Java 当抛出未检查的异常时，ExceptionHandler是否有方法输出参数？

我想要一种能够在

handleUncaughtException

中输出变量

url

的方法。但是，由于异常来自另一个类

Fetcher

，我是否可以知道

url

参数可能导致此未检查异常

---

如果我错了，请纠正我，但我相信我不能将

url

作为字段存储在

TestController

中，因为我同时调用

/content

端点。

您可以做的是在请求属性中设置

url

，然后在

@ExceptionHandler

中检索它

@Component
public class Fetcher {
    private void sendGet(String url) {
        someMethodThatCanThrowUncheckedException();
    }
}

@Controller
public class TestController {
    @Autowired
    private Fetcher fetcher;

    @RequestMapping(value = {"/content"}, method = RequestMethod.GET)
    @ResponseBody
    public String getContent(@RequestParam(value = "url", required = true) String url) {
        fetcher.sendGet(url);
        return "Success";
    }

    @ExceptionHandler({Exception.class})
    @ResponseBody
    public String handleUncaughtException(final Exception exception) {
        return "An internal error has occured trying to fetch url: " + url;
    }
}

然后取回它

@RequestMapping(value = {"/content"}, method = RequestMethod.GET)
@ResponseBody
public String getContent(@RequestParam(value = "url", required = true) String url, HttpServletRequest request) {
    request.setAttribute("url", url);
    fetcher.sendGet(url);
    return "Success";
}

---

您可能应该将属性键设置为常量。

谢谢！我决定使用request.getParameter（“url”）instead@Popcorn如果

url

来自请求参数，这将起作用。它在任何其他情况下都不起作用。

@ExceptionHandler({Exception.class})
@ResponseBody
public String handleUncaughtException(final Exception exception, HttpServletRequest request) {
    String url = (String) request.getAttribute("url");
    if (url != null) 
        return "An internal error has occured trying to fetch url: " + url;
    else 
        return "something else";
}