在Perl cript中运行sed命令时出错

Question

I am trying to run the following command in perl script :

#!/usr/bin/perl

my $cmd3 =`sed ':cycle s/^\(\([^,]*,\)\{0,13\}[^,|]*\)|[^,]*/\1/;t cycle' file1 >file2`;
 system($cmd3);

but is not producing any output nor any error. Although when I am running the command from command line it is working perfectly and gives desired output. Can you guys please help what I am doing wrong here ? Thanks

Answer 1

system()不返回输出，仅返回exit状态。

Answer 2

To see the output, print $cmd3 .

my $cmd3 = `sed ':cycle s/^\(\([^,]*,\)\{0,13\}[^,|]*\)|[^,]*/\1/;t cycle' file1 >file2`;
print "$cmd3\n";

Edit:
If you want to check for exceptional return values, use CPAN module IPC::System::Simple :

use IPC::System::Simple qw(capture);

my $result = capture("any-command"); 

Answer 3

Running sed from inside Perl is just insane.

#!/usr/bin/perl

open (F, '<', "file1") or die "$O: Could not open file1: $!\n";
while (<F>) {
    1 while s/^(([^,]*,){0,13}[^,|]*)\|[^,]*/$1/;
    print;
}

Notice how Perl differs from your sed regex dialect in that grouping parentheses and alternation are unescaped, whereas a literal round parenthesis or pipe symbol needs to be backslash-escaped (or otherwise made into a literal, such as by putting it in a character class). Also, the right-hand side of the substitution prefers $1 (you will get a warning if you use warnings and have \\1 in the substitution; technically, at this level, they are equivalent).

man perlrun has a snippet explaining how to implement the -i option inside a script if you really need that, but it's rather cumbersome. (Search for the first occurrence of " LINE: " which is part of the code you want.)

However, if you want to modify file1 in-place, and you pass it to your Perl script as its sole command-line argument, you can simply say $^I = 1; (or with use English; you can say $INPLACE_EDIT = 1; ). See man perlvar .

By the way, the comment that your code "isn't producing any output" isn't entirely correct. It does what you are asking it to; but you are apparently asking for the wrong things.

Quoting a command in backticks executes that command. So

my $cmd3 = `sed ... file1 >file2`;

runs the sed command in a subshell, there and then, with input from file1 , and redirected into file2 . Because of the redirection, the output from this pipeline is nothing, ie an empty string "" , which is assigned to $cmd3 , which you then completely superfluously attempt to pass to system .

Maybe you wanted to put the sed command in regular quotes instead of backticks (so that the sed command line would be the value of $cmd3 , which it then makes sense to pass to system ). But because of the redirection, it would still not produce any visible output; it would create file2 containing the (possibly partially substituted) text from file1 .