我开发了一个通用应用程序,它需要几个自定义异常来捕获调用webservices时遇到的错误:NoInternetAccessException,NoJSONException,UserTimeoutException,...
以下是其中一个类:
的示例 public class NoInternetAccessException : Exception
{
private DateTime time;
public DateTime Time { get { return time; } }
public NoInternetAccessException(string message, DateTime time)
: base(message)
{
this.time = time;
}
}
我在几个地方发现了这些例外情况:
在 JSONParser 中,我在其中创建URI并调用Client.GetAsync方法:
...
try
{
CancellationTokenSource cts = new CancellationTokenSource(Timeout);
response = await client.GetAsync(uri, HttpCompletionOption.ResponseHeadersRead).AsTask(cts.Token);
if (response.IsSuccessStatusCode)
return response;
}
catch (TaskCanceledException)
{
if ((_currentRetries == MaxRetries)
throw new UserTimeoutException("User Timeout Exception", DateTime.Now);
else
_currentRetries++;
}
catch (Exception e)
{
if (e.HResult == -2147012889)
throw new WrongUrlException("Wrong URL Exception", e, DateTime.Now);
}
...
在 WebService助手中,我管理所有的webservices调用:
public static async Task<Infos> GetInfos(String url, List<KeyValuePair<String, String>> parameters, String protocol)
{
var response = await JSONParser.getJSONFromUrl(url, parameters, "");
Infos infos = new Infos();
try
{
WsResponse wsResponse = JsonConvert.DeserializeObject<Infos>(response.ToString());
infosCe = JsonConvert.DeserializeObject<Infos>(wsResponse.data.ToString());
return infosCe;
}
catch (Exception e)
{
throw new DeserializeException("Deserialize exception", e, DateTime.Now, "Infos");
}
}
在调用webservices之后,所有这些例外都是最终由ViewModels 捕获:
private async Task<Infos> WebServiceGetInfos()
{
...
try
{
Infos infos = await WebServices.GetInfos(url, parameters, "");
return infosCe;
}
// Exceptions
catch (DeserializeException dE)
{
ExceptionsMsgboxHelper.MsgboxDeserialize(dE);
return null;
}
catch (NoInternetAccessException niaE)
{
ExceptionsMsgboxHelper.MsgboxNoInternetAccess(niaE, true, "");
return null;
}
catch (NoJSONException njsonE)
{
ExceptionsMsgboxHelper.MsgboxNoJSON(njsonE);
return null;
}
...
}
我希望每个异常调用 ExceptionsMsgboxHelper 帮助程序,它会显示每个异常的特定消息对话框:
public async static void MsgboxNoJSON(NoJSONException njsonE)
{
Windows.UI.Popups.MessageDialog msgbox =
new Windows.UI.Popups.MessageDialog("There is a problem when retrieving data. If the problem persists, please contact your administrator.",
"Unexpected data received");
await msgbox.ShowAsync();
}
=&GT;但这不起作用,因为Message Dialog在Try-Catch子句中不起作用......
我也在stackoverflow上寻找解决方案:
Message dialog not showing in catch clause
Try-Catch doesn't show Message Dialog box with await
但我不知道如何使这个解决方案适应我的情况,因为我使用了几个自定义例外:
public static async Task Foo()
{
Exception e = null;
try
{
//just something to throw an exception
int a = 0;
int n = 1 / a;
}
catch (Exception ex)
{
e = ex;
}
if (e != null)
await ShowDialog();
}
=&GT;这样做是否比将此代码复制到每个自定义异常更好?
...
DeserializeException dEx = null;
NoInternetAccessException niaEx = null;
NoJSONException njsonEx = null;
try
{
Infos infos = await WebServices.GetInfos(url, parameters, "");
return infosCe;
}
// Exceptions
catch (DeserializeException dE)
{
dEx = de;
}
catch (NoInternetAccessException niaE)
{
niaEx = niaE;
}
catch (NoJSONException njsonE)
{
njsonEx=njsonE;
}
...
if (dEx != null)
{
ExceptionsMsgboxHelper.MsgboxDeserialize(dEx);
return null;
}
if (niaEx != null)
{
ExceptionsMsgboxHelper.MsgboxNoInternetAccess(niaEx);
return null;
}
if (njsonEx != null)
{
ExceptionsMsgboxHelper.MsgboxNoJSON(njsonEx);
return null;
}
...
=&GT;它似乎并不强大或可维护......
答案 0 :(得分:0)
我终于通过捕获一个通用异常并测试它来解决我的问题:
...
Exception ex = null;
try
{
Infos infos = await WebServices.GetInfos(url, parameters, "");
return infosCe;
}
// Exceptions
catch (Exception e)
{
ex = e;
}
if (ex != null)
{
if (ex is DeserializeException)
{
DeserializeException dE = (DeserializeException)ex;
ExceptionsMsgboxHelper.MsgboxDeserialize(dE);
}
if (ex is NoInternetAccessException)
{
NoInternetAccessException niaE = (NoInternetAccessException)ex;
ExceptionsMsgboxHelper.MsgboxNoInternetAccess(niaE);
}
if (ex is NoJSONException)
{
NoJSONException njsonE = (NoJSONException)ex;
ExceptionsMsgboxHelper.MsgboxNoJSON(njsonE);
}
...
}
return null;
<强> =&GT;如果您认为有更好的方法,请指出我。