Re: Simple regex to remove all but a list of words

The right way to prevent SQL injection is to use DBI with placeholders instead of interpolating values into SQL.

If you still want to use a regex, it might work like this:

# with regex:
my $regex = join '|', @good_words;
if (param('evil') =~ m/\A(?:$re)\z/){
   print "good param\n";
}

# with a hash:
my %whitelist;
@whitelist{@good_words} = (1) x @good_words;
if ($whitelist{param('evil')}){
  print "good param\n";
}
[download]

Comment on Re: Simple regex to remove all but a list of words Download Code

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Re^2: Simple regex to remove all but a list of words by FunkyMonk (Bishop) on Jun 04, 2008 at 21:53 UTC
`my %whitelist; @whitelist{@good_words} = (1) x @good_words;` [download] `my %whitelist = map { $_ => 1 } @good_words;` [download] Is a better idiom (IMO, of course), because I hate having to use a declaration followed by an initialisation. Unless I state otherwise, all my code runs with strict and warnings	[reply] [d/l] [select]
Re^3: Simple regex to remove all but a list of words by moritz (Cardinal) on Jun 04, 2008 at 22:16 UTC
So we can start the battle of style: would you rather have a clumsy, useless loop (or map), or rather two clumsy, lines, one with declaration, one with initialization? Perl 6 to the help, if you don't mind wrapping your head around function style a bit, you can use the zip operator `Z`: `pugs> my @good = <a b c> ("a", "b", "c") pugs> @good Z 1 xx @good (("a", 1), ("b", 1), ("c", 1))` [download] Pugs doesn't flatten the list correctly, so you can't assign it to a hash in pugs, but a correct implementation will allow that. (Notice also the beauty of not needing any grouping parenthesis, because the precedence levels are just right(tm)). `# Perl 6: my %hash = @good Z 1 xx @good` [download]	[reply] [d/l] [select]