[英]How RestTemplate parse response
在开发spring REST Client时,出现一个问题:
{
"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"
}
]
}
保存用户的模型类:
@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,我使用“ RestTemplate”获取结果。
RestTemplate restTemplate = new RestTemplate();
ResponseEntity<User[]> response = restTemplate.exchange(URL_GET,HttpMethod.GET,request, User[].class);
并得到错误。 我知道这是因为主键是“结果”,但是我可以指定restTemplate从哪里解析此JSON?
并且有可能在文件名上显示“ emailId”以获取直接值吗? 一些模板?
从“主键是结果”开始:
一种。 如果您只处理以下一种Web服务,我将为实际的有效负载创建一个包装类:
public class Return{ // Class property cannot be called "return" because it is Java reserved name. @JsonProperty("return") private User[] array; .... getter and setter }
b。 如果您处理多个Web服务,而实际负载在“返回”字段中,则将创建一个通用包装器类:
public class Return<T>{ // Class property cannot be called "return" because it is Java reserved name. @JsonProperty("return") private T[] array; .... getter and setter }
致电RestRemplate
:
ResponseEntity<Return<User>> response = restTemplate.exchange(URL_GET, HttpMethod.GET, request, new ParameterizedTypeReference<Return<User>>(){}); User[] usersArray = response2.getBody().getArray();
从名为“ value”的JSON属性中的属性值开始,我将创建两个自定义JsonDeserializer
(一个):一个用于单个值,一个用于值数组,并在适用的地方使用@JsonDeserialize
注释每个属性:
单值解串器:
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; } }
值序列化器数组:
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; } }
这是新的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 }
祝好运!
您可以使用注释@JsonRootName在响应中指定根元素。 所以试试这个:
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonRootName(value ="result")
public class User {
public User(){
}
private boolean admin;
....
}
您还可以使用JsonPath库浏览json:
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>>() {});
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