Question

鉴于以下json：

{
   "User":{
      "firstname":"john",
      "gender":"female",
      "verified":"no"
   }
}

有没有办法删除/忽略“用户”节点，所以我可以使用jackson绑定？你看，如果我尝试这样做的话：

User user= mapper.readValue(content, User.class); //where content is the json above

用户类是：

public class User
{
     String _firstname, _gender, _verified;

     [getters and setters]
}

它出错了：

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException：无法识别的字段“用户”

我知道我可以初始化User类并且只是手动设置setter但是我想知道是否可以或更好地删除/忽略“User”节点？

Answer 1

在JSON的根目录中，您有另一个名为User的JSON对象。您不能忽略它，因为它包含您要提取的对象。您需要做的是从嵌套的JSON对象而不是根JSON对象中提取对象。

您可以通过获取嵌套的JSON对象来实现。

ObjectMapper mapper = new ObjectMapper(); 
JsonNode node = mapper.readTree(content);
User user = mapper.readValue(node.get("User").traverse(), User.class);

此外，我不确定杰克逊是否支持像{/ 1>}那样悬挂,

"verified":"no",

Answer 2

此解决方案通过在rootName上设置ObjectReader来使用纯杰克逊：

import java.io.IOException;

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.ObjectReader;

public class User {

    public static void main(String[] args) throws JsonProcessingException, IOException {

        String json = "{\"User\":{\"firstname\":\"john\",\"gender\":\"female\",\"verified\":\"no\"}}";
        ObjectMapper mapper = new ObjectMapper();
        ObjectReader reader = mapper.reader(User.class).withRootName("User");
        User user = reader.readValue(json);

        System.out.println(user.getFirstname());
    }

    private String firstname;
    private String lastname;
    private String verified;
    private String gender;

    public String getFirstname() {
        return firstname;
    }

    public void setFirstname(String firstname) {
        this.firstname = firstname;
    }

    public String getLastname() {
        return lastname;
    }

    public void setLastname(String lastname) {
        this.lastname = lastname;
    }

    public String getVerified() {
        return verified;
    }

    public void setVerified(String verified) {
        this.verified = verified;
    }

    public String getGender() {
        return gender;
    }

    public void setGender(String gender) {
        this.gender = gender;
    }

}

在Java中删除json中的节点

2 个答案: