i need to output the text of lines 1,4,5 to standard error stream.lines 2 & 3 to standard output.at end i need to output how many errors it got.
example of a log file may be:
ERROR - no error found.
I found ERROR in line 9.
Nothing to do.
ERROR not excepted.
ERROR: your file is not available.
several input files might be called like this
cat logfile1 logfile2 | myprogram.pl

here is my code:

use strict;
sub trim($)
{
my $string = shift;
$string =~ s/^\s+//;
return $string;
}

my $line;
my $count = 0;

open (LOG,"logdata") or die "Unable to open logfile";

while($line = <LOG>){
$line = trim($line);
if($line !~ /^ERROR/ ){
print STDOUT "$line.\n";
next;
}
else{
print STDERR "$line.\n";
$count++;
}

}
print "There were $count errors.\n";
close(LOG);

i dont know how to take the filename from argument

Which part of http://perldoc.perl.org/ have you already read ? If none, I suggest to start from the links in the leftmost column.

Regarding "argument" - if you mean command line arguments, start from http://perldoc.perl.org/perlvar.html and look for occurrences of "argument", "command" substrings.

ARGV but really cant figure how to use it

@ARGV is an array. So what is exactly the problem using the array - compared to other arrays ?

This are different subjects.
Your example with cat would mean that the data come from a pipe which has nothing to do with arguments

Filenames as arguments require indeed the @ARGV array.

Hint: in the first case the perlscript "doesn't know" about the filenames, in the second case the script handles the filenames.

Markus

Regardless of Perl in case of

'bar' can not know what <some_files> are - unless this info is explicitly passed through STDOUT by 'foo'.

...

Well, probably it can - in a very unportable and convoluted way, looking into process table, etc.

My intention was to point the OP to the difference between @ARGV and <STDIN>. He would be better off to consider before writing the script how the logfile(s) should be passed to it.

Markus

It's a good point.

Obviously your program is required to handle the case in which no arguments are provided and the input comes through stdin.

Some program need to also handle the case in which the arguments provide the input filename. It is easy enough to check whether you have an input filename specified as an argument and if not, default to taking input from stdin.

But start by making sure you understand the assignment. Are you only required to handle input from stdin, or are you required to handle input either way (from stdin or from a specified file)?

i am required to handle input from stdin and not a specified file. i have provided the code for a specified file but it will not be a specified file.it will be the files that the user will provide..

I updated my code to take multiple input

Here is my updated code
#!/usr/bin/perl -w
use strict;
sub trim($)
{
my $string = shift;
$string =~ s/^\s+//;
return $string;
}

FILE:foreach(@ARGV){

my $line;
my $count=0;

open (FILE,$_) or die "Unable to open logfile";

while($line = <FILE>){
$line = trim($line);
if($line !~ /^ERROR/ ){
print STDOUT "$line.\n";
next;
}
else{
print STDERR "$line.\n";
$count++;
}


}
print "There were $count errors.\n";
close(FILE);
}

But When I execute, cat logfile1 logfile2 | myprogram.pl

I am getting error
-bash: zprogram11.pl: command not found

Thanks for all the help...

You say that, but then you act like you don't know it.
I think you should reread some of the answers you already received.

There is nothing to get from ARGV. Your program is not supposed to know or care about the filename that was the source for stdin.

You don't open any file to read stdin. STDIN is effectively a file that is already open when your program starts. You just pull text from <STDIN>

stdin is one input. The fact that it might have come from multiple files through cat is invisible to your program.

I have no clue why that command would give that error. I'm not an expert in either bash or perl, so maybe I'm misunderstanding some key detail, but it looks to me like you are giving us incorrect or incomplete information.

Is 'zprogram11.pl' executable ? Is current working directory in $PATH ? Why won't you call your script as

?

over here myprogram.pl = zprogram11.pl

I can call ./zprogram11.pl but i need to process multiple log file. That is why i need to pass those file name.

I need to pass it like this

cat logFile1 logFile2 | zprogram11.pl

Normally when your homefolder is not in $PATH, you will have to executeMarkus