perl - perl 用普通括号替换大括号

Question

我需要编写一个脚本来查看文件并用普通括号替换任何大括号。到目前为止，我有：

use strict;
use warnings;
open(INFILE,"<rscore") || die "Couldn't open rscore for reading!\n";
open(OUTFILE,">rscore.new") || die "Couldn't open rscore.new for writing!\n";
while(<INFILE>){
  $_ =~ s/{/(/gi; #g for every occurrence, i for case-insensitive
  print OUTFILE $_;
}
close INFILE;
close OUTFILE;
rename("rscore.new","rscore") || die "Couldn't rename the new file!\n"; 

并收到以下错误：

syntax error near line 10 near insensitive print.

这可能是一件愚蠢的事情。此外，如果有更有效的方法来做到这一点（我确信有），我愿意接受建议。

Answer 1

这是一个 perl 单行代码。

$ perl -lpe 'tr |{}|()|' <infile >outfile

Answer 2

试试这个：

$string =~ tr/{}/()/;

Answer 3

This is a perl one liner for inplace editing the file

perl -pi -e 's/{/(/g;s/}/)/g' your_file

This below will print the output on console

perl -p -e 's/{/(/g;s/}/)/g' your_file