Hi Alan , All
Just in case the data structures and the sort routines intimidate you , the
following hack works too :
#usage : [perl.exe \ perl] reorder.pl infilepath reorderedOutFilePath
open INFILE, @ARGV[0] or print "can't open INFILE!" ;
open OUTFILE, ">@ARGV[1]" or print "can't create OUTFILE!" ;
while(<INFILE>)
{
if(/^[0-9]{2}:/)
{
print OUTFILE "$_";
}
else
{
push @temp , $_ ;
}
}
close INFILE;
foreach $element(@temp)
{
print OUTFILE "$element";
}
close OUTFILE;
Just copy paste the above program and it'll do the trick :)
-aman.
-----Original Message-----
From: Alan C. [mailto:acummingAT@;cwnetDOT.com]
Sent: Friday, November 08, 2002 2:53 PM
To: [EMAIL PROTECTED]
Subject: sort two patterns, control structure help needed
Hello,
This must be easy. But I've not yet enough experience with Perl's control
structures.
perl mysort.pl infile.txt > outfile.txt
The stack of numbers with colons below reside within infile.txt
120:2
126:2
13:15
140:3
14:3
141:3
14:3
15:11
My task or goal is to get each of them lines over into outfile.txt (as re
ordered lines) with all of the \d\d: (2 digits then colon) up at top of file
then with the 3 digits colon underneath
How to set up a twice through instead of only 1 pass through of the
infile.txt? (and get all of the 2digits on 1st pass, then get all of the 3
digits on second pass)? Or, same task/goal done even yet a different way is
ok too.
#!/perl/bin/perl -w
# mysort.pl
while ($line = <>) {
if ($line =~ m/\b\d\d(?=:)/) {
print $line;
if ($line =~ m/\b\d\d\d(?=:)/) {
print $line;
}
}
}
That above gets/snags all the 2 digits only. There's no 3 digits to
outfile.txt
--
Thanks. Alan.
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