'How to get substring between two strings in DART?

How can i achieve similar solution to: How to get a substring between two strings in PHP? but in DART

For example I have a String:
String data = "the quick brown fox jumps over the lazy dog"
I have two other Strings: quick and over
I want the data inside these two Strings and expecting result:
brown fox jumps



Solution 1:[1]

You can use String.indexOf combined with String.substring:

void main() {
  const str = "the quick brown fox jumps over the lazy dog";
  const start = "quick";
  const end = "over";

  final startIndex = str.indexOf(start);
  final endIndex = str.indexOf(end, startIndex + start.length);

  print(str.substring(startIndex + start.length, endIndex)); // brown fox jumps
}

Note also that the startIndex is inclusive, while the endIndex is exclusive.

Solution 2:[2]

I love regexp with lookbehind (?<...) and lookahead (?=...):

void main() {
  var re = RegExp(r'(?<=quick)(.*)(?=over)');
  String data = "the quick brown fox jumps over the lazy dog";
  var match = re.firstMatch(data);
  if (match != null) print(match.group(0));
}

Solution 3:[3]

final str = 'the quick brown fox jumps over the lazy dog';
final start = 'quick';
final end = 'over';

final startIndex = str.indexOf(start);
final endIndex = str.indexOf(end);
final result = str.substring(startIndex + start.length, endIndex).trim();

Solution 4:[4]

You can do that with the help of regex. Create a function that will return the regex matches as

Iterable<String> _allStringMatches(String text, RegExp regExp) => 
regExp.allMatches(text).map((m) => m.group(0));

And then define your regex as RegExp(r"[quick ]{1}.*[ over]{1}"))

Solution 5:[5]

The substring functionality isn't that great, and will throw errors for strings above the "end" value.

To make it simpler user this custom function that doesn't thrown an error, instead using the max length.

String substring(String original, {required int start, int? end}) {
  if (end == null) {
    return original.substring(start);
  }
  if (original.length < end) {
    return original.substring(start, original.length);
  }
  return original.substring(start, end);
}

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Dominik Palo
Solution 2
Solution 3 CopsOnRoad
Solution 4 bimsina
Solution 5 Oliver Dixon