如何为 JAXB/Moxy 注释的 POJO class 生成 JSONSchema

Question

我知道有很多与使用JSONSchema从JAXB注释类生成 JSONSchema 相关的问题，但我找不到使用Jackson注释 ZA2F2ED4F8EBC2CBB4C21A29DC40AB 生成MOXY的任何示例。

我只想知道如何为我的MOXY注释 class 生成JSONSchema ？ 到目前为止，当我为我的JAXB/Moxy注释的 class 生成JSONSchema时，我只得到一个字段：

{
  "type" : "any"
}

以下是我的 class ，我想使用Jackson和JAXB/Moxy注释为其生成JSONSchema ：（如您所见，我将name和age作为必填字段，但它们没有显示在生成的JSONSchema ）

@XmlRootElement(name = "Customer")
@XmlType(name = "Customer", propOrder = {"name", "age", "userExtensions"})
@XmlAccessorType(XmlAccessType.FIELD)
@NoArgsConstructor
@Getter
@Setter
@JsonInclude(JsonInclude.Include.NON_NULL)
@AllArgsConstructor
public class Customer extends Person {

  @XmlElement(name = "name",required = true)
  private String name;

  @XmlElement(name = "name",required = true)
  private String age;

  @JsonSerialize(using = CustomExtensionsSerializer.class)
  @XmlJavaTypeAdapter(TestAdapter.class)
  @XmlPath(".")
  @JsonValue
  private Map<String, Object> userExtensions = new HashMap<>();

  @JsonAnySetter
  public void setUserExtensions(String key, Object value) {
    userExtensions.put(key, value);
  }

  public Map<String, Object> getUserExtensions() {
    return userExtensions;
  }
}

以下是生成 JSONSchema 的Main JSONSchema ：

class JsonSchemaGenerator{

  public static void main(String[] args) throws JsonProcessingException {
    ObjectMapper objectMapper = new ObjectMapper();
    TypeFactory typeFactory = TypeFactory.defaultInstance();
    AnnotationIntrospector introspector = new JaxbAnnotationIntrospector(typeFactory);
    objectMapper.getDeserializationConfig().with(introspector);
    objectMapper.getSerializationConfig().with(introspector);

    //To force mapper to include JAXB annotated properties in Json schema
    objectMapper.registerModule(new JaxbAnnotationModule());
    SchemaFactoryWrapper visitor = new SchemaFactoryWrapper();
    objectMapper.acceptJsonFormatVisitor(objectMapper.constructType(Customer.class), visitor);

    JsonSchema inputSchema = visitor.finalSchema();
    String schemaString = objectMapper.writerWithDefaultPrettyPrinter().writeValueAsString(inputSchema);

    System.out.println(schemaString);
  }
}

Answer 1

我将@JsonValue与我的Map<String,Object>一起使用，这就是我得到type:any的原因。 我删除了它，然后尝试了它，它对我有用。

@XmlRootElement(name = "Customer")
@XmlType(name = "Customer", propOrder = {"name", "age", "userExtensions"})
@XmlAccessorType(XmlAccessType.FIELD)
@NoArgsConstructor
@Getter
@Setter
@JsonInclude(JsonInclude.Include.NON_NULL)
@AllArgsConstructor
public class Customer extends Person {

  @XmlElement(name = "name",required = true)
  private String name;

  @XmlElement(name = "name",required = true)
  private String age;

  @JsonSerialize(using = CustomExtensionsSerializer.class)
  @XmlJavaTypeAdapter(TestAdapter.class)
  @XmlPath(".")
  private Map<String, Object> userExtensions = new HashMap<>();

  @JsonAnySetter
  public void setUserExtensions(String key, Object value) {
    userExtensions.put(key, value);
  }

  public Map<String, Object> getUserExtensions() {
    return userExtensions;
  }
}

class JsonSchemaGenerator{

  public static void main(String[] args) throws JsonProcessingException {
    ObjectMapper objectMapper = new ObjectMapper();
    TypeFactory typeFactory = TypeFactory.defaultInstance();
    AnnotationIntrospector introspector = new JaxbAnnotationIntrospector(typeFactory);
    objectMapper.getDeserializationConfig().with(introspector);
    objectMapper.getSerializationConfig().with(introspector);

    //To force mapper to include JAXB annotated properties in Json schema
    objectMapper.registerModule(new JaxbAnnotationModule());
    SchemaFactoryWrapper visitor = new SchemaFactoryWrapper();
    objectMapper.acceptJsonFormatVisitor(objectMapper.constructType(Customer.class), visitor);

    JsonSchema inputSchema = visitor.finalSchema();
    String schemaString = objectMapper.writerWithDefaultPrettyPrinter().writeValueAsString(inputSchema);

    System.out.println(schemaString);
  }

}