android - 如何从异常中获取消息？

Question

以下是我的一段代码：\

public void LoadProjectFile(String Filename) throws Exception
{
  //  m_Renderer.m_Project=new Project();
   refresh_project();
    m_Renderer.m_SelectedProjectPath =  Filename;

    try {
        m_Renderer.m_Project.load_file(Filename);

    }
    catch (Exception e)

    {
        //Exception a = e.getMessage();
        String a= e.getMessage();

        throw new Exception(a);
    }

    //AppFuncs.m_GisProject=m_Renderer.m_Project;

}



        try
        {

            Map.this.mGLView.LoadProjectFile(AppFuncs.g_path);
             Map.this.mGLView.requestRender();
    }
        catch (Exception e)
        {
            br=1;
             b=e.getMessage();
        }

加载项目文件引发我在 Map 类中收到的异常。This exception contains message: Java.lang.exception: java.lang.exception: file not found.

我只想显示"File not found"消息。那么我应该如何从异常中获取此消息？

Answer 1

捕获所有类型的异常而不是基本异常：

try {
    m_Renderer.m_Project.load_file(Filename);

}
catch (FileNotFoundException e)
{
    String a= "File not found"
    throw new Exception(a);
}
catch (SomeOhterException e){ 
    String a= "Some other message"
    throw new Exception(a);
}
//at the end, but it shouldn't be necessary 
catch (Exception e){ 
    String a= "Something happend we don't know what"
    throw new Exception(a);
}

很快：对不同的异常使用不同的类来显示正确的信息，而不是使用异常消息。

Answer 2

只需抓住FileNotFoundException而不是Exception.

例如：

try {
      //whatever
} catch (FileNotFoundException e) {
      System.out.println("File not found");
}