How to find string which is started with ".."?

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"]

Do you mean strings beginning with exactly 2 (and no more) periods? — Reinstate Monica -- notmaynard, Feb 04 '13 at 20:54
Remember that . is a special character, you need to escape it in regex: \. — Haile, Feb 04 '13 at 20:54
@iamnotmaynard - Yes I want strings only 2 periods at beginning. — DoLoveSky, Feb 04 '13 at 20:56

Reinstate Monica -- notmaynard · Accepted Answer · 2013-02-04T21:04:56.627

5

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

edited Feb 04 '13 at 21:04

answered Feb 04 '13 at 20:59

Reinstate Monica -- notmaynard

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Could you explain the code of your one how works in such matching? – DoLoveSky Feb 04 '13 at 21:01
1

\. because you have to escape the dot (is like * for regexp, selects anything). So basically is "string_start:dot dot (not dot) anything else" read some regexp basics is easy. Also some regexp negation: http://stackoverflow.com/questions/3977455/how-do-i-turn-any-regex-into-an-complement-of-itself-without-complex-hand-editin – MGP Feb 04 '13 at 21:03

score 2 · Answer 2 · answered Feb 04 '13 at 20:58

2

Try selecting on /^\.\.[^\.]/ (starts with two dots and then not a dot).

ss = ["..ab","...cc","..ps","....kkls"]
ss.select { |x| x =~ /^\.\.[^\.]/ } # => ["..ab", "..ps"]

answered Feb 04 '13 at 20:58

maerics

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can you explain your code,how it ignoring strings having started with more than 2 `.` periods? – DoLoveSky Feb 04 '13 at 21:08

score 2 · Answer 3 · answered Feb 04 '13 at 21:02

Try using /^\.{2}\w/ as the regular expression.

A quick explanation: ^ means the start of the string. Without this, it can match dots that are found in the middle of the string.

\. translates to . -- if you use the dot on its own, it will match any non-newline character

{2} means that you're looking for two of the dots. (you could rewrite /\.{2}/ as /\.\./)

Finally, the \w matches any word character (letter, number, underscore).

A really good place to test Ruby regular expressions is http://rubular.com/ -- it lets you play with the regex and test it right in your browser.

mu is too short · Answer 4 · 2019-03-23T17:45:29.143

1

You don't need regex for this at all, you can just extract the appropriate leading chunks using String#[] or String#slice and do simple string comparisons:

>> a = ["..ab", "...cc", "..ps", "....kkls", ".", "..", "..."]
>> a.select { |s| s[0, 2] == '..' && s[0, 3] != '...' }
=> ["..ab", "..ps", ".."]

edited Mar 23 '19 at 17:45

answered Feb 04 '13 at 21:41

mu is too short

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The point is that you don't have to use a regex when simple string operations will do. – mu is too short Feb 04 '13 at 21:50

score 0 · Answer 5 · answered Feb 04 '13 at 20:58

0

Maybe this:

["..ab","...cc","..ps","....kkls"].each {|x| puts x if /^\.{2}\w/.match(x) }

Or if you want to make sure the . doesn't match:

["..ab","...cc","..ps","....kkls"].each {|x| puts x if /^\.{2}[^\.]/.match(x) }

answered Feb 04 '13 at 20:58

rainkinz

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How to find string which is started with ".."?

5 Answers5