Regular expression hell

[chrisjasp]

Hello All,
I'm trying to replace some specific text in like 600 files in a directory. I've been working with sed but i'm open to anything that works. Basically I want to replace a period or periods that occur between to other certain characters on every line in the file. For example this line:

[5.6h>VistaRelease5.6h] | blah blah
[6.0.1>VistaRelease6.0.1] | blah blah .

So I want to replace only the periods that are between the ">" and the "]" with an underscore. I have been able to replace the ">" to the "]" with an underscore producing [6.0.1_ | blah blah. But obviously this is not what i'm looking for. I used ..sed 's/>.*[]]/_/'

Any help would be appreciated.

[wje_lf]

I don't think this can be done with a regular expression. Try this Perl script as a filter:

#!/usr/bin/perl

while($in_line=<STDIN>)
{
  chomp($in_line);

  # Don't change any lines which do not contain > followed somewhere by ].

  if($in_line!~/\>.*\]/)
  {
    print("$in_line\n");

    next;
  }

  $first_part=$in_line;

  $first_part=~s/^([^>]*)(.*)$/$1/;
  $in_line   =~s/^([^>]*)(.*)$/$2/;

  $second_part=$in_line;

  $second_part=~s/^([^\]]*)(.*)$/$1/;
  $in_line    =~s/^([^\]]*)(.*)$/$2/;

  $second_part=~s/\./\_/g;

  print("$first_part$second_part$in_line\n");
}

I've briefly tested it, and it seems to work.

If you prefer awk to Perl, there's a good chance that an awk expert will be along to give an awk alternative.

And, of course, there's also a good chance that someone will come along with a regular expression that does the job handily.

But in the meantime, this script will work.

Depending on your distribution, you may have to change its first line, though:

#!/usr/bin/perl

to reflect the location of Perl on your system. To find that, do this at the command line:

which perl

Hope this helps.

[chrisjasp]

Thank you. It works perfectly. I made a little bash script to iterate all 600 files applying your perl filter and it works great. Thank you very much.