我有一个从休息Web服务请求返回的JSON字符串, 我想将这个字符串解析为确定类的对象数组, 这是JSON字符串
[
{
"validationCode": null,
"FirstName": "Samer",
"LastName": "Shame",
"MobileNumber": "0991992993",
"SimNumber": null,
"Email": null,
"PhoneNumber": "0991992994",
"Name": "Abo Alshamat",
"ID": 1
},
{
"validationCode": null,
"FirstName": "Ahmad",
"LastName": "Ali",
"MobileNumber": "0992993994",
"SimNumber": null,
"Email": null,
"PhoneNumber": "0992993995",
"Name": "AL-Kamal",
"ID": 2
},
{
"validationCode": null,
"FirstName": null,
"LastName": null,
"MobileNumber": "0993377800",
"SimNumber": null,
"Email": null,
"PhoneNumber": null,
"Name": "Abo-MAhmoud",
"ID": 12
},
{
"validationCode": null,
"FirstName": "William",
"LastName": "Ammar",
"MobileNumber": "0993994995",
"SimNumber": null,
"Email": null,
"PhoneNumber": "0993994996",
"Name": "Four Season",
"ID": 3
},
{
"validationCode": null,
"FirstName": "Ammar",
"LastName": "William",
"MobileNumber": "0999555777",
"SimNumber": null,
"Email": null,
"PhoneNumber": null,
"Name": "uuuuu",
"ID": 20
},
{
"validationCode": null,
"FirstName": null,
"LastName": null,
"MobileNumber": "0999888777",
"SimNumber": null,
"Email": null,
"PhoneNumber": null,
"Name": "NewOneFromI2",
"ID": 18
},
{
"validationCode": null,
"FirstName": null,
"LastName": null,
"MobileNumber": "0999998997",
"SimNumber": null,
"Email": null,
"PhoneNumber": "0999999998",
"Name": "JOURY",
"ID": 4
},
{
"validationCode": null,
"FirstName": null,
"LastName": null,
"MobileNumber": "202020",
"SimNumber": null,
"Email": null,
"PhoneNumber": null,
"Name": "TestTestRestaurant,Ammar,Hamed",
"ID": 19
}
]
我想从中获取实例的类是:
@interface Restaurant : NSObject
@property (nonatomic,strong) NSString *ID;
@property (nonatomic,strong) NSString* FirstName;
@property (nonatomic,strong) NSString* LastName;
@property (nonatomic,strong) NSString* MobileNumber;
@property (nonatomic,strong) NSString* simNumber;
@property (nonatomic,strong) NSString* PhoneNumber;
@property (nonatomic,strong) NSString* Name;
@end
最好的方法是什么,请原谅我的问题可能来自基础知识,但我是新客观的C
谢谢你的时间。答案 0 :(得分:8)
我建议为Restaurant
类实现一个init方法。
-(instancetype) initWithParameters:(NSDictionary*)parameters
{
self = [super init];
if (self) {
//initializations
_validationCode = parameters[@"validationCode"]; // may be NSNull
_firstName = [parameters[@"FirstName"] isKindOfClass:[NSNull class]] ? @""
: parameters[@"FirstName"];
...
}
return self;
}
注意:您可能拥有JSON Null,这使您的初始化有点复杂。当相应的JSON值为Null时,您需要决定如何初始化属性。
您的parameters
字典将是您从服务器获取的JSON数组中的第一级字典。
首先,创建一个JSON表示,即JSON中的NSArray对象:
NSError* localError;
id restaurantsObjects = [NSJSONSerialization JSONObjectWithData:data
options:0
error:&localError];
IFF没有失败,您的restaurantsObjects
现在应该是一个NSArray
对象,其中包含餐馆NSDictionary
s。
现在,可以直接创建NSMutableArray
,其中将填充Restaurant
个对象:
NSMutableArray* restaurants = [[NSMutableArray alloc] init];
for (NSDictionary* restaurantParameters in restaurantsObjects) {
Restaurant* restaurant = [Restaurant alloc] initWithParameters: restaurantParameters];
[restaurants addObject:restaurant];
}
最后,您可以在某个控制器中设置属性restaurants
:
self.restaurants = [restaurants copy];
答案 1 :(得分:6)
你的JSON有一系列词典...... 首先将您的数据转换为NSArray。
NSError *jsonError = nil;
NSArray *jsonArray = (NSArray *)[NSJSONSerialization JSONObjectWithData:jsonData options:nil error:&jsonError];
现在你有一系列JSON字典迭代。
for (NSDictionary *dic in jsonArray){
// Now you have dictionary get value for key
NSString *firstName = (NSString*) [dic valueForKey:@"FirstName"];//We are casting to NSString because we know it will return a string. do this for every property...
}