iphone - 我的应用使用 URLRequest 登录两次

Question

当我在我的应用程序中按下登录时，它在 API 中登录两次而不是一次..，这有问题但我找不到什么，因为它只执行此代码一次。

   NSUserDefaults *defaults =[NSUserDefaults standardUserDefaults];     
    NSHTTPURLResponse   * response;
                NSError             * error;
                NSMutableURLRequest * request;
                NSString            * params;
                NSString *urlAddress = [NSString stringWithFormat:@"%@/?action=request&api=json&module=ManagementModule&function=startSession&instance=0",[ConnectServer returnserverip]];
                NSLog(@"UPX %@",[ConnectServer returnserverip]);
                NSLog(@"IP %@",[ConnectServer returnclientip]);
                if([defaults boolForKey:@"enablePincode"]){
                    NSString *account = [defaults stringForKey:@"myAccount"];
                    NSString *username =[defaults stringForKey:@"myUsername"];
                    NSString *password = [defaults stringForKey:@"myPassword"];
                    NSString *clientip = [ConnectServer returnclientip];
                    NSString *clientname = [ConnectServer returnclientname];
                    params = [[[NSString alloc] initWithFormat:@"params=&auth[password]=%@&auth[mode]=%@&auth[account]=%@&auth[user]=%@&auth[rights]=%@&auth[user_ip]=%@&auth[client_name]=%@",password,@"password",account,username,@"user",clientip,clientname] autorelease];
                }
                else {
                    NSString *clientip = [ConnectServer returnclientip];
                    NSString *clientname = [ConnectServer returnclientname];
                    params = [[[NSString alloc] initWithFormat:@"params=&auth[password]=%@&auth[mode]=%@&auth[account]=%@&auth[user]=%@&auth[rights]=%@&auth[user_ip]=%@&auth[client_name]=%@",[myPassword text],@"password",[myAccount text],[myUsername text],@"user",clientip,clientname] autorelease];
                }
                request = [[[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:urlAddress] cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:60] autorelease];
                NSData *myRequestData = [params dataUsingEncoding:NSUTF8StringEncoding];
                [NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[[NSURL URLWithString: urlAddress] host]];
                [request setHTTPMethod:@"POST"];
                [request setHTTPBody:myRequestData];
                [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Accept"];
                [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
                [request setValue:[NSString stringWithFormat:@"%d", [myRequestData length]] forHTTPHeaderField:@"Content-Length"];
                [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error]; 
     NSLog(@"RESPONSE HEADERS: \n%@", [response allHeaderFields]);
    request.URL = [NSURL URLWithString:urlAddress];
    error       = nil;
    response    = nil;
    NSData * data = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
    NSLog(@"The server saw:\n%@", [[[NSString alloc] initWithData:data encoding: NSASCIIStringEncoding] autorelease]);
    NSLog(@"Parameters: %@", params);
    NSLog(@"Actual sended parameters to the server: %@", myRequestData);
    NSString *Sresponse;
    Sresponse = [[[NSString alloc] initWithData:data encoding: NSASCIIStringEncoding] autorelease];

Answer 1

代码中有两个请求:

[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error]; 

往下五行

NSData * data = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

提示:在这种情况下，使用 wireshark 或我最喜欢的 Charles 来解码 SSL 连接。