238

我想Accept:在我使用 Spring 的RestTemplate.

这是我的 Spring 请求处理代码

@RequestMapping(
    value= "/uom_matrix_save_or_edit", 
    method = RequestMethod.POST,
    produces="application/json"
)
public @ResponseBody ModelMap uomMatrixSaveOrEdit(
    ModelMap model,
    @RequestParam("parentId") String parentId
){
    model.addAttribute("attributeValues",parentId);
    return model;
}

这是我的 Java REST 客户端：

public void post(){
    MultiValueMap<String, String> params = new LinkedMultiValueMap<String, String>();
    params.add("parentId", "parentId");
    String result = rest.postForObject( url, params, String.class) ;
    System.out.println(result);
}

这对我有用；我从服务器端得到一个 JSON 字符串。

我的问题是：当我使用 RestTemplate 时，如何指定Accept:标头（例如application/json, application/xml, ... ）和请求方法（例如GET, , ... ）？POST

4

6 回答 6

427

我建议使用其中一种exchange接受 的方法，HttpEntity您也可以为其设置HttpHeaders. （您还可以指定要使用的 HTTP 方法。）

例如，

RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setAccept(Collections.singletonList(MediaType.APPLICATION_JSON));

HttpEntity<String> entity = new HttpEntity<>("body", headers);

restTemplate.exchange(url, HttpMethod.POST, entity, String.class);

我更喜欢这个解决方案，因为它是强类型的，即。exchange期望一个HttpEntity.

但是，您也可以将其HttpEntity作为request参数传递给postForObject.

HttpEntity<String> entity = new HttpEntity<>("body", headers);
restTemplate.postForObject(url, entity, String.class);

RestTemplate#postForObjectJavadoc中提到了这一点。

该request参数可以是 aHttpEntity以便向请求添加额外的 HTTP 标头。

于 2013-10-08T04:19:23.140 回答
150

您可以在 RestTemplate 中设置一个拦截器“ClientHttpRequestInterceptor”，以避免每次发送请求时都设置标头。

public class HeaderRequestInterceptor implements ClientHttpRequestInterceptor {

        private final String headerName;

        private final String headerValue;

        public HeaderRequestInterceptor(String headerName, String headerValue) {
            this.headerName = headerName;
            this.headerValue = headerValue;
        }

        @Override
        public ClientHttpResponse intercept(HttpRequest request, byte[] body, ClientHttpRequestExecution execution) throws IOException {
            request.getHeaders().set(headerName, headerValue);
            return execution.execute(request, body);
        }
    }

然后

List<ClientHttpRequestInterceptor> interceptors = new ArrayList<ClientHttpRequestInterceptor>();
interceptors.add(new HeaderRequestInterceptor("Accept", MediaType.APPLICATION_JSON_VALUE));

RestTemplate restTemplate = new RestTemplate();
restTemplate.setInterceptors(interceptors);
于 2015-02-06T10:31:08.770 回答
28

如果像我一样，你很难找到一个使用带有基本身份验证的标头和其余模板交换 API 的示例，这就是我最终解决的问题......

private HttpHeaders createHttpHeaders(String user, String password)
{
    String notEncoded = user + ":" + password;
    String encodedAuth = Base64.getEncoder().encodeToString(notEncoded.getBytes());
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON);
    headers.add("Authorization", "Basic " + encodedAuth);
    return headers;
}

private void doYourThing() 
{
    String theUrl = "http://blah.blah.com:8080/rest/api/blah";
    RestTemplate restTemplate = new RestTemplate();
    try {
        HttpHeaders headers = createHttpHeaders("fred","1234");
        HttpEntity<String> entity = new HttpEntity<String>("parameters", headers);
        ResponseEntity<String> response = restTemplate.exchange(theUrl, HttpMethod.GET, entity, String.class);
        System.out.println("Result - status ("+ response.getStatusCode() + ") has body: " + response.hasBody());
    }
    catch (Exception eek) {
        System.out.println("** Exception: "+ eek.getMessage());
    }
}
于 2017-04-24T14:11:47.080 回答
20

