当抛出未经检查的异常时，ExceptionHandlers 有没有办法输出参数？

Question

@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;
    }
}

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，有什么方法可以知道url 参数可能导致这个未经检查的异常吗？

如果我错了，请纠正我，但我相信我不能只将url 作为字段存储在TestController 中，因为我同时调用了/content 端点。

Answer 1

您可以做的是在请求属性中设置url 并在您的@ExceptionHandler 中检索它。

@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";
}

然后检索它

@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";
}

您可能应该将属性键设为常量。