用于检查不同异常条件的大量 if/else 块的替代方案？

Question

除了像这样的大量 if/else 块之外，是否有更优雅的方法来检查所有可能的异常类型？

public ModelAndView resolveException(HttpServletRequest request, 
HttpServletResponse response, Object obj, Exception e) {

    if (e instanceof BadException)
    {
      displayMessage("That was bad.");
    }

    else if (e instanceof ReallyBadException)
    {
      displayMessage("That was really bad.");
    }

    else if (e instanceof ReallyReallyBadException)
    {
      displayMessage("That was really really bad.");
    }

    // ...
    // and so on
    // ...

    return null;
}

Answer 1

如果您控制此方法中最终出现的异常，则可以利用其中嵌入的消息：

public ModelAndView resolveException(HttpServletRequest request, HttpServletResponse response, Object obj, Exception e) {
    displayMessage(e.getMessage());
    return null;
}

但只有当消息对最终用户有意义时才应该这样做。通常，您应该以用户不会注意到任何错误的方式处理异常。

Answer 2

public ModelAndView resolveException(HttpServletRequest request, 
        HttpServletResponse response, Object obj, Exception exception)
{
    try
    { 
        throw exception;
    }
    catch(BadException e)
    {
        displayMessage("That was bad.");
    } catch (ReallyBadException e)
    {
      displayMessage("That was really bad.");
    } catch (ReallyReallyBadException e)
    {
      displayMessage("That was really really bad.");
    }

    // ...
    // and so on
    // ...

    return null;
}

Answer 3

使用多个 catch 块，请参阅此处。

Answer 4

您可以为要显示的消息创建特定异常类型的预定义映射：

private Map<Class<?>, String> exceptionMessages = ...;

public ModelAndView resolveException(HttpServletRequest request, HttpServletResponse response, Object obj, Exception e) {

    if exceptionMessages.containsKey(e.getClass()) {
        displayMessage(exceptionMessages.get(e.getClass()));
    }
    else {
         // what to do by default?
    }
    return ...;
}

或者更好的是，您可以将映射外部化到属性文件或 ResourceBundle，以便您可以从代码以外的某个位置加载用户可见的字符串（和/或全球化）。

此代码示例无法对异常类型的层次结构使用相同的消息，但如果在映射中未找到直接类，则遍历继承链应该非常简单。