如何将 Location 标头添加到 http 响应？答案

【问题标题】：How to add Location header to the http response?如何将 Location 标头添加到 http 响应？
【发布时间】：2021-02-17 08:30:15
【问题描述】：

我有一个 Java 项目，我正在使用 Servlet 来处理 http 请求。我也使用 Spring

当我收到创建新对象（例如帐户）的请求时，我还想返回带有新创建对象的 GET URL 的“位置”标头。例如：位置：/accounts/1000

我了解标头已添加到 Servlet 过滤器（如果我错了，请纠正我）

public class ApiLogFilter implements Filter {

    private static final Logger LOGGER = LoggerFactory.getLogger("apilogger");

    @Override
    public void init(FilterConfig filterConfig) throws ServletException {

    }

    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {
        HttpServletRequest httpServletRequest = (HttpServletRequest) servletRequest;
        HttpServletResponse httpServletResponse = ((HttpServletResponse) servletResponse);
       
        httpServletResponse.addHeader( "Location","the location value");
        
        try {
          
            filterChain.doFilter(servletRequest, servletResponse);
          
        } finally {
            String queryString = httpServletRequest.getQueryString() != null ? httpServletRequest.getQueryString() : "N/A";
            String logMessage = "URL: " + httpServletRequest.getRequestURL() + ", Query String: " + queryString + ", Response Status: " + httpServletResponse.getStatus() ;
            LOGGER.info(logMessage);
        }
    }

    @Override
    public void destroy() {

    }
}

但我不明白如何从 API 获取位置值

@RequestMapping("/accounts")
public class IgnoreRuleController {

    private AccountService accountService;

    public void setIgnoreRuleService(IgnoreRuleService ignoreRuleService) {
        this.accountService = ignoreRuleService;
    }

    @RequestMapping(method = RequestMethod.POST)
    @ResponseBody
    public String createAccount(@RequestBody Account account) {
        return new Gson().toJson(accountService.createAccount(account));
    }
    
}

【问题讨论】：

标签： java spring servlets http-headers servlet-filters

【解决方案1】：

我在这里找到了解决方案 http://learningviacode.blogspot.com/2013/07/post-with-location.html

您无需对过滤器执行任何操作。在 api 本身中：

  @RequestMapping(method = RequestMethod.POST)
    @ResponseBody
    public ResponseEntity<String> createIgnoreRule(@RequestBody IgnoreRule ignoreRule) {
        String response = new Gson().toJson(ignoreRuleService.createIgnoreRule(ignoreRule));

        final URI location = ServletUriComponentsBuilder
                .fromCurrentServletMapping().path("/ignore_rules/{id}").build()
                .expand(ignoreRule.getId()).toUri();

        final HttpHeaders headers = new HttpHeaders();
        headers.setLocation(location);


        final ResponseEntity<String> entity = new ResponseEntity<>(response, headers, HttpStatus.CREATED);
        return entity;
    }

【讨论】：

【解决方案2】：

很简单，直接传header丢你的方法签名即可：

@RequestMapping(value="/create-account", method = RequestMethod.POST)
@ResponseBody
public String createAccount(@RequestHeader HttpHeaders httpHeader, @RequestBody Account account) {
    var s = httpHeader.get("Location");

    System.out.println(s.get(0));
    return ...
}

事实上，你也可以传递包含所有内容（标题、正文、...）的整个请求：

@RequestMapping(value="/create-account", method = RequestMethod.POST)
@ResponseBody
public String createAccount(HttpServletRequest httpRequest, @RequestBody Account account) {
    var s = httpRequest.getHeader("Location");

    System.out.println(s);

    return ....
}

【讨论】：