创建模型类以在没有任何库的情况下进行解析答案

【问题标题】：create Model class to parse without any library创建模型类以在没有任何库的情况下进行解析
【发布时间】：2018-04-26 10:44:42
【问题描述】：

我想为 json 创建一个模型类。下面给出了我的 JSON 示例

来自 API 的 json 响应：

msg = '{"type":"TYPE_NM","payload":{"responseCode":0,"nextCheckTime":30}}';

我想创建一个可编码（Swift）的属性，就像在 Objective-C 中一样。

我创建了两个 nsobject 接口作为“type”和“payload”。下面我给我的课sn-ps。

      //msg model 
      @interface MessageModel : NSObject

      @property (nonatomic) NSString *type;
      @property (nonatomic) Payload *payload;
      @end
      //for payload
      @interface Payload : NSObject

      @property (nonatomic) NSUInteger responseCode;
      @property (nonatomic) NSUInteger nextCheckTime;

     @end

【问题讨论】：

标签： objective-c json parsing model

【解决方案1】：

您可以将json字符串转换为NSDictionary对象并使用它来创建MessageModel

有效载荷

@interface Payload : NSObject

@property (nonatomic) NSUInteger responseCode;
@property (nonatomic) NSUInteger nextCheckTime;

@end

@implementation Payload

- (instancetype)initWithDictionary:(NSDictionary *)dict {
  self = [super init];

  if (self) {
    _responseCode = [dict[@"responseCode"] integerValue];
    _nextCheckTime = [dict[@"nextCheckTime"] integerValue];
  }

  return self;
}

@end

消息模型

@interface MessageModel : NSObject

@property (nonatomic) NSString *type;
@property (nonatomic) Payload *payload;
@end

@implementation MessageModel

- (instancetype)initWithDictionary:(NSDictionary *)dict {
  self = [super init];

  if (self) {
    _type = dict[@"type"];
    _payload = [[Payload alloc] initWithDictionary:dict[@"payload"]];
  }

  return self;
}

- (instancetype)initWithJson:(NSString *)json {
  self = [super init];

  if (self) {
    NSData *data = [json dataUsingEncoding:NSUTF8StringEncoding];
    NSDictionary *dict = [NSJSONSerialization JSONObjectWithData:data options:0 error:nil];

    _type = dict[@"type"];
    _payload = [[Payload alloc] initWithDictionary:dict[@"payload"]];
  }

  return self;
}

@end

用法

NSString *input = @"{\"type\":\"TYPE_NM\",\"payload\":{\"responseCode\":0,\"nextCheckTime\":30}}";
MessageModel *model = [[MessageModel alloc] initWithJsonString:input];

【讨论】：