Regex match count of characters that are separated by non

 

I want to count characters, but they may be separated by characters that do not match.

Here is an example. I want to match a text that has 10 or more word-characters. It may include spaces but i don't want to count the spaces.

Should not match: "foo bar baz" (should count 9)
Should not match: "a a" (should count 2)
Should match: "foo baz bars" (should count 10, match whole string)

This is what i came up with, but it counts the whole thing: 

((?<=s)*w(?=s)*){10}

Edit I do not want to include spaces for counting. Sorry I edited this a few times, I didn't describe it correctly.

Any ideas on this?

Hey I think this would a simple but working one: 

( *?[0-9a-zA-Z] *?){10,}

Breaking the regex down:

  • ( *? --------It can start with space(s)
  • [0-9a-zA-Z] -Followed with the alphanumeric values
  • *?) ---------It can end with space(s)
  • {10,} -------Matches this pattern 10 or more times

    • Key: When I look at the count for regexes, it applies to the group, ie, the things in the brackets " () ", this case, multiple spaces followed ONE from the alphanumeric values followed by spaces are still counted as one match. Hope it helps. :)

    使用每个单个字符占用空格的组，并计算组： 

    ^(s*w){10,}s*$

    Using JS: Remove the spaces, then do the w check 

    'foo baz barz'.replace(/ /g,'').match(/w{10,}/) != null //true
'foo bar baz'.replace(/ /g,'').match(/w{10,}/) != null //false

    Match phone numbers in text: 

    var test = 'something foo baz barz 07999-777-111 and 01234 567890 01234567890 some more'.match(/(((?0d{4})?[ -]?d{3}[ -]?d{3})|((?0d{3})?[ -]?d{3}[ -]?d{4})|((?0d{2})?[ -]?d{4}[ -]?d{4}))([ -]?#(d{4}|d{3}))?/g);
//result: ["07999-777-111", "01234 567890", "01234567890"]
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