0

我正在尝试使用 AFNetworking 调用 Rest API,但没有得到正确的响应字符串。这是我的代码:

NSURL *url = [[NSURL alloc] initWithString:@"https://www.ez-point.com/api/v1/ezpoints"];
    NSURLRequest *request = [[NSURLRequest alloc] initWithURL:url];

    AFJSONRequestOperation *operation =
    [AFJSONRequestOperation JSONRequestOperationWithRequest:request
        success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
            NSLog(@"%@",@"testing");
            NSLog(@"%@",operation.responseData);

        }
            failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
                NSLog(@"%@", @"Error");
        }];

    [operation start];

但我把它打印出来:

2013-10-29 08:31:08.175 EZ-POINT[4004:c07] testing
2013-10-29 08:31:08.175 EZ-POINT[4004:c07] (null)

正如你所看到的,它返回 null,我期待这个:

{"status": "user_invalid", "data": "", "token": "", "errors": ["user not found"], "user_id": ""}
4

1 回答 1

1

我更习惯这种设置请求的方式:

NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:@"https://www.ez-point.com/api/v1/ezpoints"]];
AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request 
    success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
        NSLog(@"%@",@"testing");
        NSLog(@"%@",operation.responseData);

    }
    failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
        NSLog(@"%@", @"Error");
    }];

[operation start];
于 2013-10-29T04:51:06.860 回答