regex help 3

[travs69]

my %month = ( 'JAN' => 1, 'FEB' => 2, 'MAR' => 3, 'APR' => 4, 'MAY' => 5, 'JUN' => 6, 'JUL' => 7, 'AUG' => 8, 'SEP' => 9, 'OCT' => 10, 'NOV' => 11, 'DEC' => 12);
$col[5] = 'MAY052010';

($c{month}, $c{day}, $c{year}) = $col[5] =~ /(.{3})(\d{2})(\d{4})/;
$c{month} = $month{$c{month}} #or $month{$1}
print "$c{month}\n";

I'd really like to have that done in one line.. just seems like I'm missing something easy.

Thanks

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Travis - Those who say it cannot be done are usually interrupted by someone else doing it; Give the wrong symptoms, get the wrong solutions;

 

[rharsh]

I think you're just missing:

($c{month}, $c{day}, $c{year}) = [red][b]([/b][/red]$col[5] =~ /(.{3})(\d{2})(\d{4})/[red][b])[/b][/red];

 

[travs69]

the regex part works,

but I am having to do the hash convert in the next line.. I'm trying to figure out a way to do it all in one line.

Thanks

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Travis - Those who say it cannot be done are usually interrupted by someone else doing it; Give the wrong symptoms, get the wrong solutions;

 

[stevexff]

Given that the date is fixed format, why use a regex at all? Just substring it...

Steve

[small]"Every program can be reduced by one instruction, and every program has at least one bug. Therefore, any program can be reduced to one instruction which doesn't work." (Object:erlDesignPatterns)[/small]

 

[travs69]

Never thought of it truthfully, can you do a substring like a regex and get multiple things back or just run 3 seperate substrings?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Travis - Those who say it cannot be done are usually interrupted by someone else doing it; Give the wrong symptoms, get the wrong solutions;

 

[prex1]

Not sure to understand what you want to do. Assuming this is:
1)extract month,day,year from date
2)if month is in 3 letters, translate it into numeric.
If that is correct, supposing the month would be in 2 figures (when in this version, however your regex is not correct in this case), then you can test the length of the variable to determine which one you have and do

($c{month},$c{day},$c{year})=(length($col[5])>8?$month{substr($col[5],0,3)}:substr($col[5],0,2),substr($col[5],-6,2),substr($col[5],-4));

If month, when numeric, is in 3 figures, then you can do

($c{month},$c{day},$c{year})=($col[5]=~/^(\d{3})/?$1:$month{substr($col[5],0,3)},substr($col[5],3,2),substr($col[5],5));

Could not find a solution in one line with regexes, but substr should also be faster.
Of course I assume you don't need any error checking on the sanity of your string...

Franco

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[rharsh]

Oh, that's what you meant by one line. How about something like:

($c{month}, $c{day}, $c{year}) = (map {defined $month{$_} ? $month{$_} : $_} $col[5] =~ /(.{3})(\d{2})(\d{4})/);
print "$c{month}\n";

 

[travs69]

Thanks guys.. I'm going to play with that some tonight..

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Travis - Those who say it cannot be done are usually interrupted by someone else doing it; Give the wrong symptoms, get the wrong solutions;