[英]Jackson, deserialize plain JSON array to a single Java object
An external service is providing a JSON array with plain/primitive elements (so without field names, and without nested JSON objects). 外部服务提供了一个具有简单/原始元素的JSON数组(因此没有字段名,也没有嵌套的JSON对象)。 For example:
例如:
["Foo", "Bar", 30]
I would like to convert this to an instance of the following Java class using Jackson: 我想使用Jackson将其转换为以下Java类的实例:
class Person {
private String firstName;
private String lastName;
private int age;
Person(String firstName, String lastName, int age) {
this.firstName = firstName;
this.lastName = lastName;
this.age = age;
}
}
(This class can be adapted if needed.) (如果需要,可以修改该类。)
Question: is it possible to deserialize this JSON to Java using something like this? 问题:是否可以使用类似的方法将此JSON反序列化为Java?
Person p = new ObjectMapper().readValue(json, Person.class);
Or is this only possible by writing a custom Jackson deserializer for this Person class? 还是只能通过为此Person类编写自定义的Jackson解串器来实现?
I did try the following, but that didn't work: 我确实尝试了以下方法,但是没有用:
import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;
import java.io.IOException;
public class Person {
private String firstName;
private String lastName;
private int age;
@JsonCreator
public Person(
@JsonProperty(index = 0) String firstName,
@JsonProperty(index = 1) String lastName,
@JsonProperty(index = 2) int age) {
this.firstName = firstName;
this.lastName = lastName;
this.age = age;
}
public static void main(String[] args) throws IOException {
String json = "[\"Foo\", \"Bar\", 30]";
Person person = new ObjectMapper().readValue(json, Person.class);
System.out.println(person);
}
}
Result: Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: Argument #0 of constructor [constructor for Person, annotations: {interface com.fasterxml.jackson.annotation.JsonCreator=@com.fasterxml.jackson.annotation.JsonCreator(mode=DEFAULT)}] has no property name annotation; must have name when multiple-parameter constructor annotated as Creator at [Source: (String)"["Foo", "Bar", 30]"; line: 1, column: 1]
结果:
Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: Argument #0 of constructor [constructor for Person, annotations: {interface com.fasterxml.jackson.annotation.JsonCreator=@com.fasterxml.jackson.annotation.JsonCreator(mode=DEFAULT)}] has no property name annotation; must have name when multiple-parameter constructor annotated as Creator at [Source: (String)"["Foo", "Bar", 30]"; line: 1, column: 1]
Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: Argument #0 of constructor [constructor for Person, annotations: {interface com.fasterxml.jackson.annotation.JsonCreator=@com.fasterxml.jackson.annotation.JsonCreator(mode=DEFAULT)}] has no property name annotation; must have name when multiple-parameter constructor annotated as Creator at [Source: (String)"["Foo", "Bar", 30]"; line: 1, column: 1]
You don't need @JsonCreator
, just use @JsonFormat(shape = JsonFormat.Shape.ARRAY)
您不需要
@JsonCreator
,只需使用@JsonFormat(shape = JsonFormat.Shape.ARRAY)
@JsonFormat(shape = JsonFormat.Shape.ARRAY)
public static class Person {
@JsonProperty
private String firstName;
@JsonProperty
private String lastName;
@JsonProperty
private int age;
}
And use @JsonPropertyOrder({"firstName", "lastName", "age" } )
if you need to preserve some alternative field declaration order in your bean. 如果需要在bean中保留一些替代字段声明顺序,请使用
@JsonPropertyOrder({"firstName", "lastName", "age" } )
。
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