Json加速器和AFNetworking

Question

當操作正在運行時，我無法在JSON模型CREATED BY ACCELERATOR中輸入數據。 你能告訴我我做錯了什麼嗎？

{ 
    [super viewDidLoad]; 

NSLog(@"you are in a tableViewController"); 
self.title = @"NavigationOrdini"; 


NSURLRequest *req = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://www.stampa6x3.com/json.php?azione=ordini"]]; 
AFJSONRequestOperation* operation; 

operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:req 
                  success:^(NSURLRequest *request, NSURLResponse *response, id JSON) 
{ 

    [[ordiniModel alloc] initWithDictionary:JSON]; 



} 
      failure:^(NSURLRequest *request, NSURLResponse *response, NSError 
         *error, id JSON) { 
       [self setTitle:@"Dictionary"]; 
       NSLog(@"failed! %d",[error code]); 
      }]; 
[operation start]; 


ordiniModel*test; 

NSLog(@"il valore è %@",test.ordini.description); 

} 

Answer 1

AFJSONRequestOperation是異步的，這意味着代碼將在應用程序的其餘部分運行時繼續執行。代碼實際完成時運行完成塊。

所以嘗試：

NSLog(@"you are in a tableViewController"); 
self.title = @"NavigationOrdini"; 
ordiniModel *test; // <-- create variable here 

NSURLRequest *req = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://www.stampa6x3.com/json.php?azione=ordini"]]; 
AFJSONRequestOperation* operation; 

operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:req 
                  success:^(NSURLRequest *request, NSURLResponse *response, id JSON) 
{ 

    test = [[ordiniModel alloc] initWithDictionary:JSON]; // <-- Assign here 
    NSLog(@"il valore è %@",test.ordini.description); 


} 
      failure:^(NSURLRequest *request, NSURLResponse *response, NSError 
         *error, id JSON) { 
       [self setTitle:@"Dictionary"]; 
       NSLog(@"failed! %d",[error code]); 
      }]; 
[operation start];