04-11-2009
Date format checker
Hi
I want the user to enter a date in the format 16-APR-2000 .
I need to put validations for that in my script ..
Please help
Thanks and Regards
Ultimatix
10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
hi,
for reading a cobol indexed file i need to convert "mmddyy" date format to "ccyyddd" format.
i checked the datecalc and other scripts but couldnt modify them to cater to my need:(...
The datecalc gives an output which i believe is the total days till that date, but i want to convert it... (2 Replies)
Discussion started by: Bhups
2 Replies
2. UNIX for Advanced & Expert Users
Hi ,
I have written a shell script that takes the current date on the server and stores it in a file.
echo get /usr/home/data-`date '+%Y%d'`.xml> /usr/local/sandeep/GetFILE.ini
I call this GetFILE.ini file from an sftp program to fetch a file from /usr/home/ as location. The file is in... (3 Replies)
Discussion started by: bsandeep_80
3 Replies
3. Shell Programming and Scripting
I have a comma delimited log file which has the date as MM/DD/YY in the 2nd column, and HH:MM:SS in the 3rd column.
I need to change the date format to YYYY-MM-DD and merge it with the the time HH:MM:SS. How will I got about this?
Sample input
02/27/09,23:52:31
02/27/09,23:52:52... (3 Replies)
Discussion started by: hazno
3 Replies
4. UNIX for Dummies Questions & Answers
I have a list of dates in the following format: mm/dd/yyyy and want to change these to the MySQL standard format: yyyy-mm-dd.
The dates in the original file may or may not be zero padded, so April is sometimes "04" and other times simply "4".
This is what I use to change the format:
sed -i '' -e... (2 Replies)
Discussion started by: figaro
2 Replies
5. Shell Programming and Scripting
I need to increment a date value through shell script.
Input value consist of start date and end date in DATE format of unix.
For eg.
I need increment a date value of 1/1/09 to 31/12/09 i.e for a whole yr.
The output must look like
1/1/09
2/2/09
.
.
.
31/1/09
.
.
1/2/09
.
28/2/09... (1 Reply)
Discussion started by: sunil087
1 Replies
6. Shell Programming and Scripting
Hello Experts,
How do i get date after 5 days from current date in YYYYMMDD format?
How do you compare date in YYYYMMDD format?
Thanks (8 Replies)
Discussion started by: needyourhelp10
8 Replies
7. UNIX for Dummies Questions & Answers
Hello,
I am trying to show today's date and time in a better format than ‘date' (Using positional parameters). I found a command mktime and am wondering if this is the best command to use or will this also show me the time elapse since 1/30/70? Any help would be greatly appreciated, Thanks... (3 Replies)
Discussion started by: citizencro
3 Replies
8. Shell Programming and Scripting
hi there
I have file names in different format as below
triss_20111117_fxcb.csv
triss_fxcb_20111117.csv
xpnl_hypo_reu_miplvdone_11172011.csv
xpnl_hypo_reu_miplvdone_11-17-2011.csv
xpnl_hypo_reu_miplvdone_20111117.csv
xpnl_hypo_reu_miplvdone_20111117xfb.csv... (10 Replies)
Discussion started by: manas_ranjan
10 Replies
9. UNIX for Dummies Questions & Answers
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
10. Shell Programming and Scripting
i try to set linux date & time in specific format but it keep giving me error
Example :
date "+%d-%m-%C%y %H:%M:%S" -d "19-01-2017 00:05:01"
or
date +"%d-%m-%C%y %H:%M:%S" -d "19-01-2017 00:05:01"
keep giving me this error :
date: invalid date ‘19-01-2017 00:05:01'
Please use CODE tags... (7 Replies)
Discussion started by: umen
7 Replies
LEARN ABOUT OSX
apr::bucketalloc
apache_mod_perl-108~358::mod_perl-2.0.7::docs::api::APR:UserkContributed Peapache_mod_perl-108~358::mod_perl-2.0.7::docs::api::APR::BucketAlloc(3)
NAME
APR::BucketAlloc - Perl API for Bucket Allocation
Synopsis
use APR::BucketAlloc ();
$ba = APR::BucketAlloc->new($pool);
$ba->destroy;
Description
"APR::BucketAlloc" is used for bucket allocation.
"new"
Create an "APR::BucketAlloc" object:
$ba = APR::BucketAlloc->new($pool);
class: "APR::BucketAlloc"
arg1: $pool ( "APR::Pool object" )
The pool used to create this object.
ret: $ba ( "APR::BucketAlloc object" )
The new object.
since: 2.0.00
This bucket allocation list (freelist) is used to create new buckets (via "APR::Bucket->new") and bucket brigades (via
"APR::Brigade->new").
You only need to use this method if you aren't running under httpd. If you are running under mod_perl, you already have a bucket
allocation available via "$c->bucket_alloc" and "$bb->bucket_alloc".
Example:
use APR::BucketAlloc ();
use APR::Pool ();
my $ba = APR::BucketAlloc->(APR::Pool->pool);
my $eos_b = APR::Bucket::eos_create($ba);
"destroy"
Destroy an "APR::BucketAlloc object":
$ba->destroy;
arg1: $ba ( "APR::BucketAlloc object" )
The freelist to destroy.
ret: no return value
since: 2.0.00
Once destroyed this object may not be used again.
You need to destroy $ba only if you have created it via "APR::BucketAlloc->new". If you try to destroy an allocation not created by this
method, you will get a segmentation fault.
Moreover normally it is not necessary to destroy allocators, since the pool which created them will destroy them during that pool's cleanup
phase.
See Also
mod_perl 2.0 documentation.
Copyright
mod_perl 2.0 and its core modules are copyrighted under The Apache Software License, Version 2.0.
Authors
The mod_perl development team and numerous contributors.
perl v5.16.2 2011-02apache_mod_perl-108~358::mod_perl-2.0.7::docs::api::APR::BucketAlloc(3)