如何使正则表达式用于多种条件？

Question

I have regexp code like below (I'm using VerbalExpression dart plugin ), My purpose is to check that a string starts with "36", followed by "01", "02", or "03". After that can be anything as long as the whole string is 16 characters long.

  var regex = VerbalExpression()
  ..startOfLine()
  ..then("36")
  ..then("01")
  ..or("02")
  ..anythingBut(" ")
  ..endOfLine();
String nik1 = "3601999999999999";
String nik2 = "3602999999999999";
String nik3 = "3603999999999999";

print('result : ${regex.hasMatch(nik1)}');
print('Hasil : ${regex.hasMatch(nik2)}');
print('Hasil : ${regex.hasMatch(nik3)}');

my code only true for nik1 and nik2, however i want true for nik3, I noticed that i can't put or() after or() for multiple check, it just give me all false result, how do i achieve that?

Answer 1

I'm not familiar with VerbalExpression , but a RegExp that does this is straightforward enough.

const pattern = r'^36(01|02|03)\S{12}$';

void main() {
  final regex = RegExp(pattern);
  print(regex.hasMatch('3601999999999999'));    // true
  print(regex.hasMatch('3602999999999999'));    // true
  print(regex.hasMatch('3603999999999999'));    // true
  print(regex.hasMatch('360199999999999'));     // false
  print(regex.hasMatch('3600999999999999'));    // false
  print(regex.hasMatch('36019999999999999'));   // false
}

Pattern explanation:

• The r prefix means dart will interpret it as a raw string ("$" and "\\" are not treated as special).
• The ^ and $ represent the beginning and end of the string, so it will only match the whole string and cannot find matches from just part of the string.
• (01|02|03) "01" or "02" or "03". | means OR. Wrapping it in parentheses lets it know where to stop the OR.
• \\S matches any non-whitespace character.
• {12} means the previous thing must be repeated 12 times, so \\S{12} means any 12 non-whitespace characters.