GMT" -- old rfc850 HTTP format "Tuesday, 08-Feb-1994 14:15:29 GMT" -- broken rfc850 HTTP format "03/Feb/1994:17:03:55 -0700" -- common logfile format "09 Feb 1994 22:23:32 GMT" -- HTTP format (no weekday) "08-Feb-94 14:15:29 GMT" -- rfc850 format (no weekday) "08-Feb-1994 14:15:29 GMT" -- broken rfc850 format (no weekday) "1994-02-03 14:15:29 -0100" -- ISO 8601 format "1994-02-03 14:15:29" -- zone is optional "1994-02-03" -- only date "1994-02-03T14:15:29" -- Use T as separator "19940203T141529Z" -- ISO 8601 compact format "19940203" -- only date "08-Feb-94" -- old rfc850 HTTP format (no weekday, no time) "08-Feb-1994" -- broken rfc850 HTTP format (no weekday, no time) "09 Feb 1994" -- proposed new HTTP format (no weekday, no time) "03/Feb/1994" -- common logfile format (no time, no offset) "Feb 3 1994" -- Unix 'ls -l' format "Feb 3 17:03" -- Unix 'ls -l' format "11-15-96 03:52PM" -- Windows 'dir' format The parser ignores leading and trailing whitespace. It also allow the seconds to be missing and the month to be numerical in most formats. If the year is missing, then we assume that the date is the first matching date before current month. If the year is given with only 2 digits, then parse_date() will select the century that makes the year closest to the current date. time2iso( [$time] ) Same as time2str(), but returns a "YYYY-MM-DD hh:mm:ss"-formatted string representing time in the local time zone. time2isoz( [$time] ) Same as time2str(), but returns a "YYYY-MM-DD hh:mm:ssZ"-formatted string representing Universal Time. SEE ALSO
"time" in perlfunc, Time::Zone COPYRIGHT
Copyright 1995-1999, Gisle Aas This library is free software; you can redistribute it and/or modify it under the same terms as Perl itself. perl v5.16.3 2012-03-30 HTTP::Date(3)
All times are GMT -4. The time now is 06:08 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy