CODEWITHSUNDEEP
regexdartflutter
Nitneuq
Nitneuq

Reputation: 5012

dart regex remove space phone

I tried all this regex solution but no match REGEX Remove Space

I work with dart and flutter and I tried to capture only digit of this type of string :

case 1

     aaaaaaaaa 06 12 34 56 78  aaaaaa

case 2

    aaaaaaaa 0612345678 aaaaaa

case 3

   aaaaaa +336 12 34 56 78 aaaaa

I search to have only 0612345678 with no space and no +33. Just 10 digit in se case of +33 I need to replace +33 by 0

currently I have this code \D*(\d+)\D*? who run with the case 2

Upvotes: 3

Views: 2898

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626903

You may match and capture an optional +33 and then a digit followed with spaces or digits, and then check if Group 1 matched and then build the result accordingly.

Here is an example solution (tested):

var strs = ['aaaaaaaaa 06 12 34 56 78  aaaaaa', 'aaaaaaaa 0612345678 aaaaaa', 'aaaaaa +336 12 34 56 78 aaaaa', 'more +33 6 12 34 56 78'];
  for (int i = 0; i < strs.length; i++) {
    var rx = new RegExp(r"(?:^|\D)(\+33)?\s*(\d[\d ]*)(?!\d)");
    var match = rx.firstMatch(strs[i]);
    var result = "";
    if (match != null) {
      if (match.group(1) != null) {
        result = "0" + match.group(2).replaceAll(" ", "");
      } else {
        result = match.group(2).replaceAll(" ", "");
      }
      print(result);
  }
}

Returns 3 0612345678 strings in the output.

The pattern is

(?:^|\D)(\+33)?\s*(\d[\d ]*)(?!\d)

See its demo here.

  • (?:^|\D) - start of string or any char other than a digit
  • (\+33)? - Group 1 that captures +33 1 or 0 times
  • \s* - any 0+ whitespaces
  • (\d[\d ]*) - Group 2: a digit followed with spaces or/and digits
  • (?!\d) - no digit immediately to the right is allowed.

Spaces are removed from Group 2 with a match.group(2).replaceAll(" ", "") since one can't match discontinuous strings within one match operation.

Upvotes: 4

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