IOS : JSONModel : NSInvalidArgumentException

Question

我有这个json数据

{"data": [{"id": 3,"nameAr": "Test","nameEn": "Test","active": 1,
"subCategories": [{"id": 27,"nameAr": "Test",
"nameEn": "Test","active": 0,
"sections": [{"id": 53,"nameAr": "Test",
"nameEn": "Test","active": 0},
{"id": 52,"nameAr": "Test","nameEn": "Test","active": 0
}]}]}],"code": "1001","message": "success"
}

当使用JSONModel https://github.com/icanzilb/JSONModel解析此json时

解析此代码的代码

self.categoriesModels = [CategoryModel arrayOfModelsFromDictionaries: [results objectForKey:@"data"]];

并试图访问嵌套的json

categoryModel.subCategories

我犯了这个错误

-CategoryModel subCategories:未识别的选择器发送到实例0x17404b910 2015-05-03 20:38:15.019 AkshefFeen2268:786267 *终止应用程序由于未被识别的异常“NSInvalidArgumentException”，原因是：‘-分级模式subCategories:未识别的选择器发送给实例0x17404b910’*第一次抛出堆栈：(0x185fb42d8 0x1977800e4 0xfbb3a4 0x185fb8154 0x185ebaccc 0x27608 0x18ab174 0x18abd7790 0xa78a240xa240x18a18a604 0x18abd9ec0fx6xc4x4x6fx10)

我的模特

1-类别模型.h

#import "JSONModel.h"
#import "SubCategoryModel.h"

@protocol CategoryModel
@end

@interface CategoryModel : JSONModel

@property (assign, nonatomic) int id;
@property (strong, nonatomic) NSString* nameAr;
@property (strong, nonatomic) NSString* nameEn;
@property (assign, nonatomic) int active;
@property (strong, nonatomic) NSArray<SubCategoryModel>* subCategories;

@end

2-次级类别模型.h

@protocol SubCategoryModel
@end

@interface SubCategoryModel : JSONModel

@property (assign, nonatomic) int id;
@property (strong, nonatomic) NSString* nameAr;
@property (strong, nonatomic) NSString* nameEn;
@property (assign, nonatomic) int active;
@property (strong, nonatomic) NSArray<SectionModel,Optional>* subCategories;

@end

3-SectionModel.h

#import "JSONModel.h"

@protocol SectionModel
@end

@interface SectionModel : JSONModel

@property (assign, nonatomic) int id;
@property (strong, nonatomic) NSString* nameAr;
@property (strong, nonatomic) NSString* nameEn;
@property (assign, nonatomic) int active;

@end

为什么我会得到这个错误，以及如何解决它？

Answer 1

@ahmed :可能有两个拥有性的原因: 1.可能是因为您的任何JSONModel类中都使用了@合成关键字。

1. 这可能是由于父对象和子对象之间的命名相同。

请参考此链接获取JSONModel：单击此处中的首选命名约定。

或者你也可以在这里提出同样的问题，你将得到正式的答案或决议：单击此处