我可以有条件地使用@JsonUnwrapped吗?我不想在序列化期间使用它,但希望在反序列化对象时使用它。
一种方法是创建两个不同的类或创建一个子类,覆盖仅需要在序列化和反序列化时表现不同的属性。这听起来不太合适。还有其他替代方案或Jackson解决该问题的方法吗?
POJO
类与Jackson
注释解耦。你可以在运行时使用MixIn
添加必要的注释。请参见以下示例:import com.fasterxml.jackson.annotation.JsonUnwrapped;
import com.fasterxml.jackson.databind.ObjectMapper;
public class JacksonTest {
private static final String UNWRAPPED_JSON = "{\n" +
" \"age\" : 13,\n" +
" \"first\" : \"Huckleberry\",\n" +
" \"last\" : \"Finn\"\n" +
"}";
@Test
public void test() throws IOException {
System.out.println("### Serialize without unwrapped annotation ###");
ObjectMapper serializer = new ObjectMapper();
System.out.println(serializer.writerWithDefaultPrettyPrinter().writeValueAsString(createParent()));
System.out.println("### Deserialize with unwrapped annotation ###");
ObjectMapper deserializer = new ObjectMapper();
deserializer.addMixInAnnotations(Parent.class, ParentMixIn.class);
System.out.println(deserializer.readValue(UNWRAPPED_JSON, Parent.class));
}
private Parent createParent() {
Name name = new Name();
name.first = "Tom";
name.last = "Sawyer";
Parent parent = new Parent();
parent.setAge(12);
parent.setName(name);
return parent;
}
}
class Parent {
private int age;
private Name name;
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public Name getName() {
return name;
}
public void setName(Name name) {
this.name = name;
}
@Override
public String toString() {
return "Parent{" +
"age=" + age +
", name=" + name +
'}';
}
}
interface ParentMixIn {
@JsonUnwrapped
Name getName();
}
class Name {
public String first, last;
@Override
public String toString() {
return "Name{" +
"first='" + first + '\'' +
", last='" + last + '\'' +
'}';
}
}
### Serialize without unwrapped annotation ###
{
"age" : 12,
"name" : {
"first" : "Tom",
"last" : "Sawyer"
}
}
### Deserialize with unwrapped annotation ###
Parent{age=13, name=Name{first='Huckleberry', last='Finn'}}
@JsonUnwrapped(prefix = "name_")
@JsonProperty(access = READ_ONLY)
private Name nameObject;
// method will deserialize original JSON and set it to unwrapped field
@JsonSetter
public void setName(Name name) {
this.nameObject = name;
}
// simple getter
@JsonIgnore
public Name getName() {
return this.nameObject;
}
mapper.addMixIn(compA.class, compAMixIn.class); mapper.addMixIn(compB.class, compBMixIn.class); mapper.convertValue(...)'
假设compAMixIn解开了compB,而compBMixIn解开了compC。 - experiment unit 1998X