Question

我正在尝试设置一个方案来统一处理Spring中的异常。因此，我需要一种方法将上下文信息传递到@ExceptionHandler - 带注释的方法，例如考虑以下内容：

@ExceptionHandler(Exception.class)
public void handleException(Exception ex, HttpServletRequest request) {
    // Need access to myContext from login()
}

@RequestMapping(value = "{version}/login", method = RequestMethod.POST)
public void login(HttpServletRequest request, @PathVariable String version, @RequestParam("userName") String userName, @RequestParam("password") String password, ModelMap model) throws Exception {
    ...
    myContext = "Some contextual information"
    ...
    i_will_always_throw_an_exception();
}

由于Spring负责将抛出的异常转换为handleException()的调用，因此我很难找到将myContext传递给处理程序的方法。我有一个想法是创建一个HttpServletRequest的子类。如果这种方法有效，我会得到这样的代码：

@ExceptionHandler(Exception.class)
public void handleException(Exception ex, MyCustomHttpServletRequest request) {
    // I now have access to the context via the following
    String myContext = request.getContext();
}

@RequestMapping(value = "{version}/login", method = RequestMethod.POST)
public void login(MyCustomHttpServletRequest request, @PathVariable String version, @RequestParam("userName") String userName, @RequestParam("password") String password, ModelMap model) throws Exception {
    ...
    myContext = "Some contextual information"
    request.setContext(myContext);
    ...
    i_will_always_throw_an_exception();
}

但是，如果我遵循这种方法，我如何正确使用我自己的任意子类HttpServletRequest来使其工作？

Answer 1

难道你不能把它放到异常中（如果需要的话 - 用新的异常包装原始异常）？

@ExceptionHandler(MyContextualException.class)
public void handleException(MyContextualException ex) {
    // Need access to myContext from login()
}

@RequestMapping(value = "{version}/login", method = RequestMethod.POST)
public void login(HttpServletRequest request, @PathVariable String version, @RequestParam("userName") String userName, @RequestParam("password") String password, ModelMap model) throws Exception {
    ...
    myContext = "Some contextual information"
    ...
    try {
        i_will_always_throw_an_exception();
    } catch (Exception ex) {
        throw new MyContextualException(myContext, ex);
    }
}

另一种方法是将上下文作为请求属性传递：

request.setAttribute("myContext", myContext);

在Spring 3.0异常处理程序中获取上下文信息

1 个答案: