使用Jackson将JSON数组反序列化为单个Java对象

Question

我的想法是，我想将JSON数组["foo", "bar"]转换为Java对象，因此我需要通过索引将每个数组元素映射到property。

假设我有以下JSON：

{
  "persons": [
    [
      "John",
      "Doe"
    ],
    [
      "Jane",
      "Doe"
    ]
  ]
}

正如您所看到的，每个人只是一个数组，其中第一个名称是索引为0的元素，而最后一个名称是索引为1的元素。

我想将它反序列化为List<Person>。我使用mapper如下：

mapper.getTypeFactory().constructCollectionType(List.class, Person.class)

其中Person.class是：

public class Person {
    public final String firstName;
    public final String lastName;

    @JsonCreator
    public Person(@JsonProperty() String firstName, @JsonProperty String lastName) {
        this.firstName = firstName;
        this.lastName = lastName;
    }
}

我想知道我是否可以某种方式将数组索引指定为@JsonProperty参数而不是它的关键名称？

Answer 1

感谢bureaquete建议使用自定义反序列化器。但它更适合我用SimpleModule而不是@JsonDeserialize注释注册它。下面是完整的JUnit测试示例：

@RunWith(JUnit4.class)
public class MapArrayToObjectTest {
    private static ObjectMapper mapper;

    @BeforeClass
    public static void setUp() {
        mapper = new ObjectMapper();
        SimpleModule customModule = new SimpleModule("ExampleModule", new Version(0, 1, 0, null));
        customModule.addDeserializer(Person.class, new PersonDeserializer());
        mapper.registerModule(customModule);
    }

    @Test
    public void wrapperDeserializationTest() throws IOException {
        //language=JSON
        final String inputJson = "{\"persons\": [[\"John\", \"Doe\"], [\"Jane\", \"Doe\"]]}";
        PersonsListWrapper deserializedList = mapper.readValue(inputJson, PersonsListWrapper.class);
        assertThat(deserializedList.persons.get(0).lastName, is(equalTo("Doe")));
        assertThat(deserializedList.persons.get(1).firstName, is(equalTo("Jane")));
    }

    @Test
    public void listDeserializationTest() throws IOException {
        //language=JSON
        final String inputJson = "[[\"John\", \"Doe\"], [\"Jane\", \"Doe\"]]";
        List<Person> deserializedList = mapper.readValue(inputJson, mapper.getTypeFactory().constructCollectionType(List.class, Person.class));
        assertThat(deserializedList.get(0).lastName, is(equalTo("Doe")));
        assertThat(deserializedList.get(1).firstName, is(equalTo("Jane")));
    }
}

class PersonsListWrapper {
    public List<Person> persons;
}

class Person {
    final String firstName;
    final String lastName;

    Person(String firstName, String lastName) {
        this.firstName = firstName;
        this.lastName = lastName;
    }
}

class PersonDeserializer extends JsonDeserializer<Person> {
    @Override
    public Person deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException {
        JsonNode node = jp.readValueAsTree();
        return new Person(node.get(0).getTextValue(), node.get(1).getTextValue());
    }
}

请注意，如果您不需要包装器对象，则可以使用mapper将JSON数组[["John", "Doe"], ["Jane", "Doe"]]直接反序列化为List<Person>，如下所示：

List<Person> deserializedList = mapper.readValue(inputJson, mapper.getTypeFactory().constructCollectionType(List.class, Person.class));

Answer 2

它很容易序列化，但不容易以这种方式反序列化;

可以使用@JsonValue将以下类序列化为字符串数组;

public class Person {

    private String firstName;
    private String lastName;

    //getter,setter,constructors

    @JsonValue
    public List<String> craeteArr() {
        return Arrays.asList(this.firstName, this.lastName);
    }
}

但是要反序列化，我必须创建一个包装类，并使用@JsonDeserialize自定义反序列化;

public class PersonWrapper {

    @JsonDeserialize(using = CustomDeserializer.class)
    private List<Person> persons;

    //getter,setter,constructors
}

和自定义反序列化器本身;

public class CustomDeserializer extends JsonDeserializer<List<Person>> {

    @Override
    public List<Person> deserialize(JsonParser jsonParser, DeserializationContext context) throws IOException {
        JsonNode node = jsonParser.readValueAsTree();
        ObjectMapper mapper = new ObjectMapper();
        return IntStream.range(0, node.size()).boxed()
                .map(i -> {
                    try {
                        List<String> values = mapper.readValue(node.get(i).toString(), List.class);
                        return new Person().setFirstName(values.get(0)).setLastName(values.get(1));
                    } catch (IOException e) {
                        throw new RuntimeException();
                    }
                }).collect(Collectors.toList());
    }
}

您需要在反序列化器逻辑中进行适当的验证，以检查每个迷你数组是否包含正好两个值，但这很有效。

我宁愿使用这些步骤，也许隐藏@JsonDeserialize，我会做以下事情;

@Retention(RetentionPolicy.RUNTIME)
@JacksonAnnotationsInside
@JsonDeserialize(using = CustomDeserializer.class)
public @interface AcceptPersonAsArray {}

所以你可以在PersonWrapper中使用一些自定义注释

public class PersonWrapper {

    @AcceptPersonAsArray
    private List<Person> persons;

    //getter,setter,constructors
}