Question

我在Spring Boot应用程序中使用Spring Rest Template。

即使我通过了凭据，我也总是收到401 Unauthorized错误。

我可以通过Chrome REST Web服务客户端访问此服务。

是否有一种简化的方法来访问SpringBoot中的REST模板。

以下是到目前为止完成的代码段，导致401错误

 private DetailsBean invokeDetailsRestService(UserParam userParam){
    ResponseEntity<DetailsBean> responseEntity = null;
    String url = "https://dev.com/app/identifyuser/";
    RestClientConfig restClientConfig =new RestClientConfig("user123","pass123");
    responseEntity= restClientConfig.postForEntity(url, userParam, DetailsBean.class);
    log.debug("User Details : {} ", responseEntity.getBody());
    return responseEntity.getBody();
}

public ClientHttpRequestFactory getRequestFactory(String userName,String password){
    CredentialsProvider credsProvider = new BasicCredentialsProvider();
    credsProvider.setCredentials( new AuthScope(null, -1), new UsernamePasswordCredentials(userName,password) );
    HttpClient httpClient = HttpClients.custom().setDefaultCredentialsProvider(credsProvider).build();
    return new HttpComponentsClientHttpRequestFactory(httpClient);
}

RestClientConfig类

  public RestClientConfig(String username, String password) {
    CredentialsProvider credsProvider = new BasicCredentialsProvider();
    credsProvider.setCredentials(
            new AuthScope(null, -1),
            new UsernamePasswordCredentials(username, password));
    HttpClient httpClient = HttpClients.custom().setDefaultCredentialsProvider(credsProvider).build();
    setRequestFactory(new HttpComponentsClientHttpRequestFactory(httpClient));
}

错误：

WARN c7af55b5-1cac-4db6-a202-202416c27ba4 
 12612 --- [apr-8082-exec-8] o.a.http.impl.auth.HttpAuthenticator    
: NEGOTIATE authentication error: 
No valid credentials provided (Mechanism level: 
No valid credentials provided (Mechanism level: 
Failed to find any Kerberos tgt))

Answer 1

使用以下代码解决了授权问题..

应使用以下代码将凭据传递给Spring REST模板：

   String userAndPass = "Test:Test123";

   HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_XML);
    headers.add("Authorization", "Basic " +  Base64Utility.encode(userAndPass.getBytes()));

Answer 2

当我试图打电话给webservice时，我遇到了类似的问题，这解决了我的问题：

restTemplate.getInterceptors().add(new BasicAuthorizationInterceptor("userName", "password"));
restTemplate.postForObject('','',''');

传递这样的凭据，它应该解决问题。

Answer 3

我使用了spring boot 2.2.4.RELEASE版本。然后我按照下面的方式工作。

RestTemplate restTemplate = new RestTemplate();
restTemplate.getInterceptors().add(new BasicAuthenticationInterceptor(username, password));
RequestDto requestDto = new RequestDto();
// set parameter 
ResponseDto response = restTemplate.postForObject(URL, requestDto, ResponseDto.class);

或

RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setBasicAuth(username, password);
RequestDto requestDto = new RequestDto();
// set parameter 
HttpEntity<RequestDto> request = new HttpEntity<>(requestDto, headers);
ResponseDto response = restTemplate.postForObject(URL, request, ResponseDto.class);

Spring REST模板 - 401未经授权的错误

3 个答案: