【问题标题】:iOS JSON Parsing Objective-C and pass POST method parameteriOS JSON解析Objective-C并传递POST方法参数
【发布时间】:2016-08-31 05:11:57
【问题描述】:

我想使用 volleylibrary 解析 Android 中的 JSON 进行登录。在登录活动中,通过 post 和响应发送两个参数(用户名、密码),如下所示。

参数:用户名、密码

for success:

{
"status": 1,
"message": "successfully login" ,  
    "result": [
{
    "name": "abc",
    "email": "abc@gmail.com",
    "Img": "http://img.com/img.png"
}      
       ]
}

对于错误:

{
"status": 0,
"message": "email or password doesn’t exist"    
}

【问题讨论】:

  • android me krna hai ya iOS me? (你想在 iOS 或 Android 上做吗?)
  • 你能出示你的代码吗

标签: ios objective-c json afnetworking


【解决方案1】:

API 调用在 iOS 中使用 AFNetworking 和 POST 方法

NSMutableDictionary *parameters = [NSMutableDictionary dictionary];
[parameters setObject:username forKey:@"username"];
[parameters setObject:password forKey:@"password"];

然后调用方法

- (void)apiRequestForPOSTWithServiceName:(NSString *)serviceName andParameters:(NSDictionary *)parameter withCompletion:(void(^)(id response, BOOL isSuccess))completion
{
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.requestSerializer = [AFJSONRequestSerializer serializer];
NSString *url = [NSString stringWithFormat:@"%@%@", BASE_URL, serviceName];
[manager POST:url parameters:parameter success:^(AFHTTPRequestOperation *operation, id responseObject) {
    //NSLog(@"JSON: %@", responseObject);

    completion(responseObject,YES);

} failure:^(AFHTTPRequestOperation *operation, NSError *error) {
    NSLog(@"Error: %@", error);

    completion(error,NO);
}];
}

从服务器获取数据后需要解析响应对象

- (void)parseDataWithDict:(id) responeDict
{
NSMutableDictionary *dict = (NSMutableDictionary *) responeDict;
NSString *status = dict[@“status”];
NSString *message = dict[@“message”];
NSDictionary *dictRes = [responeDict[@"result"] firstObject];  // If the result having more data in array, need to iterate the array
NSString *name = dictRes[@“name”];
NSString *email = dictRes[@“email”];
NSString *Img = dictRes[@“Img”];
// If you have any model object class, then add to it and return the object
}

谢谢...愉快的编码。

【讨论】:

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