好的所以我被要求创建一个新的问题我现在已经让我的json返回正常并进入我的城市级别和花花公子现在使用以下结构。但我的问题是如何将其作为List返回,以便可以绑定到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当我处理我刚刚在xaml中获取popcorn.dal.city时,如何将列表中的对象返回到listview原因
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>
问题的完整性等级
public class City
{
public string id { get; set; }
public string timing_title { get; set; }
}
public class Citys
{
public List<City> city { get; set; }
}
我的助手功能
public static T Deserialize<T>(this string SerializedJSONString)
{
var stuff = JsonConvert.DeserializeObject<T>(SerializedJSONString);
return stuff;
}
编辑显示基于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;
}
}
同样在调用web服务时,错误检查的最佳方法是什么,我不能使用try catch并发出消息说服务无法联系,但是我应该捕获它们的特定消息。
cinema_city方法返回此json
{ “城市”:[{ “ID”: “5521”, “timing_title”: “拉合尔”},{ “ID”: “5517”, “timing_title”: “卡拉奇”},{ “ID”: “5538”, “timing_title”: “伊斯兰堡”},{ “ID”: “5535”, “timing_title”: “拉瓦尔品第”},{ “ID”: “5518”, “timing_title”: “海得拉巴”},{ “ID”: “5512”, “timing_title”: “费萨拉巴特”},{ “ID”: “8028”, “timing_title”: “古杰朗瓦拉”},{ “ID”: “8027”, “timing_title”:“古吉拉特“}]}
答案 0 :(得分:0)
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到您的Citys类(顺便说一下,将它重命名为Cities?)
所以你的代码将成为:
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List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用这种方式编写代码。 json.net async mehods只是同步方法的包装器。所以我将您的代码简化为:
public async Task<Citys> GetCityListAsync(){
var json = await WCFRESTServiceCall("GET", "cinema_city");
return json.Deserialize(Citys);
}