Regular Expressions 101

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An explanation of your regex will be automatically generated as you type.
Detailed match information will be displayed here automatically.
  • All Tokens
  • Common Tokens
  • General Tokens
  • Anchors
  • Meta Sequences
  • Quantifiers
  • Group Constructs
  • Character Classes
  • Flags/Modifiers
  • Substitution
  • A single character of: a, b or c
    [abc]
  • A character except: a, b or c
    [^abc]
  • A character in the range: a-z
    [a-z]
  • A character not in the range: a-z
    [^a-z]
  • A character in the range: a-z or A-Z
    [a-zA-Z]
  • Any single character
    .
  • Alternate - match either a or b
    a|b
  • Any whitespace character
    \s
  • Any non-whitespace character
    \S
  • Any digit
    \d
  • Any non-digit
    \D
  • Any word character
    \w
  • Any non-word character
    \W
  • Non-capturing group
    (?:...)
  • Capturing group
    (...)
  • Zero or one of a
    a?
  • Zero or more of a
    a*
  • One or more of a
    a+
  • Exactly 3 of a
    a{3}
  • 3 or more of a
    a{3,}
  • Between 3 and 6 of a
    a{3,6}
  • Start of string
    ^
  • End of string
    $
  • A word boundary
    \b
  • Non-word boundary
    \B

Regular Expression
No Match

/
/
g

Test String

Code Generator

Language

Generated Code

re = /(?:(?:^|\G(?<!^)[^'"]*")[^"]*+(?:"[^"']*"[^"]*)*+"|\G(?<!^))[^'"]*+(')/ str = '" This is a random sentence \' with \'a quote, the quote should match" " There is no quote here thats the problem" Anything here should not match but now it does: \' . " Some text " some more text " this is a quote : \' that should match" " When there is a quote (\') here, the other one does not work : " \' and that\'s perfect " even if you remove the first one this : " (\') " will make it work because of the greedy ( I think ) but ifyou remove those between parenthesis, the other one is matching as of now, which I do not want to happen." Another example would be this one : The following should not work, but it does "This is being quoted" not this: (\') " and this is also being quoted " "\'" ' # Print the match result str.scan(re) do |match| puts match.to_s end

Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Ruby, please visit: http://ruby-doc.org/core-2.2.0/Regexp.html