Convert DD-MMM-YYYY to MM-DD-YYYY Post: 302304473

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Change Date from dd-mmm-yyyy to mm/dd/yyyy

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

2. Shell Programming and Scripting

converting the date field from dd/mm/yyyy to yyyy/mm/dd

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...

3. Shell Programming and Scripting

Sed: zero-padding dates (or: convert d/m/yyyy to dd/mm/yyyy)

4. Shell Programming and Scripting

convert d/m/yyyy to YYYY-MM-DD

My csv has data like this x,x,3452,2/18/1986,abc x,g,19711,1/24/1986,abc i want to replace date in the following format YYYY-mm-dd how do i convert using awk script ?

5. Shell Programming and Scripting

change date format from yyyy/mm/dd to dd/mm/yyyy

(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. Shell Programming and Scripting

Converting date DD MM YYYY to DD MON YYYY

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 ...

7. UNIX for Dummies Questions & Answers

Epoch date to YYYY/MM/DD or MM/DD/YYYY

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...

8. Shell Programming and Scripting

Bash to convert to m-d-yyyy

I am using bash that when run downloads a file a verifies that there is data in it. What I am not able to do is have a user enter a date in any format they wish and have it converted to m-d-yyyy. Thank you :). Bash printf " Welcome to NGS analysis, checking for new files and creating a...

9. Shell Programming and Scripting

Date format YYYY/MM/DD to DD/MM/YYYY

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...

10. UNIX for Advanced & Expert Users

How to convert mmm-yy to mm/dd/yyyy format in UNIX ?

How to convert mmm-yy to mm/dd/yyyy format in unix ? example: Jan-99 to 01/01/1999 Jan-00 to 01/01/2000 Jan-25 to 01/01/2025 Dec-99 to 01/12/1999 Dec-00 to 01/12/2000 Dec-25 to 01/12/2025 YY anything between 00-50 should be 2000-2050 YY anything between 51-99 should be 1951-1999 ...

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)

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Change Date from dd-mmm-yyyy to mm/dd/yyyy

Discussion started by: stringzz

2. Shell Programming and Scripting

converting the date field from dd/mm/yyyy to yyyy/mm/dd

Discussion started by: pareshan

3. Shell Programming and Scripting

Sed: zero-padding dates (or: convert d/m/yyyy to dd/mm/yyyy)

Discussion started by: jgrogan

4. Shell Programming and Scripting

convert d/m/yyyy to YYYY-MM-DD

Discussion started by: nuthalapati