如何在Spring RestTemplate请求上设置"Accept:“头？
如何在Spring RestTemplate请求上设置"Accept:“头？
提问于 2013-10-08 11:45:36
我想在使用Spring的RestTemplate发出的请求中设置Accept:的值。
下面是我的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;
}下面是我的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);
}这对我来说很有效；我从服务器端得到一个JSON字符串。
我的问题是:如何指定Accept:标头(例如application/json、application/xml等)和请求方法(例如GET、POST等)当我使用RestTemplate时？
回答 5
Stack Overflow用户
发布于 2015-02-06 18:31:09
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报头。
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);Stack Overflow用户
发布于 2017-04-24 22:11:47
如果您像我一样，很难找到一个使用标头和基本身份验证以及rest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());
}
}Stack Overflow用户
发布于 2018-08-17 04:13:06
使用RestTemplate调用RESTful接口
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()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;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/19238715
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