【WEB系列】RestTemplate之非200状态码信息捕获

SpringBootWEB系列RestTemplate
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文章目录
  1. I. 项目环境
    1. 1. 鉴权端点
  2. II. 异常捕获
    1. 1. 未捕获场景
    2. 2. 异常捕获
  3. II. 其他
    1. 0. 项目&系列博文
    2. 1. 一灰灰Blog

前面介绍的RestTemplate的使用，都是接口正常返回200的状态码case，当返回非200状态码时，会直接抛异常，如果我希望能捕获异常，并针对正常获取返回的message，可以如何处理呢？

I. 项目环境

博文测试项目完全基于【WEB系列】RestTemplate基础用法小结的项目环境，建议配合查看

基本环境：IDEA + maven + SpringBoot 2.2.1.RELEASE

测试的REST服务借助前一篇的鉴权，如果鉴权失败，则返回401状态码，具体实现如下

1. 鉴权端点

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private String getHeaders(HttpServletRequest request) {
    Enumeration<String> headerNames = request.getHeaderNames();
    String name;

    JSONObject headers = new JSONObject();
    while (headerNames.hasMoreElements()) {
        name = headerNames.nextElement();
        headers.put(name, request.getHeader(name));
    }
    return headers.toJSONString();
}

private String getParams(HttpServletRequest request) {
    return JSONObject.toJSONString(request.getParameterMap());
}

private String getCookies(HttpServletRequest request) {
    Cookie[] cookies = request.getCookies();
    if (cookies == null || cookies.length == 0) {
        return "";
    }

    JSONObject ck = new JSONObject();
    for (Cookie cookie : cookies) {
        ck.put(cookie.getName(), cookie.getValue());
    }
    return ck.toJSONString();
}

private String buildResult(HttpServletRequest request) {
    return buildResult(request, null);
}

private String buildResult(HttpServletRequest request, Object obj) {
    String params = getParams(request);
    String headers = getHeaders(request);
    String cookies = getCookies(request);

    if (obj != null) {
        params += " | " + obj;
    }

    return "params: " + params + "\nheaders: " + headers + "\ncookies: " + cookies;
}

/**
 * 标准的http auth验证
 *
 * @param request
 * @param response
 * @return
 * @throws IOException
 */
@GetMapping(path = "auth")
public String auth(HttpServletRequest request, HttpServletResponse response) throws IOException {
    String auth = request.getHeader("Authorization");
    if (StringUtils.isEmpty(auth)) {
        response.setStatus(401);
        response.setHeader("WWW-Authenticate", "Basic realm=\"input username and password\"");
        return buildResult(request) + "\n>>>no auth header";
    }

    String[] userAndPass = new String(new BASE64Decoder().decodeBuffer(auth.split(" ")[1])).split(":");
    if (userAndPass.length < 2) {
        response.setStatus(401);
        response.setHeader("WWW-Authenticate", "Basic realm=\"input username and password\"");
        return buildResult(request) + "\n>>>illegal auth: " + auth;
    }

    if ("user".equalsIgnoreCase(userAndPass[0]) && "pwd".equalsIgnoreCase(userAndPass[1])) {
        return buildResult(request) + "\n>>>auth: success!";
    }

    response.setStatus(401);
    response.setHeader("WWW-Authenticate", "Basic realm=\"input username and password\"");
    return buildResult(request) + "\n>>>illegal user or pwd!";
}

一个简单的鉴权逻辑如上，从请求头中拿到Authorization对应的value，并解析用户名密码，如果满足则正确返回；如果不存在or不满足，则返回http状态码为401，并携带对应的提示信息

II. 异常捕获

1. 未捕获场景

当我们直接像之前一样使用RestTemplate时，看一下效果如何

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try {
    // 如果返回状态码不是200，则直接抛异常，无法拿到responseBody
    RestTemplate restTemplate = new RestTemplate();
    HttpEntity<String> ans =
            restTemplate.getForEntity("http://127.0.0.1:8080/auth?name=一灰灰&age=20", String.class);
    log.info("exception with no auth res: {}", ans);
} catch (Exception e) {
    log.info("exception with no auth error: {}", e);
}

输出如下，走入了catch逻辑，从异常堆栈中，也只能看到401 Unauthorized，拿不到返回的Response body

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(exception with no auth error: {}

org.springframework.web.client.HttpClientErrorException$Unauthorized: 401 Unauthorized
	at org.springframework.web.client.HttpClientErrorException.create(HttpClientErrorException.java:81)
	at org.springframework.web.client.DefaultResponseErrorHandler.handleError(DefaultResponseErrorHandler.java:123)
	at org.springframework.web.client.DefaultResponseErrorHandler.handleError(DefaultResponseErrorHandler.java:102)
	at org.springframework.web.client.ResponseErrorHandler.handleError(ResponseErrorHandler.java:63)
	at org.springframework.web.client.RestTemplate.handleResponse(RestTemplate.java:785)
	at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:743)
	at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:677)
	at org.springframework.web.client.RestTemplate.getForEntity(RestTemplate.java:345)
	at com.git.hui.boot.resttemplate.rest.RestTemplateDemo.exception(RestTemplateDemo.java:354)
	...

2. 异常捕获

更详细原理定位请参考：【WEB系列】RestTemplate 4xx/5xx 异常信息捕获

为了处理上面的问题，我们可以设置自定义的ResponseErrorHandler来处理

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RestTemplate restTemplate = new RestTemplate();
restTemplate.setErrorHandler(new ResponseErrorHandler() {
    @Override
    public boolean hasError(ClientHttpResponse clientHttpResponse) throws IOException {
        return false;
    }

    @Override
    public void handleError(ClientHttpResponse clientHttpResponse) throws IOException {
        log.info("some error!");
    }
});
HttpEntity<String> ans = restTemplate.getForEntity("http://127.0.0.1:8080/auth?name=一灰灰&age=20", String.class);
log.info("exception with no auth after errorHandler res: {}", ans);

输出如下, 401为返回的状态码，其中也包含了ResponseBody，然后再业务中根据状态码和返回结果进行处理即可

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(exception with no auth after errorHandler res: <401,params: {"name":["一灰灰"],"age":["20"]}
headers: {"host":"127.0.0.1:8080","connection":"keep-alive","accept":"text/plain, application/json, application/*+json, */*","user-agent":"Java/1.8.0_171"}
cookies: 
>>>no auth header,[WWW-Authenticate:"Basic realm="input username and password"", Content-Type:"text/plain;charset=UTF-8", Content-Length:"227", Date:"Mon, 29 Jun 2020 09:57:06 GMT"]>

II. 其他

0. 项目&系列博文

博文

源码

1. 一灰灰Blog

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