This sounds amazingly powerful and flexible. Thanks a lot. I will try it
asap.
The only problem is that I will need to learn Perl if I want to be able to
write such scripts…
Le samedi 6 octobre 2012 01:25:58 UTC+2, eremita a écrit :
>
>
> On 5 Oct 2012, at 16:18, jmichel <jmi.m...@gmail.com <javascript:>>
> wrote:
>
> > I have a file consisting of groups of lines (unknown number of lines in
> each group).
> > Each line begins by a 6 digit number, followed by an unknown sequence of
> words and numbers.
> > Consecutive lines starting with the same number form a group.
> > My problem is to combine lines from each group into a single line,
> keeping only the first occurrence of the distinctive number.
> > I have been able to "find" groups using the pattern
> > (\d{6})(.+)(?:\r\1(.+))+
> > However, this does not appear to store the expressions matching the
> inner parentheses into separate variables.
> > Is there a way to achieve the desired replacement using grep?
>
> Using regular expressions yes, but you need a routine. If you put a file
> containing
> this Perl Script in ~/LibraryApplication Support/BBEdit/Text Filters, it
> will do what
> you want. Open the Text Filters palette from the Window menu and you will
> see the
> filter. Double-click it or click on Run or, if its a frequent task,
> assign a shortcut to the
> script.
>
> Save this as ???.pl
>
> #!/usr/bin/perl
> my %hash;
> my $six_digits = "[0-9]{6}";
> my $remaining_text = ".*";
> my $delimiter = ""; # or ", " for example
> while (<>) {
> if ( /^($six_digits)($remaining_text)/ ) {
> $hash{$1} .= $2 # append the text after the 6 digits
> }
> }
> for (sort {$a<=>$b} keys %hash) {
> print "$_$delimiter$hash{$_}\n"
> }
>
> #JD
>
>
>
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