Question

我有一个 JSON 字符串需要转换为 JAVA 对象。我需要将一些字段包装到不同的 JAVA 类中。我面临的问题是我无法包装它，并且我将 Java 字段设为 null。请看下面的JSON

{
    "first_name": "John",
    "last_name": "DCosta",
    "age": "29",
    "phone": "+173341238",
    "address_line_1": "43 Park Street",
    "address_line_2": "Behind C21 Mall",
    "city": "Cario",
    "country": "UK",
    "child1": {
        "name": "Peter",
        "age": "5"
    },
    "child2": {
        "name": "Paddy",
        "age": "2"
    },
    "child3": {
        "name": "Premus",
        "age": "1"
    }
}

请看下面我的 JAVA 类 -

Details.java

public class Details {
    
    private Person person;
    private Address address;
    private Child[] children;
    
    public Person getPerson() {
        return person;
    }
    public void setPerson(Person person) {
        this.person = person;
    }
    public Address getAddress() {
        return address;
    }
    public void setAddress(Address address) {
        this.address = address;
    }
    public Child[] getChildren() {
        return children;
    }
    public void setChildren(Child[] children) {
        this.children = children;
    }

}

Person.java

import com.fasterxml.jackson.annotation.JsonProperty;

    public class Person {
        
        @JsonProperty("first_name") 
        private String firstName;
        
        @JsonProperty("last_name")
        private String lastName;
        
        @JsonProperty("age")
        private Integer age;
    
    
        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 Integer getAge() {
            return age;
        }
    
        public void setAge(Integer age) {
            this.age = age;
        }
        
        
    
    }

Address.java

import com.fasterxml.jackson.annotation.JsonProperty;

public class Address {
    
    @JsonProperty("phone")
    private String phone;
    
    @JsonProperty("address_line_1")
    private String addressLine1;
    
    @JsonProperty("address_line_2")
    private String addressLine2;
    
    @JsonProperty("city")
    private String city;
    
    @JsonProperty("country")
    private String country;

    public String getPhone() {
        return phone;
    }

    public void setPhone(String phone) {
        this.phone = phone;
    }

    public String getAddressLine1() {
        return addressLine1;
    }

    public void setAddressLine1(String addressLine1) {
        this.addressLine1 = addressLine1;
    }

    public String getAddressLine2() {
        return addressLine2;
    }

    public void setAddressLine2(String addressLine2) {
        this.addressLine2 = addressLine2;
    }

    public String getCity() {
        return city;
    }

    public void setCity(String city) {
        this.city = city;
    }

    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

}

Child.java

public class Child {
    
    private String name;
    private Integer age;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Integer getAge() {
        return age;
    }
    public void setAge(Integer age) {
        this.age = age;
    }

}

我的将 JSON 转换为 JAVA 对象的代码 -

    String filePath = "test.json";
    File file = new File(filePath);
    ObjectMapper mapper = new ObjectMapper();
    mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
    Details details = mapper.readValue(file, Details.class);
    
    System.out.println(details.getPerson());

我面临的问题是我得到详细信息对象中的所有值都是null。如果我删除 mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false); 然后我得到以下异常

Exception in thread "main" com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "first_name" (class learn.springboot.model.Details), not marked as ignorable (3 known properties: "address", "person", "children"])
 at [Source: (File); line: 2, column: 17] (through reference chain: learn.springboot.model.Details["first_name"])
    at com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.ja

Answer 1

我认为不可能有一些包装类并期望 Jackson 从包装类中扁平化和提取所有这些字段并将扁平的 JSON 字段映射到它们。

基于 Details 类，Jackson 希望有一个像下面这样的 JSON 对象（省略了内部字段）：

{
  "person": {},
  "address": {},
  "children": []
}

因此，您必须将 Details 类更改为如下所示：

public class Details {
    @JsonProperty("first_name") 
    private String firstName;
        
    @JsonProperty("last_name")
    private String lastName;
        
    @JsonProperty("age")
    private Integer age;

    ...
}

无法在 Jackson 的自定义 Java 类中包装 JSON 属性

1 个答案: