返回列表而不是项目Webservice Phone 8.1 UAP

本文关键字:Phone UAP Webservice 项目 列表 返回 | 更新日期: 2023-09-27 18:07:28

Ok,所以我已经被要求创建一个新的问题,我已经得到我的json现在返回到我的类的城市fine和dandy现在使用以下结构。但我的问题是我如何返回它作为一个列表,以便可以绑定到一个列表视图例如,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我如何返回对象列表到我的列表视图因为当我处理我只是得到爆米花。dal。city在xaml

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>

问题完备性的类

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

My Helper Function here

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

编辑显示webservice调用,这是基于php mysql的,我不需要改变,只是。net侧到一个列表

   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;
    }
}

当调用webservice时,最好的错误检查方法是什么?我不,我可以使用try catch并放置消息说服务无法联系,但我应该捕获它们的特定消息。

cinema_city方法返回json

{"城市":[{" id ": " 5521 "、"timing_title":"拉合尔"},{" id ": " 5517 ", " timing_title ": "卡拉奇"},{" id ": " 5538 "、"timing_title":"巴基斯坦"},{" id ": " 5535 "、"timing_title":"拉瓦尔品第"},{" id ": " 5518 ", " timing_title ":海德拉巴"},{" id ": " 5512 "、"timing_title":"费萨尔巴德"},{" id ": " 8028 ", " timing_title ": "古吉兰瓦拉"},{" id ": " 8027 "、"timing_title":"Gujrat"}]}

返回列表而不是项目Webservice Phone 8.1 UAP

你又问同样的问题了。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"}]}

你需要将它反序列化到你的城市类(顺便把它重命名为城市也许?)

所以你的代码应该是:

  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;
 }

或者如果您想只返回List而不返回包装器:

 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;
 }

我也不知道你为什么这样写代码。.net异步方法只是同步方法的包装。所以我将你的代码简化为:

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