How do I parse a string into a number with Dart?

Question

I would like to parse strings like 1 or 32.23 into integers and doubles. How can I do this with Dart?

Answer 1

You can parse a string into an integer with int.parse(). For example:

var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345

Note that int.parse() accepts 0x prefixed strings. Otherwise the input is treated as base-10.

You can parse a string into a double with double.parse(). For example:

var myDouble = double.parse('123.45');
assert(myDouble is double);
print(myDouble); // 123.45

parse() will throw FormatException if it cannot parse the input.

Answer 2

In Dart 2 int.tryParse is available.

It returns null for invalid inputs instead of throwing. You can use it like this:

int val = int.tryParse(text) ?? defaultValue;

Answer 3

Convert String to Int

var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
print(myInt.runtimeType);

Convert String to Double

var myDouble = double.parse('123.45');
assert(myInt is double);
print(myDouble); // 123.45
print(myDouble.runtimeType);

Example in DartPad

Answer 4

 void main(){
  var x = "4";
  int number = int.parse(x);//STRING to INT

  var y = "4.6";
  double doubleNum = double.parse(y);//STRING to DOUBLE

  var z = 55;
  String myStr = z.toString();//INT to STRING
}

int.parse() and double.parse() can throw an error when it couldn't parse the String

Answer 5

As per dart 2.6

The optional onError parameter of int.parse is deprecated. Therefore, you should use int.tryParse instead.

Note: The same applies to double.parse. Therefore, use double.tryParse instead.

  /**
   * ...
   *
   * The [onError] parameter is deprecated and will be removed.
   * Instead of `int.parse(string, onError: (string) => ...)`,
   * you should use `int.tryParse(string) ?? (...)`.
   *
   * ...
   */
  external static int parse(String source, {int radix, @deprecated int onError(String source)});

The difference is that int.tryParse returns null if the source string is invalid.

  /**
   * Parse [source] as a, possibly signed, integer literal and return its value.
   *
   * Like [parse] except that this function returns `null` where a
   * similar call to [parse] would throw a [FormatException],
   * and the [source] must still not be `null`.
   */
  external static int tryParse(String source, {int radix});

So, in your case it should look like:

// Valid source value
int parsedValue1 = int.tryParse('12345');
print(parsedValue1); // 12345

// Error handling
int parsedValue2 = int.tryParse('');
if (parsedValue2 == null) {
  print(parsedValue2); // null
  //
  // handle the error here ...
  //
}

Answer 6

you can parse string with int.parse('your string value');.

Example:- int num = int.parse('110011'); print(num); // prints 110011 ;

Answer 7

Above solutions will not work for String like:

String str = '123 km';

So, the answer in a single line, that works in every situation for me will be:

int r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), '')) ?? defaultValue;
or
int? r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), ''));

But be warned that it will not work for the below kind of string

String problemString = 'I am a fraction 123.45';
String moreProblem = '20 and 30 is friend';

If you want to extract double which will work in every kind then use:

double d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), '')) ?? defaultValue;
or
double? d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), ''));

This will work for problemString but not for moreProblem.