Question

我是Java领域的新手。我正在尝试使用Spring Data REST及其Android客户端构建REST服务。目前我正在尝试测试（JUnit）该服务。

当我致电服务获取用户列表（http://localhost:8080/..../api/users/）

时，我在Postman中收到以下回复

{
  "_embedded": {
    "users": [{
      "firstName" : "Maharaj",
      "lastName" : "NaamChakrawarti",
      "email" : "maharaj@mast.in.naam",
      "_links" : {
      "self" : {
        "href" : "http://localhost:8080/.../api/users/2"
       }
      }
    },{
      "firstName" : "Ji",
      "lastName" : "NaamChakrawartiDas",
      "email" : "G@Naam.masti",
      "_links" : {
      "self" : {
        "href" : "http://localhost:8080/.../api/users/3"
       }
      }
   }]
   },
 "_links" : {
   "self" : {
     "href" : "http://localhost:8080/com.feiswrox.webserviceRS/api/users"
    },
    .
    .
   },
   "page" : {
     "size" : 20,
     .
     .
   }
  }

我尝试使用以下代码来解析此响应 -

private static HttpHeaders getHeaders() {
    String plainCredentials = "G@Naam.masti:ShreeRam1974";
    String base64Credentials = new String(Base64.encode(plainCredentials.getBytes()));

    HttpHeaders headers = new HttpHeaders();
    headers.add("Authorization", "Basic " + base64Credentials);
    headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    return headers;
}
@Test
// @SuppressWarnings("unchecked")
public void listAllUsersHal() {
    System.out.println("\nTesting listAllUsersHal API-----------");
    RestTemplate restTemplate = new RestTemplate();

    HttpEntity<String> request = new HttpEntity<String>(getHeaders());

    Collection<Resource<User>> userList = null;

    ResponseEntity<Resources<Resource<User>>> responseResource = restTemplate.exchange(REST_SERVICE_URI + "/users/", HttpMethod.GET,
            request, new ParameterizedTypeReference<Resources<Resource<User>>>() {}, Collections.emptyMap());

    if (responseResource.getStatusCode() == HttpStatus.OK) {
        System.out.println(responseResource.getBody().toString());
        Resources<Resource<User>> userResource = responseResource.getBody();
        userList = userResource.getContent();
        for(Resource<User> userRsrc : userList){
            System.out.println(userRsrc.toString());
        }
    }
}

我在userList变量中获得了一些值，但是代码执行会跳过循环使用以下代码

    ResponseEntity<String> responseString = restTemplate.exchange(REST_SERVICE_URI + "/users/", HttpMethod.GET,
            request, String.class);

我得到一个与上面给出的字符串相同的字符串，即

{ "_embedded":{....

但我不知道如何将其解析为用户列表。

我使用类似的代码从json转换单个用户并且它起作用 RestTemplate restTemplate = new RestTemplate（）;

    HttpEntity<String> request = new HttpEntity<String>(getHeaders());
    ResponseEntity<Resource<User>> response = restTemplate.exchange(REST_SERVICE_URI + "/users/2", HttpMethod.GET,
            request, new ParameterizedTypeReference<Resource<User>>() {});

    User user = null;

    if (response.getStatusCode() == HttpStatus.OK) {
        Resource<User> userResource = response.getBody();
        user = userResource.getContent();
    }

我在输入以下代码时犯了什么错误吗？

 Collection<Resource<User>> userList = null;

ResponseEntity<Resources<Resource<User>>> responseResource = restTemplate.exchange(REST_SERVICE_URI + "/users/", HttpMethod.GET,
        request, new ParameterizedTypeReference<Resources<Resource<User>>>() {}, Collections.emptyMap());

请帮我构建REST服务返回的json用户列表。

我也可以在andriod应用程序中使用相同的代码吗？

将application / hal + json转换为pojo

0 个答案: