topic Re: shell script help in Operating System - HP-UX

shell script help

Cem Tugrul — Mon, 20 Dec 2010 13:44:12 GMT

Hello Forum,

i have to manipulate many files which rows (cat -vn) always change.But what i have to do;

1 -All these files start (first records start with 811)

2- it is like an array when starts with 811*
then must end with an another record 819*

so this means this is a meaningfull record THEN I have to produce a random file from the original file...Here is an example;

My original_file records like;

811020000060810000056970000100001101213 DE811125281
8120201202504101213000000000000000000001716909789501202504000 0000000171690
8130105077820101213 00 B80037 00000000000000000000000000 000104275
8140381.06225-6007 81062256007 00000000970009900000000017700000000171690 004 0000Y000010 B80037
819030000001000000100000010000001000000100000001716900000000171690000000000000000000000000000000000000000000000
811020000060810000056970000100001101213 DE811125281
8120201202511101213000000000000000000001807509789501202511000 0000000180750
8130105077819101213 00 B80037 00000000000000000000000000 000090000
8140381.15205-5307 81152055307 00000000150009900000000120500000000180750 004 0000Y000010 B80037
819030000001000000100000010000001000000100000001807500000000180750000000000000000000000000000000000000000000000
811020000060810000056970000100001101213 DE811125281
8120201202506101213000000000000000000004044309789501202506000 0000000404430
8130105077815101213 00 B80037 00000000000000000000000000 000178500
8140381.46730-6097 81467306097 00000000170009900000000237900000000404430 004 0000Y000010 B80037
819030000001000000100000010000001000000100000004044300000000404430000000000000000000000000000000000000000000000
811020000060810000056970000100001101213 DE811125281
8120201202505101213000000000000000000000019909789501202505000 0000000001990
8130105077821101213 00 B80037 00000000000000000000000000 000000725
8140382.97485-0825 82974850825 00000000050009900000000003980000000001990 AT 0000Y000010 B80037
819030000001000000100000010000001000000100000000019900000000001990000000000000000000000000000000000000000000000
811020000060810000056970000100001101213 DE811125281
8120201202514101213000000000000000000000024549789501202514000 0000000002454
8130105078047101213 00 B80037 00000000000000000000000000 000000120
8140381.25503-6471 81255036471 00000000060009900000000004090000000002454 004 0000Y000010 B80037
819030000001000000100000010000001000000100000000024540000000002454000000000000000000000000000000000000000000000

SO I have to produced a file witha random filename and this file includes records from my orginal file but it smust be like;

811020000060810000056970000100001101213 DE811125281
8120201202504101213000000000000000000001716909789501202504000 0000000171690
8130105077820101213 00 B80037 00000000000000000000000000 000104275
8140381.06225-6007 81062256007 00000000970009900000000017700000000171690 004 0000Y000010 B80037
81903000000100000010000001000000100000010000000

Any suggestions would be appreciated...

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 13:50:22 GMT

So another file has to be produced again form my orginal file as below with the records; and so on...

811020000060810000056970000100001101213 DE811125281
8120201202506101213000000000000000000004044309789501202506000 0000000404430
8130105077815101213 00 B80037 00000000000000000000000000 000178500
8140381.46730-6097 81467306097 00000000170009900000000237900000000404430 004 0000Y000010 B80037
819030000001000000100000010000001000000100000004044300000000404430000000000000000000000000000000000000000000000

Re: shell script help

James R. Ferguson — Mon, 20 Dec 2010 14:05:03 GMT

Hi Cem:

If what you want extracted is the groups of lines beginning with "811" and ending with "819" you could do:

# perl -ne '/^811/../^819/ and print' file

If you want only the first group matching this range, do:

# perl -ne '?^811?..?^819? and print' file

Regards!

...JRF...

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 14:19:45 GMT

James,
Thank you...yes i want extracted is the groups of lines beginning with "811" and ending with "819" so this means I will produced may files from my orginal_file BUT

perl -ne '/^811/../^819/ my_orginal_file

not worked!!! Did I do anything wrong?

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 14:25:12 GMT

I want to produce many files from my orginal_file starts with 811 and ends 819(with end of line 819)

Re: shell script help

James R. Ferguson — Mon, 20 Dec 2010 14:37:49 GMT

Hi (again):

> yes i want extracted is the groups of lines beginning with "811" and ending with "819" so this means I will produced may files from my orginal_file BUT ...

Since you don't have anything other than successive ranges in your original input you could do:

# perl -ne 'if (?^811?..?^819?) {print};reset,print "\n" if /^819/' file

This would inject a blank line between the ranges. Is that your objective?

