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 ...
//
}