Question

 {
    "response": [
        {
            "id": "1",
            "name": "xx"
        },
        {
            "id": "2",
            "name": "yy"
        }
    ],
    "errorMsg": "",
    "code": 0
}

如何使用jackson解析器单独解析“响应”。我收到错误

Unrecognized field "errorMsg", not marked as ignorable.

我的模型类Response.java

public class Response {
@JsonProperty("id")
private Integer id;
@JsonProperty("name")
private String name;
}

Answer 1

您的数据模型有点不完整，这正是杰克逊指出的。要改善这种情况，您应该映射更多字段。

public class Response {
    @JsonProperty("id")
    private Integer id;
    @JsonProperty("name")
    private String name;
    // getter/setter...
}
public class Data {
    @JsonProperty("response")
    private List<Response> response;
    @JsonProperty("errorMsg")
    private String errorMsg;
    @JsonProperty("code")
    private int code;
    // getter/setter...
}

Answer 2

您可以创建父对象并使用@JsonIgnoreProperties。或者，您可以使用ObjectMapper's convertValue()方法（例如

）获取节点并将其转换为响应对象

try {
    String json = "{\"response\":[{\"id\":\"1\",\"name\":\"xx\"},{\"id\":\"2\",\"name\":\"yy\"}],\"errorMsg\":\"\",\"code\":0}";
    ObjectMapper mapper = new ObjectMapper();
    JsonNode node = mapper.readTree(json);
    List<Response> responses = mapper.convertValue(node.findValues("response").get(0), new TypeReference<List<Response>>() {});
    System.out.println(responses);
} catch (JsonProcessingException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
}

使用JACKSON解析器解析特定的JSON数组

2 个答案: