flatter-json_serializable fromJson:null调用了方法“[]”

Question

在我的应用程序中，我将数据导出到JSON。但是，当我尝试导入JSON文件时，它会抛出一个错误，如下所示：

E/flatter:[错误：flatter/shell/common/shell.cc（184）]省道错误：未处理的异常：NoSuchMethodError:对null调用了方法“[]”。接收方：null尝试调用：#0对象。noSuchMethod（dart:core/runtime/libobject_patch.dart:50:5）#1$RecipesModelFromJson(file:///home/stube/IdeaProjects/Flutter_Recipe/lib/JSON/recipes.g.dart:11:11)#2个新模式。fromJson（包：Time2Eat/JSON/recipes.dart:20:63）#3$RecipesModelFromJson(file:///home/stube/IdeaProjects/Flutter_Recipe/lib/JSON/recipes.g.dart:32:22)#4种新型配方。fromJson（包：Time2Eat/JSON/recipes.dart:20:63）#5 RecipebookState。创建RecipeJSON（包：Time2Eat/recipe/recipebook.dart:111:32）#6 RecipebookState。getPath（包：Time2Eat/recipe/recipebook.dart:123:9）#7 RecipebookState。建筑（包装：Time2Eat/recipe/recipebook.dart:185:19）#8动画儿童_表演（套装：颤振、速度、拨号、src、动画、儿童、飞镖：60:24）9_把手（包装：颤振/src/材料/墨水井。省道：507:14）#10#。建筑（包装：颤振/src/material/ink_well.省道：562:30）#11手势识别器。invokeCallback（包：flatter/src/signatures/recognizer.dart:102:24）#12 TapGestureCongnizer_检查（程序包：颤振/src/signatures/tap.dart:242:9）#13 TapGestureRecognizer。handlePrimaryPointer（软件包：flatter/src/signatures/tap.dart:175:7）#14 PrimaryPointer检测识别器。handleEvent（套装：颤振/src/手势/识别器。省道：315:9）#15点外部_调度（包：颤振/src/手势/指针路由器。dart:73:12）#16点路由器。路线（包：颤振/src/signatures/pointer_router.dart:101:11）#17_WidgetsFlutterBinding

recipes.g.dart，这是由json_serializable产生的：

// **************************************************************************
// JsonSerializableGenerator
// **************************************************************************

RecipesModel _$RecipesModelFromJson(Map<String, dynamic> json) {
  return RecipesModel(
      json['name'] as String,
      json['image'] as String,
      json['description'] as String,
      json['favorite'] as int,
      json['timestamp'] as String,
      json['preperation'] as String,
      json['creation'] as String,
      json['resting'] as String,
      json['people'] as String,
      json['backgroundColor'] as String,
      (json['zutaten'] as List)
          ?.map((e) => e == null
              ? null
              : ZutatenModel.fromJson(e as Map<String, dynamic>))
          ?.toList(),
      (json['zubereitung'] as List)
          ?.map((e) => e == null
              ? null
              : ZubereitungModel.fromJson(e as Map<String, dynamic>))
          ?.toList())
    ..recipeModel =
        RecipesModel.fromJson(json['recipeModel'] as Map<String, dynamic>);
}

Map<String, dynamic> _$RecipesModelToJson(RecipesModel instance) =>
    <String, dynamic>{
      'name': instance.name,
      'image': instance.image,
      'description': instance.description,
      'favorite': instance.favorite,
      'timestamp': instance.timestamp,
      'preperation': instance.preperation,
      'creation': instance.creation,
      'resting': instance.resting,
      'people': instance.people,
      'backgroundColor': instance.backgroundColor,
      'zutaten': instance.zutaten,
      'zubereitung': instance.zubereitung,
      'recipeModel': instance.recipeModel
    };

ZutatenModel _$ZutatenModelFromJson(Map<String, dynamic> json) {
  return ZutatenModel(
      json['zutat'] as String,
      json['number'] as String,
      json['measure'] as String
  );
}

Map<String, dynamic> _$ZutatenModelToJson(ZutatenModel instance) =>
    <String, dynamic>{
      'zutat': instance.zutat,
      'number': instance.number,
      'measure': instance.measure
    };

ZubereitungModel _$ZubereitungModelFromJson(Map<String, dynamic> json) {
  return ZubereitungModel(json['number'] as String, json['steps'] as String);
}

Map<String, dynamic> _$ZubereitungModelToJson(ZubereitungModel instance) =>
    <String, dynamic>{'number': instance.number, 'steps': instance.steps};

我在一个函数中调用RecipeModel.fromJson（）方法，它应该能够将这些数据保存在数据库中。功能看起来像这样：

createJson() async{
      FlutterDocumentPickerParams params = FlutterDocumentPickerParams(
          allowedFileExtensions: ['json'],
          invalidFileNameSymbols: ['/']
      );
      final path = await FlutterDocumentPicker.openDocument(params: params);
      File file = new File(path);
      createRecipeJson(file);
      Map<String,dynamic> jSON = json.decode(path.readAsStringSync());
      var model = RecipesModel.fromJson(jSON);
      showBottomSnack("Name: ${model.name}", ToastGravity.BOTTOM);
    }

我希望有人能帮我解决问题。那太棒了。

Answer 1

检查类成员中的@JsonSerializable（nullable： false）（从ZutatenModel和RecipeModel开始）。

如果希望它们有时为null，请不要在注释其模型类时标记null:false。

如果是这种情况，请将其更改为@JsonSerializable（nullable:true）。顺便说一句，默认设置为true。

当一些类属性被提供null但是（nullable： false）向JsonSerializer建议这些属性不会为空时，通常会发生此异常。