Re: help regex replacement

[zilore mumba]

---

From: zilore mumba <zmu...@yahoo.com>
To: "Thurn, Martin (TASC)" <martin.th...@tasc.com>
Sent: Monday, April 30, 2012 10:29 PM
Subject: Re: help regex replacement

My mail should have read
sample input

start_date = '2012-04-29_00:00:00', '2012-04-29_00:00:00', '2012-04-29_00:00:00',
 end_date   = '2012-05-01_00:00:00', '2012-05-01_00:00:00', '2012-05-01_00:00:00',

The model would have run on 29th May,  to produce a 48hour forecast to 1st May 2012.

When I run it on 30th May it should produce a 48hour forecast to 2nd May, i.e. change the entries above so that start_date is 30th and end_date is 2nd May as below.

sample output
start_date = '2012-04-30_00:00:00', '2012-04-30_00:00:00', '2012-04-30_00:00:00',
 end_date   = '2012-05-02_00:00:00', '2012-05-02_00:00:00', '2012-05-02_00:00:00',

---

From: "Thurn, Martin (TASC)" <martin.th...@tasc.com>
To: zilore mumba <zmu...@yahoo.com>
Cc: "Thurn, Martin (TASC)" <martin.th...@tasc.com>
Sent: Monday, April 30, 2012 7:45 PM
Subject: RE: help regex replacement

1.  Reduce your sample program to the MINIMUM amount of code that reproduces your problem.  Especially, get rid of all file-open-read-write and get rid of looping. 

2.  Include some sample INPUT data (and clearly indicate).

3.  Include some sample desired OUTPUT data (and clearly indicate).

4.  What is your intention when writing “start_date .._00” in your regex?  I think you don’t understand what “.” means in a regex. 

5.  What is your intention when writing “end_date * _00” in your regex?  I think you don’t understand what “*” means in a regex. 

 

 

 - - Martin

 

From: activeperl-boun...@listserv.activestate.com [mailto:activeperl-boun...@listserv.activestate.com] On Behalf Of zilore mumba
Sent: Monday, April 30, 2012 13:15
To: activeperl@listserv.activestate.com
Subject: help regex replacement

 

Hello Kind folks,
I have to automate a modle namelist by replacing the start and end
date of the model run as per lines below.

start_date = '2012-04-26_00:00:00','2012-

04-26_00:00:00',
end_date   = '2012-04-26_09:00:00','2012-04-26_09:00:00',

I have been checking the documentation on how to te replace a regex
starting with "^x" and ending in $y", no success.
I tried as below:

 my @now = localtime;
 my $shour = strftime "%H",@now;
 my $sday = strftime "%d",@now;
 my $smonth = strftime "%m",@now;
 my $syear = strftime "%Y",@now;

 my ($eday, $emonth, $eyear) = (localtime (time() + 172800)) [ 3,4,5 ];
 $eyear = sprintf( "%04d", $eyear + 1900 );
 $emonth = sprintf( "%02d", ++$emonth);

 open (IN, "<$in_fn") or die "open '$in_fn: failed $! ($^E)";
  my @fields = <IN>;
  close IN;
for (@fields)
    {
     s/^start_date .._00:00:00',$/start_date =
'${syear}-${smonth}-${sday}_00:00:00','${syear}-${smonth}-${sday}_00:00:00',
     '${syear}-${smonth}-${sday}_00:00:00',/g;
      exit;

     s/^end_date * _00:00:00',$/end_date =
'${eyear}-${emonth}-${eday}_00:00:00','${eyear}-${emonth}-${eday}_00:00:00',
     '${eyear}-${emonth}-${eday}_00:00:00',/g;
    }
  open (OUT, ">$out_fn") or die "open '$out_fn: failed $! ($^E)";
   print OUT "@fields\n";
  close OUT;

Does such a way of replacing exist?
Help will be appreciated.

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