无法显示来自 responsebody HttpClient get 请求的数据

Question

在我之前的 post 中，我询问了有关使用 C# 中的主体从 Http Get Request 获取数据的问题。现在我面临另一个错误，Android.Util.AndroidRuntimeException: 'Only the original thread that created a view hierarchy can touch its views.'有谁知道如何解决这个问题？

代码

  var client = new HttpClient();

    var request = new HttpRequestMessage
    {
        Method = HttpMethod.Get,
        RequestUri = new Uri("my url"),
        Content = new StringContent("my json body content", Encoding.UTF8, "application/json"),
    };

    var response = await client.SendAsync(request).ConfigureAwait(false);
    response.EnsureSuccessStatusCode();
    var responsebody = await response.Content.ReadAsStringAsync().ConfigureAwait(false);
    string text = responsebody.ToString();
    string[] str = text.Split(new[] { ',', ':' }, StringSplitOptions.RemoveEmptyEntries);
    string result = str[10];
    labelTxt.Text = result;

Answer 1

试试下面的方法（假设这整个事情是一个名为“SomeFunction”的函数）

private void SomeFunction() {
 Device.BeginInvokeOnMainThread(() => {
  var client = new HttpClient();

  var request = new HttpRequestMessage {
   Method = HttpMethod.Get,
    RequestUri = new Uri("my url"),
    Content = new StringContent("my json body content", Encoding.UTF8, "application/json"),
  };

  var response = await client.SendAsync(request).ConfigureAwait(false);
  response.EnsureSuccessStatusCode();
  var responsebody = await response.Content.ReadAsStringAsync().ConfigureAwait(false);
  string text = responsebody.ToString();
  string[] str = text.Split(new [] {
   ',',
   ':'
  }, StringSplitOptions.RemoveEmptyEntries);
  string result = str[10];
  labelTxt.Text = result;
 });
}

Answer 2

您好，请尝试此代码。

public async static Task<T> GetResultAsync<T>(string Url)
{
    try
    {
        if (CrossConnectivity.Current.IsConnected)
        {
            using (var client = new HttpClient())
            {
                client.BaseAddress = new Uri(BaseUrl);
                client.DefaultRequestHeaders.Clear();
                client.SetBearerToken(Helpers.Settings.ServiceToken);
                client.DefaultRequestHeaders.Add("Language", Helpers.Settings.AppLanuageSetting);
                client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
                HttpResponseMessage response = await client.GetAsync($"{Url}").ConfigureAwait(false);
                if (response.IsSuccessStatusCode)
                {
                    var result = await response.Content.ReadAsStringAsync();
                    return JsonConvert.DeserializeObject<T>(result,new JsonSerializerSettings { PreserveReferencesHandling = PreserveReferencesHandling.None });
                }
                else
                {
                    return default(T);
                }
            }
        }
        else
        {
            Acr.UserDialogs.UserDialogs.Instance.Toast("Please check your internet connection");
            return default(T);
        }
    }
    catch (Exception Ex)
    {
        Logging.ErrorLog(Ex, "GetResultAsync");
        return default(T);
    }
}

并使用这个方法

var results = await APIService.GetResultAsync>($"{ApiEndpoints}/{"Paramters1"}/{pageNo}");

Answer 3

正如 curiousBoy 上面所说，您只能从 UI 线程（主线程）修改 UI 个元素。

所以你可以尝试将 labelTxt.Text = result; 放在 Device.BeginInvokeOnMainThread 方法中。

var client = new HttpClient();

var request = new HttpRequestMessage
{
    Method = HttpMethod.Get,
    RequestUri = new Uri("my url"),
    Content = new StringContent("my json body content", Encoding.UTF8, "application/json"),
};

var response = await client.SendAsync(request).ConfigureAwait(false);
response.EnsureSuccessStatusCode();
var responsebody = await response.Content.ReadAsStringAsync().ConfigureAwait(false);
string text = responsebody.ToString();
string[] str = text.Split(new[] { ',', ':' }, StringSplitOptions.RemoveEmptyEntries);
string result = str[10];
Device.BeginInvokeOnMainThread(() => {labelTxt.Text = result;});

无法显示来自 responsebody HttpClient get 请求的数据

Cannot display data from responsebody HttpClient get request

c#

api

httpclient

xamarin.forms