awk 命令使用正则表达式替换字符串

Question

I have a file with content as below, there are no new lines:

1234%#@@!#12346@!#4562366@!#@!#@!#+++++456789%#@@!#12346@!#4562366@!#@!#@!#

i want the file to be modified as

1234%#@@!#@!#@!#@!#@!#+++++456789%#@@!@!#@!#@!#@!#

ie anything coming between delimiter @!# and the next delimiter @!# should be made null.

Any pointers to this will be useful. Thanks a lot

Answer 1

Here's some verbose perl:

use strict;

# read the file, the filename is given as a parameter to the perl script
open my $f, "<", shift; 
my $data = <$f>; 
close $f;

# the possible delimeters
my @delims = qw( @!# %#@ +++++ );

# split the data into an array of intermixed fields and delimiters
my $delim_re = join "|", map {quotemeta} @delims;
my @fields_and_delims = split /($delim_re)/, $data;;

for (my $i=2; $i < @fields_and_delims; $i+=2) {
    # empty this element if between "@!#" delimiters
    $fields_and_delims[$i] = "" if $fields_and_delims[$i-1] eq $delims[0] 
                               and $fields_and_delims[$i+1] eq $delims[0];
}

# the output, with no trailing newline
print join "", @fields_and_delims;

Then

$ printf "%s" "1234%#@@!#12346@!#4562366@!#@!#@!#+++++456789%#@@!#12346@!#4562366@!#@!#@!#+++++‌​++201546987@!#123456@!#4562366@!#@!#@!#+++++++" > file
$ perl script.pl file
1234%#@@!#@!#@!#@!#@!#+++++456789%#@@!#@!#@!#@!#@!#+++++‌​++201546987@!#@!#@!#@!#@!#+++++++