Regards!

...JRF...

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 14:46:58 GMT

My original file Ok from your command;
perl -ne 'if (?^811?..?^819?) {print};reset,print "\n" if /^819/' my_orignal_file

BUT this time i need i want to write out these records to new files.That's what i want.

From your command the output is seen;

811020000060810000056970000100001101213 DE811125281
8120201202504101213000000000000000000001716909789501202504000 0000000171690
8130105077820101213 00 B80037 00000000000000000000000000 000104275
8140381.06225-6007 81062256007 00000000970009900000000017700000000171690 004 0000Y000010 B80037
819030000001000000100000010000001000000100000001716900000000171690000000000000000000000000000000000000000000000

811020000060810000056970000100001101213 DE811125281
8120201202511101213000000000000000000001807509789501202511000 0000000180750
8130105077819101213 00 B80037 00000000000000000000000000 000090000
8140381.15205-5307 81152055307 00000000150009900000000120500000000180750 004 0000Y000010 B80037
819030000001000000100000010000001000000100000001807500000000180750000000000000000000000000000000000000000000000

811020000060810000056970000100001101213 DE811125281
8120201202506101213000000000000000000004044309789501202506000 0000000404430
8130105077815101213 00 B80037 00000000000000000000000000 000178500
8140381.46730-6097 81467306097 00000000170009900000000237900000000404430 004 0000Y000010 B80037
819030000001000000100000010000001000000100000004044300000000404430000000000000000000000000000000000000000000000

811020000060810000056970000100001101213 DE811125281
8120201202505101213000000000000000000000019909789501202505000 0000000001990
8130105077821101213 00 B80037 00000000000000000000000000 000000725
8140382.97485-0825 82974850825 00000000050009900000000003980000000001990 AT 0000Y000010 B80037
819030000001000000100000010000001000000100000000019900000000001990000000000000000000000000000000000000000000000

811020000060810000056970000100001101213 DE811125281
8120201202514101213000000000000000000000024549789501202514000 0000000002454
8130105078047101213 00 B80037 00000000000000000000000000 000000120
8140381.25503-6471 81255036471 00000000060009900000000004090000000002454 004 0000Y000010 B80037
819030000001000000100000010000001000000100000000024540000000002454000000000000000000000000000000000000000000000

AND GOES ON...But i need that each [811-819] block has to be >> a new filename

Re: shell script help

James R. Ferguson — Mon, 20 Dec 2010 14:53:41 GMT

Hi (abain):

> But i need that each [811-819] block has to be >> a new filename

Oh, OK, that's easily done. Try this:

# cat ./rangedivide
#!/usr/bin/perl
use strict;
use warnings;
my ( $fh, $file, $nbr );
$file = "myoutput";
$nbr = 1;
while (<>) {
if (?^811?..?^819?) {
if (/^811/) {
open( $fh, '>', "$file.$nbr" ) or die;
}
print $fh $_;
}
if (/^819/) {
close $fh;
reset;
$nbr++;
}
}
1;

...run as :

# ./rangedivide file

As written, your output file(s) will be named "myoutput.n" where 'n' ranges from one to the number of ranges seen.

Regards!

...JRF...

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 15:12:28 GMT

./rangedivide.sh: 0403-057 Syntax error at line 1 : `(' is not expected.

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 15:16:08 GMT

oppps...sorry..my fault.
That's what i want.
Thank you very much....

Re: shell script help

James R. Ferguson — Mon, 20 Dec 2010 15:19:38 GMT

Hi:

> ./rangedivide.sh: 0403-057 Syntax error at line 1 : `(' is not expected.

Did you cut-and-paste what I posted? Also, don't name the script "*.sh". That infers a shell script as opposed to a Perl one. If you want to use a file extension, use '*.pl'.

Regards!

...JRF...

Re: shell script help

Cem Tugrul — Mon, 20 Dec 2010 15:23:12 GMT

Thank's so much James...

Re: shell script help

Steven Schweda — Mon, 20 Dec 2010 16:20:07 GMT

> [...] That infers a shell script [...]

No, it implies a shell script. _You_ might
infer that it's a shell script. (But what's
in the file matters more than the name in
this case.)

Re: shell script help

James R. Ferguson — Mon, 20 Dec 2010 17:00:31 GMT

Hi:

@Steven: > [...] That infers a shell script [...]
No, it implies a shell script.

Well, there is the usage notes for the word 'infer' here:

http://dictionary.reference.com/browse/infer

But then this would buttress your suggested distinction :-)

http://www.wsu.edu/~brians/errors/imply.html

Regards!

...JRF...