DISCLAIMER: This is UNTESTED ... It's just something to try...
$OrgLine = "adshe ms0e sad qweoic,m qwod x0 vndu qiudb siu .... ";
While length of $OrgLine > 70 {
Match on =~ /(.{0,70})\s/
Print $1
Remove $1 from $OrgLine ;
}
Print remainder of $OrgLine;
So, the regex would return (again, this is untested) $1 with up to 70
characters that terminated with a white space.
Just a thought ...
-rjf
> -----Original Message-----
> From: M z [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, March 06, 2002 10:58
> To: [EMAIL PROTECTED]
> Subject: help me with a better backstepper?
>
>
> Hi
>
> I've written a little program to analyze lines that
> are longer than 70 characters, and if so, break at
> position 70 (if it is whitespace), if not whitespace
> (i.e. is in the middle of a word) it steps back to the
> first whitespace and breaks there.
>
> However, I think there is probably a better way to do
> this. Can someone either suggest an alternate route
> or improve this snippet?
>
> ps. if you do suggest something cryptic, please detail
> for me, as I am still new!!! thanks
>
> while(<F>) {
> chomp;
> if (length $_ < 70){
> print F1 "$_\n";
> }
>
> else {
>
> split(//, $_);
> if ($_[69] eq " ") {
> s/^(.{69}) (.*)$/$1\n+ $2\n/;
> print F1;
> }
> elsif($_[69] ne " " && $_[68] eq " ")
> {
> @_ = ();
> split(//, $_);
> s/^(.{68}) (.*)$/$1\n+ $2\n/;
> print F1 $_;
> }
> elsif($_[69] ne " " && $_[68] ne " "&&
> $_[67] eq " ") {
> @_ = ();
> split(//, $_);
> s/^(.{67}) (.*)$/$1\n+ $2\n/;
> print F1 $_;
> }
>
> .....
>
> # this is the strategy to back up 30 spaces
> eventually,
> but I don't want to have 30 elsif statements, although
> this works, it seems that a more experienced perl
> programmer could write this with much less code...
>
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