无法在Xamarin android的Portable类库中使用wcf服务POST方法

时间:2016-03-17 10:17:20

标签: c# android wcf xamarin

我正在使用Visual Studio中的可移植类库中的wcf webservices(POST方法)for xamarin android.App在POST方法中崩溃.Below是我的代码

IN PCL

的Class1.cs

namespace XamarinServiceCall
{
    public sealed class Class1 : IRestService
    {

        public async Task<string> GetData(UserModel model)
        {
            using (var client = new HttpClient())
            {
                var content = new StringContent(JsonConvert.SerializeObject(model), Encoding.UTF8, "application/json");

                var result = await client.PostAsync("http://xamarin-rest-service/LoginService/ValidateLogin", content);
                return await result.Content.ReadAsStringAsync();

            }
        }
    }
}

IRestService.cs

namespace XamarinServiceCall
{
    public interface IRestService
    {
        Task<string> GetData(UserModel model);
    }
}

UserModel.cs

public class UserModel
    {

        public string loginFrom { get; set; }

        public string domainName { get; set; }
        public string userName { get; set; }
        public string password { get; set; }

        //I get response of below parameters
        public int DomainType { get; set; }
        public string FirstName { get; set; }

        public int Status { get; set; }
        public int UserId { get; set; }
      }

在Xamarin Android APp中

MainActivity.cs

namespace ServiceSample
{
    [Activity(Label = "LayoutSample", MainLauncher = true, Icon = "@drawable/icon")]
    public class MainActivity : Activity
    {
            protected async override void OnCreate(Bundle bundle)
        {
            base.OnCreate(bundle);

            // Set our view from the "main" layout resource
            SetContentView(Resource.Layout.Main);

            // Get our button from the layout resource,
            // and attach an event to it
            Button button = FindViewById<Button>(Resource.Id.MyButton);

            XamarinServiceCall.Class1 serviceClass = new XamarinServiceCall.Class1();
           var model = new XamarinServiceCall.UserModel { domainName = "domainname" ,userName = "username" , password = "password" ,loginFrom = "App" };

            var result = await serviceClass.GetData(model);
            button.Text = "get call says: " + result;

        }
    }
}
邮递员答复:

enter image description here

当我调试时,我在结果变量中获得此响应。

enter image description here

当我指向postAsync Function时,它显示Expression无效

enter image description here

1 个答案:

答案 0 :(得分:0)

您不应该尝试在MainActivity的OnCreate函数内执行任何重要工作,例如调用Web服务。

当然,你不应该让它异步并在其中进行异步调用。请查看Xamarin的例子。

通常,在OnCreate中,您可以进行一些初始化,然后为Application类调用LoadApplication(可以在PCL项目中定义)。