RestTemplate如何解析响应

Question

在开发Spring REST客户端时，遇到了一个问题：

1. 我有以下JSON：

{
 "return": [
 {
    "admin": false,
    "alias": "",
    "email": "",
    "emailId": {"value": 0},
    "groups": [],
    "id": {"value": 1},
    "locked": false,
    "loggedInCount": 0,
    "master": true,
    "sms": "",
    "smsId": {"value": 0},
    "type": "POWER",
    "username": "NGCP"
 },
 {
    "admin": false,
    "alias": "",
    "email": "",
    "emailId": {"value": 0},
    "groups": [{"value": 2}],
    "id": {"value": 3},
    "locked": false,
    "loggedInCount": 0,
    "master": false,
    "sms": "",
    "smsId": {"value": 0},
    "type": "POWER",
    "username": "POLICY"
 }
        ]
}

用户模型类的保存：

@JsonIgnoreProperties(ignoreUnknown = true)
public class User {

   public User(){

   }

   private boolean admin;

   private String alias;

   private String email;

   private String emailId;

   private ArrayList<String> groups;

   private String id;

   private boolean locked;

   private int loggedInCount;

   private boolean master;

   private String sms;

   private String smsId;

   private String type;

   private String userName;

//getter and setters
}

现在我正在使用“RestTemplate”来获取结果。

RestTemplate restTemplate = new RestTemplate();
ResponseEntity<User[]> response = restTemplate.exchange(URL_GET,HttpMethod.GET,request, User[].class);

我遇到了一个错误。我知道这是因为主键是“result”，但我能否指定从哪里应该restTemplate解析JSON？

是否可以在类似于“emailId”的字段上指示直接获取值？有些模板可以做到吗？

- Vladlen Gladis

---

“get direct value” 是什么意思？ - Michal Foksa

在这个领域中，我再次拥有一对“键-值”。我只需要取值。 - Vladlen Gladis

什么是错误？“主键是结果”是什么意思？ - Michael Wiles

@Michael Wiles 问题已经解决，请查看答案。当我谈到“主键”时，我指的是JSON中的第一个键。错误是因为我没有正确地注释我的模型类。 - Vladlen Gladis

Answer 1

1. As of the "main key is result":

   a. I would create a wrapper class for the actual payload if you deal with just one of this kind of web service:

   public class Return{
       // Class property cannot be called "return" because it is Java reserved name.
       @JsonProperty("return")
       private User[] array;
       .... getter and setter
   }
   

   b. If you deal with multiple webservices where actual payload is in "return" field I would create a generic wrapper class :

   public class Return<T>{
       // Class property cannot be called "return" because it is Java reserved name.
       @JsonProperty("return")
       private T[] array;
       .... getter and setter
   }
   

   Call to RestRemplate:

   ResponseEntity<Return<User>> response = restTemplate.exchange(URL_GET, 
           HttpMethod.GET, request, new ParameterizedTypeReference<Return<User>>(){});
   User[] usersArray = response2.getBody().getArray();
   

2. As of the property value in JSON attribute called "value" I would create two custom JsonDeserializer(s): one for single value and one for array of values and annotate each property with @JsonDeserialize where it applies:

   Single value deserializer:

   public class StringValueDeserializer  extends JsonDeserializer<String>{
   
       @Override
       public String deserialize(JsonParser parser, DeserializationContext ctxt)
               throws IOException, JsonProcessingException {
           ObjectCodec codec = parser.getCodec();
           TreeNode node = codec.readTree(parser);
           JsonNode value = (JsonNode)node.get("value");
   
           if (value != null){
               return value.asText();
           }
           return null;
       }
   }
   

   Array of values derializer:

   public class StringArrayValueDeserializer  extends JsonDeserializer<List<String>>{
   
       @Override
       public List<String> deserialize(JsonParser parser, DeserializationContext ctxt)
           throws IOException, JsonProcessingException {
   
           List<String> ret = new ArrayList<>();
   
           ObjectCodec codec = parser.getCodec();
           TreeNode node = codec.readTree(parser);
   
           if (node.isArray()){
               for (JsonNode n : (ArrayNode)node){
                   JsonNode value = n.get("value");
                   if (value != null){
                       ret.add(value.asText());
                   }
               }
           }
           return ret;
       }
   }
   

   Here you are new User.class:

   public class User {
   
       private boolean admin;
   
       private String alias;
   
       private String email;
   
       @JsonDeserialize(using = StringValueDeserializer.class)
       private String emailId;
   
       @JsonDeserialize(using = StringArrayValueDeserializer.class)
       private ArrayList<String> groups;
   
       @JsonDeserialize(using = StringValueDeserializer.class)
       private String id;
   
       private boolean locked;
   
       private int loggedInCount;
   
       private boolean master;
   
       private String sms;
   
       @JsonDeserialize(using = StringValueDeserializer.class)
       private String smsId;
   
       private String type;
   
       private String username;
       .... getter and setter
   }
   

祝你好运！

- Michal Foksa

Answer 2

您也可以使用JsonPath库来浏览json数据：

String json =  restTemplate.exchange(URL_GET,HttpMethod.GET,request, String.class);
DocumentContext document = JsonPath.parse(content, json);
List<User> users = document.read("$.return.*", new TypeRef<List<User>>() {});

- romanvintonyak

Answer 3

您可以使用注释@JsonRootName来指定响应中的根元素。因此，请尝试这样做：

@JsonIgnoreProperties(ignoreUnknown = true)
@JsonRootName(value ="result")
public class User {

   public User(){

   }

   private boolean admin;

   ....
}

- Mohsan Shakir

---

如果您使用User对象进行创建或更新，json根名称注释会产生什么影响？或者这只是用于读取（获取）数据。 - Heril Muratovic