Re: Parsing list

[ww]

I think I understand that you have some data (we'll discuss list later) comprised of something like this:

user:pass user1:pass1 use:pas
         ^           ^

What is the separator (ie, the character my caret's point to)? A space, semi-colon, LF, null or something else (including no separator at all)?

Now, as to "list" -- what's shown is a list. But is that what your source looks like (minus the assignment and the rest of the Perl-ish elements, or does your source look like this?

user:pass
user1:pass1
use:pas

or, perhaps, like this (markup typo fixed here; thanks, kcott!)

user:pass,user1:pass1,use:pas

For the first, a simple split statement will do very nicely; the second begs that you read the data into an array and use for $_(@array){.... while the third looks like an invitation to use the appropriate csv module to put you data into a format you know how to mung.

Update (illustration):

#!/usr/bin/perl
use 5.016;
use warnings;
# NOT JUST BTW: this will not work without modification for some possi
+ble forms of data!

my $list = "user:passer use:pass user1:pass1";

say $list;
say "That was \$list; these are the elements of \@arr:";

my @arr= split / /, $list;

for (@arr) {
    say $_;
    my $current_element = $_;
    my ($first, $second) = split /:/, $current_element;
    say "\$first: $first & \$second: $second";
}

Output:

user:passer use:pass user1:pass1
That was $list; these are the elements of @arr:
user:passer
$first: user & $second: passer
use:pass
$first: use & $second: pass
user1:pass1
$first: user1 & $second: pass1