ios - 如何在 Json 请求中的参数中传递一个 ID 数组

Question

我有一个 JSON 请求和一些参数...我需要在这些参数之一中传递一个数组...如何？

这是我当前的请求...我有 3 个参数:纬度、经度和数组:数组必须是 ids (int) 数组

数组:NSArray arrayIds = [NSArrayWithObjects:@"1", @"2", @"3", nil];

  -(void) listarAtracoesBusca
{
    AppDelegate *appDelegate = (AppDelegate *)[[UIApplication sharedApplication]delegate];
    self.listaAtracoes = [[NSMutableArray alloc]init];
    self.indiceAtracaoAtual = 0;
    NSString *urlStr = @"";
    float latitude = 0.0;
    float longitude = 0.0;

    if(appDelegate.latitudeAtual)
        latitude = appDelegate.latitudeAtual;
    if(appDelegate.longitudeAtual)
        longitude = appDelegate.longitudeAtual;

    [self validarCamposBusca];

    urlStr = [NSString stringWithFormat:@"https://host:3000/api/ params?latitude=%f&longitude=%f&array=", latitude, longitude, arrayIds];


    NSURL *URL = [NSURL URLWithString:urlStr];
    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:URL
                                                           cachePolicy:NSURLRequestUseProtocolCachePolicy
                                                       timeoutInterval:10.0];
    [request setValue:@"haadushdiuashud" forHTTPHeaderField:@"X-User-Authorization"];
    NSString *basicAuthCredentials = [NSString stringWithFormat:@"%@:%@", @"user", @"pass"];
    NSString *authValue = [NSString stringWithFormat:@"Basic %@", AFBase64EncodedStringFromString(basicAuthCredentials)];

    [request setValue:authValue forHTTPHeaderField:@"Authorization"];
    request.HTTPMethod = @"GET";
    [request addValue:@"application/json" forHTTPHeaderField:@"Accept"];
    [NSURLConnection sendAsynchronousRequest:request
                                       queue:[NSOperationQueue mainQueue]
                           completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {

                               //Do Stuff

                               if (data.length > 0 && error == nil)
                               {
}
                               else {
                                   NSLog(@"erro: %@", error.description);
                               }
                           }];
}

Answer 1

如果你正在尝试实现查询 url 之类的东西，那么你可以使用 NSURLQueryItem，它将代表 URL 查询部分的单个键/值对。

NSURLComponents *components = [NSURLComponents componentsWithString:@"http://stackoverflow.com"];
NSURLQueryItem *search = [NSURLQueryItem queryItemWithName:@"q" value:@"ios"];
NSURLQueryItem *count = [NSURLQueryItem queryItemWithName:@"count" value:@"10"];
components.queryItems = @[ search, count ];
NSURL *url = components.URL; // http://stackoverflow.com?q=ios&count=10

否则，您需要使用 NSJSONSerialization 将 JSON 作为请求的主体传递。引用here了解更多详情。