c# - 返回列表而不是项目 Webservice Phone 8.1 UAP

Question

好的 所以我被要求创建一个新问题 我现在使用以下结构让我的 json 正常返回并进入我的 city fine and dandy 类。但我的问题是如何将它作为列表返回，以便可以绑定(bind)到 ListView ，例如我的 get 调用如下

public async Task<City> GetCityListAsync()
    {

        var tcs = new TaskCompletionSource<City>();
        string jsonresult = await WCFRESTServiceCall("GET", "cinema_city");

        var list = await Task.Run(() => jsonresult.Deserialize<City>());
        tcs.SetResult(list);
        // for testing to show json being returned
        var dialog = new MessageDialog(jsonresult);
        await dialog.ShowAsync();

        return await tcs.Task;
 }

这是我调用方法的地方

private async void citys_loaded(object sender, RoutedEventArgs e)
{
        popcornpk_Dal popcorn_dal = new popcornpk_Dal();

        City _mycitys = await popcorn_dal.GetCityListAsync();

        var listView = (ListView)sender;

       listView.ItemsSource = _mycitys.ToString();
 }

所以我的主要问题是如何将对象列表返回到我的 ListView ，因为当我处理时我只是在 xaml 中获取 popcorn.dal.city

XAML 布局

<Pivot x:Name="myPivot">
        <PivotItem x:Name="pivot_item1" Header="by city" Margin="10,-76,28,-20.833">
            <StackPanel Height="505">
                <Grid Margin="0,0,0,118" Height="514"  >
                    <ListView  x:Name="listByCity" ItemsSource="{Binding}" Loaded="citys_loaded">

                        <DataTemplate>
                            <TextBlock x:Name="City" FontSize="14" Text="{Binding timing_title}"></TextBlock>

                        </DataTemplate>

                    </ListView>

                </Grid>
            </StackPanel>
        </PivotItem>

问题完整性等级

    public class City
    {
        public string id { get; set; }
        public string timing_title { get; set; }
        }
    public class Citys
    {
        public List<City> city { get; set; }
     }

这里是我的辅助函数

public static T Deserialize<T>(this string SerializedJSONString)
{
        var stuff = JsonConvert.DeserializeObject<T>(SerializedJSONString);
        return stuff;
}

编辑以显示基于 php mysql 的 webserivce 调用，我不需要将 .net 端更改为列表

   private async Task<string> WCFRESTServiceCall(string methodRequestType, string methodName, string bodyParam = "")
    {
        string ServiceURI = "/launchwebservice/index.php/webservice/" + methodName;
        HttpClient httpClient = new HttpClient();
        HttpRequestMessage request = new HttpRequestMessage(methodRequestType == "GET" ? HttpMethod.Get : HttpMethod.Post, ServiceURI);
        if (!string.IsNullOrEmpty(bodyParam))
        {
            request.Content = new StringContent(bodyParam, Encoding.UTF8, "application/json");
        }
        HttpResponseMessage response = await httpClient.SendAsync(request);
        string jsongString = await response.Content.ReadAsStringAsync();
        return jsongString;
    }
}

此外，在调用 Web 服务时，这里进行错误检查的最佳方法是什么？我不可以使用 try catch 并发送一条消息说无法联系服务，但我应该捕获它们的特定消息。

cinema_city 方法返回这个 json

{"city":[{"id":"5521","timing_title":"Lahore"},{"id":"5517","timing_title":"Karachi"},{"id":"5538","timing_title":"Islamabad"},{"id":"5535","timing_title":"Rawalpindi"},{"id":"5518","timing_title":"Hyderabad"},{"id":"5512","timing_title":"Faisalabad"},{"id":"8028","timing_title":"Gujranwala"},{"id":"8027","timing_title":"Gujrat"}]}

Answer 1

你又问了同样的问题。如果你有这个 json:

 {"city":[{"id":"5521","timing_title":"Lahore"},  
          {"id":"5517","timing_title":"Karachi"},
          {"id":"5538","timing_title":"Islamabad"},
          {"id":"5535","timing_title":"Rawalpindi"},
          {"id":"5518","timing_title":"Hyderabad"},
          {"id":"5512","timing_title":"Faisalabad"},
          {"id":"8028","timing_title":"Gujranwala"},
          {"id":"8027","timing_title":"Gujrat"}]}

您需要将其反序列化为您的 Citys 类(顺便说一句，也许将其重命名为 Cities？)

所以你的代码将是:

  public async Task<Citys> GetCityListAsync()
  {

        var tcs = new TaskCompletionSource<City>();
        string jsonresult = await WCFRESTServiceCall("GET", "cinema_city");

        var list = await Task.Run(() => jsonresult.Deserialize<Citys>());
        tcs.SetResult(list);
        // for testing to show json being returned
        var dialog = new MessageDialog(jsonresult);
        await dialog.ShowAsync();

        return await tcs.Task;
 }

或者如果您只想返回列表而不是包装器:

 public async Task<List<City>> GetCityListAsync()
    {

        var tcs = new TaskCompletionSource<List<City>>();
        string jsonresult = await WCFRESTServiceCall("GET", "cinema_city");

        var list = await Task.Run(() => jsonresult.Deserialize<Citys>());
        tcs.SetResult(list.city);
        // for testing to show json being returned
        var dialog = new MessageDialog(jsonresult);
        await dialog.ShowAsync();

        return await tcs.Task;
 }

另外我不确定你为什么要这样写代码。 json.net 异步方法只是同步方法的包装器。所以我会将您的代码简化为:

 public async Task<Citys> GetCityListAsync(){ 
    var json = await WCFRESTServiceCall("GET", "cinema_city");
    return json.Deserialize(Citys);
 }