Question

我有简单的java spring方法来创建对象

RestTemplate restTemplate = new RestTemplate();
Address address = restTemplate.getForObject(url, Address.class);

但服务器用错误的 Content-Type： text / plain 而不是 application / json （在Postman中检查）回应JSON字符串。我得到了例外：

无法提取响应：找不到合适的HttpMessageConverter用于响应类型[类地址]和内容类型[text / plain; charset = utf-8]

所以我认为，我需要将更改响应头 Content-Type 改为 application / json ，即MappingJackson2HttpMessageConverter找出JSON字符串并运行代码。

Answer 1

尝试了一个小时之后，我找到了一个简单易行的方法。

默认情况下，Json转换器仅支持＆＃34; application / json ＆＃34;。我们只是覆盖它以支持＆＃34; text / plain ＆＃34;。

MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();

// support "text/plain"
converter.setSupportedMediaTypes(Arrays.asList(TEXT_PLAIN, APPLICATION_JSON));

RestTemplate template = new RestTemplate();
template.getMessageConverters().add(converter);

// It's ok now
MyResult result = tmp.postForObject("http://url:8080/api", 
            new MyRequest("param value"), MyResult.class);

Answer 2

谢谢你的帮助！如果我无法更改响应的标题。我用右标题创建了新的响应对象。

            ClientHttpRequest clientHttpRequest = new SimpleClientHttpRequestFactory().createRequest(URI.create(str), org.springframework.http.HttpMethod.GET);
            final ClientHttpResponse clientHttpResponse = clientHttpRequest.execute();
            MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();
            Address address = new Address();
            //It always true, because service always returns 200 OK
            if (clientHttpResponse.getStatusCode() == HttpStatus.OK) {
                address = (Address) converter.read(address.getClass(), new HttpInputMessage() {
                    public InputStream getBody() throws IOException {
                        return clientHttpResponse.getBody();
                    }

                    public HttpHeaders getHeaders() {
                        HttpHeaders httpHeaders = new HttpHeaders();
                        httpHeaders.putAll(clientHttpResponse.getHeaders());
                        httpHeaders.put("Content-Type", Collections.singletonList(MediaType.APPLICATION_JSON_VALUE));
                        return httpHeaders;
                    }
                });
                busStop.setNearestAddress(address.toString());
            }

我确信这不是简单而好的解决方案，但它确实有效。

Answer 3

要为您的请求设置内容类型，您可以执行以下操作：

    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON);
    HttpEntity entity = new HttpEntity(headers);
    ResponseEntity<Address> response = restTemplate.exchange(url, HttpMethod.GET,  entity, Address.class);
    Address address = response.getBody();

如何在Spring RestTemplate的get请求中更改响应http头？

3 个答案: