Question

我的JSON采用给定格式

{
  "BL_NO": "CSV*******13",
  "LOADING_PORT": "YANTAI",
  "DISCHARGE_PORT": "YINGKOU",
  "DELIVERY_PLACE": "YINGKOU"
}

我正在尝试使用下面给出的代码：

NSString *str =[NSString stringWithFormat:@"http://wcfshiptracker.logistify.net/ShipmentTrackingService.svc/FetchData/ChinaShipping-bl_no-TGHU6571642"];
NSURL *url = [NSURL URLWithString:str];
NSData *data = [NSData dataWithContentsOfURL:url];
NSDictionary *dict = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
aryother = [dict valueForKey:@"BL_NO"];

Answer 1

你的JSON格式不正确它给出的字符串不是字典。请使用此

NSString *str =[NSString stringWithFormat:@"http://wcfshiptracker.logistify.net/ShipmentTrackingService.svc/FetchData/ChinaShipping-bl_no-TGHU6571642"];
    NSURL *url = [NSURL URLWithString:str];
    NSData *data = [NSData dataWithContentsOfURL:url];
    NSString *str3= [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:nil];
    NSError *jsonError;
    NSData *objectData = [str3 dataUsingEncoding:NSUTF8StringEncoding];
    NSMutableDictionary *dict3 = [NSJSONSerialization JSONObjectWithData:objectData
                                                            options:NSJSONReadingAllowFragments
                                                              error:&jsonError];

    NSString *a = [dict3 valueForKey:@"BL_NO"];

Answer 2

使用实体BL_NO,LODING_PORT,DISCHARGE_PORT,DELIVERY_PLACE创建一个类。 include Newtonsoft.Json dll使用JsonConvert.DeserializeObject`＆＃34;＆lt;＆＃34; YourClassName＆＃34;＆gt;＆＃34;（#String）。

    Public class YourClass
        {
        public string BL_NO{get;set;}
        public string LODING_PORT{get;set;}
        public string DISCHARGE_PORT{get;set;}
        public string DELIVERY_PLACE{get;set;}
       }
       string yourJsonString="{\"BL_NO\":\"CSV*******13\",\"LOADING_PORT\":\"YANTAI\",\"DISCHARGE_PORT\":\"YINGKOU\",\"DELIVERY_PLACE\":\"YINGKOU\"}"
     YourClass obj=JsonConvert.DeserializeObject "<"YourClass">"(yourJsonString).

如何从这种类型的json格式中获取数据？

2 个答案: