Question

I was trying to find strings out which is followed by only “..”,but couldn’t get that :

["..ab","...cc","..ps","....kkls"].each do |x|
puts x if /../.match(x)
end
..ab
...cc
..ps
....kkls
=> ["..ab", "...cc", "..ps", "....kkls"]

["..ab","...cc","..ps","....kkls"].each do |x|
puts x if /(.)(.)/.match(x)
 end
..ab
...cc
..ps
....kkls
=> ["..ab", "...cc", "..ps", "....kkls"]

Expected output:

["..ab","..ps"]

Answer 1

What you want is

/^\.\.(?!\.)/

The caret ^ at the beginning means match the beginning of the string; periods must be escaped by a backslash as \. because in regular expressions a plain period . matches any character; the (?!\.) is a negative look-ahead meaning the next character is not a period. So the expression means, “at the beginning of the string, match two periods, which must be followed by a character which is not a period.”

>> /^\.\.(?!\.)/.match "..ab"
=> #<MatchData "..">
>> /^\.\.(?!\.)/.match "...cc"
=> nil