deserialize method

  1. @override
Object? deserialize(
  1. DartType targetType,
  2. String expression,
  3. TypeHelperContextWithConfig context,
  4. bool defaultProvided,
)
override

Returns Dart code that deserializes an expression representing a JSON literal to into targetType.

If targetType is not supported, returns null.

Let's say you want to deserialize a class Foo by taking an int stored in a JSON literal and calling the Foo.fromInt constructor.

Treating expression as a opaque Dart expression representing a JSON literal, the deserialize implementation could be a simple as:

String deserialize(DartType targetType, String expression) =>
  "new Foo.fromInt($expression)";
```.

Note that [targetType] is not used here. If you wanted to support many
types of [targetType] you could write:

```dart
String deserialize(DartType targetType, String expression) =>
  "new ${targetType.name}.fromInt($expression)";
```.

Implementation

@override
Object? deserialize(
  DartType targetType,
  String expression,
  TypeHelperContextWithConfig context,
  bool defaultProvided,
) {
  final converter = _typeConverter(targetType, context);
  if (converter == null) {
    return null;
  }

  if (!converter.jsonType.isNullableType && targetType.isNullableType) {
    const converterFromJsonName = r'_$JsonConverterFromJson';
    context.addMember('''
Value? $converterFromJsonName<Json, Value>(
Object? json,
Value? Function(Json json) fromJson,
) => ${ifNullOrElse('json', 'null', 'fromJson(json as Json)')};
''');

    return _nullableJsonConverterLambdaResult(
      converter,
      name: converterFromJsonName,
      targetType: targetType,
      expression: expression,
      callback: '${converter.accessString}.fromJson',
    );
  }

  return LambdaResult(
    expression,
    '${converter.accessString}.fromJson',
    asContent: converter.jsonType,
  );
}