使用 RestTemplate 调用 RESTful API

示例 1：

RestTemplate restTemplate = new RestTemplate();
// Add the Jackson message converter
restTemplate.getMessageConverters()
                .add(new MappingJackson2HttpMessageConverter());
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
headers.set("Authorization", "Basic XXXXXXXXXXXXXXXX=");
HttpEntity<String> entity = new HttpEntity<String>("parameters", headers);
restTemplate.getInterceptors()
                .add(new BasicAuthorizationInterceptor(USERID, PWORD));
String requestJson = getRequetJson(Code, emailAddr, firstName, lastName);
response = restTemplate.postForObject(URL, requestJson, MYObject.class);

示例 2：

RestTemplate restTemplate = new RestTemplate();
String requestJson = getRequetJson(code, emil, name, lastName);
HttpHeaders headers = new HttpHeaders();
String userPass = USERID + ":" + PWORD;
String authHeader =
    "Basic " + Base64.getEncoder().encodeToString(userPass.getBytes());
headers.set(HttpHeaders.AUTHORIZATION, authHeader);
headers.setContentType(MediaType.APPLICATION_JSON);
headers.setAccept(Collections.singletonList(MediaType.APPLICATION_JSON));
HttpEntity<String> request = new HttpEntity<String>(requestJson, headers);
ResponseEntity<MyObject> responseEntity;
responseEntity =
    this.restTemplate.exchange(URI, HttpMethod.POST, request, Object.class);
responseEntity.getBody()

该getRequestJson方法创建一个 JSON 对象：

private String getRequetJson(String Code, String emailAddr, String name) {
    ObjectMapper mapper = new ObjectMapper();
    JsonNode rootNode = mapper.createObjectNode();
    ((ObjectNode) rootNode).put("code", Code);
    ((ObjectNode) rootNode).put("email", emailAdd);
    ((ObjectNode) rootNode).put("firstName", name);
    String jsonString = null;
    try {
        jsonString = mapper.writerWithDefaultPrettyPrinter()
                               .writeValueAsString(rootNode);
    }
    catch (JsonProcessingException e) {
        e.printStackTrace();
    }
    return jsonString;
}
于 2018-08-16T20:13:06.357 回答
6

HttpHeaders无需创建的简短解决方案：

RequestEntity<Void> request = RequestEntity.post(URI.create(url))
                .accept(MediaType.APPLICATION_JSON)
                .contentType(MediaType.APPLICATION_JSON)
                // any other headers
                .header("PRIVATE-TOKEN", "token")
                .build();

ResponseEntity<String> response = restTemplate.exchange(request, String.class);
return response.getBody();

更新：但如果特定的标题HttpHeaders变得简单：

RequestEntity.post(URI.create(AMOCRM_URL + url))
            .contentType(MediaType.APPLICATION_JSON)
            .headers(
                    new HttpHeaders() {{
                        setBearerAuth(getAccessToken());
                    }})
            .body(...)
于 2021-05-29T14:53:54.340 回答
5

这是一个简单的答案。希望它可以帮助某人。

import org.springframework.boot.devtools.remote.client.HttpHeaderInterceptor;
import org.springframework.http.MediaType;
import org.springframework.http.client.ClientHttpRequestInterceptor;
import org.springframework.web.client.RestTemplate;


public String post(SomeRequest someRequest) {
    // create a list the headers 
    List<ClientHttpRequestInterceptor> interceptors = new ArrayList<>();
    interceptors.add(new HttpHeaderInterceptor("Accept", MediaType.APPLICATION_JSON_VALUE));
    interceptors.add(new HttpHeaderInterceptor("ContentType", MediaType.APPLICATION_JSON_VALUE));
    interceptors.add(new HttpHeaderInterceptor("username", "user123"));
    interceptors.add(new HttpHeaderInterceptor("customHeader1", "c1"));
    interceptors.add(new HttpHeaderInterceptor("customHeader2", "c2"));
    // initialize RestTemplate
    RestTemplate restTemplate = new RestTemplate();
    // set header interceptors here
    restTemplate.setInterceptors(interceptors);
    // post the request. The response should be JSON string
    String response = restTemplate.postForObject(Url, someRequest, String.class);
    return response;
}
于 2018-07-18T14:46:38.107 回答