Xamarin管理未处理的异常

时间:2019-05-28 07:24:23

标签: c# xamarin xamarin.forms xamarin.android

我试图捕获Xamarin应用程序在发生未处理的异常时抛出的堆栈跟踪,以便可以将日志写入磁盘以供以后报告。

目前,我正在将异常和消息写入类似这样的字符串中:

    private void OutputException(StringBuilder report, Exception exception)
    {
        if (exception != null)
        {
            report.Append("\n\n");
            report.Append("\n\n");
            report.Append(exception.Message);
            report.Append("\n\n");
            report.Append(exception.StackTrace);

            OutputException(report, exception.InnerException);
        }
    }

我正在使用与此处所述相同的技术:

https://peterno.wordpress.com/2015/04/15/unhandled-exception-handling-in-ios-and-android-with-xamarin/

将事件侦听器附加到

AppDomain.CurrentDomain.UnhandledException += CurrentDomain_UnhandledException;

代码捕获了堆栈跟踪,但是当我查看记录的跟踪时,它不包含我的任何类或方法名称:

at (wrapper dynamic-method) System.Object.26(intptr,intptr,intptr)
at (wrapper native-to-managed) System.Object.26(intptr,intptr,intptr)
--- End of stack trace from previous location where exception was thrown ---
at Java.Interop.JniEnvironment+InstanceMethods.CallObjectMethod 
(Java.Interop.JniObjectReference instance, Java.Interop.JniMethodInfo method, 
 Java.Interop.JniArgumentValue* args) [0x00069] in <8acc8089c2ed40d08469fbaa6710a44c>:0 
 at Java.Interop.JniPeerMembers+JniInstanceMethods.InvokeVirtualObjectMethod (System.String encodedMember, Java.Interop.IJavaPeerable self, 
Java.Interop.JniArgumentValue* parameters) [0x0002a] in <8acc8089c2ed40d08469fbaa6710a44c>:0 
  at Java.Lang.Throwable.get_Message () [0x0000a] in <6f8818d6ffa14f4bad6fb5f5d0f0e665>:0 
  --- End of managed Java.Lang.Exception stack trace ---

java.lang.Exception:

如何处理异常并使其产生代码中的类名?我不需要行号,我只需要类和方法名。

0 个答案:

没有答案
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