java - Jackson-JsonLD 嵌套对象

Question

我想编写一个 Restful API 并使用 schema.org 注释我的数据。为此，我想使用 Jackson-Jsonld。使用 jackson-jsonld 注释简单对象没有问题，但具有嵌套对象的复杂对象让我陷入困境。在我的 jsonld 中，像 id、name 这样的简单属性被注释，但嵌套位置没有注释。

我读到了有关序列化的内容，它应该有助于获取第二个对象。然而，在实现我的序列化部分之后，序列化似乎没有改变任何东西。 这是我的示例输出，位置的类型应该是 PostalAddress，但是缺少类型:

{"@context":
    {"uri":"http://schema.org/url","name":"http://schema.org/name","location":"http://schema.org/location"},
    "@type":"http://schema.org/Organization",
    "uri":"http://localhost:8080/kangarooEvents/venue/12",
    "name":"Joondalup Library - Ground Floor Meeting Room",
    "location":{
         "address":"102 Boas Avenue",
         "city":"Joondalup",
         "zip":"6027",
         "country":"Australia",
         "state":"WA"},
    "@id":12}

我想注释一个具有单一位置的组织:

@JsonldType("http://schema.org/Organization")
public class Venue {
    @JsonldId
    private Integer id;
    @JsonldProperty("http://schema.org/url")
    private String uri;
    @JsonldProperty("http://schema.org/name")
    private String name;
    @JsonSerialize(using = CostumLocationSerializer.class)
    @JsonldProperty("http://schema.org/location")
    private Location location;

地点:

@JsonldType("http://schema.org/PostalAddress")
public class Location {
    @JsonldProperty("http://schema.org/streetAddress")
    private String address;
    @JsonldProperty("http://schema.org/addressLocality")
    private String city;
    @JsonldProperty("http://schema.org/addressRegion")
    private String state;
    @JsonldProperty("http://schema.org/addressRegion")
    private String country;
    @JsonldProperty("http://schema.org/postalCode")
    private String zipcode;

序列化:

 public class CostumLocationSerializer extends StdSerializer<Location> {
      private ObjectMapper mapper = new ObjectMapper();

      public CostumLocationSerializer(){
         this( null);

      }
      protected CostumLocationSerializer(Class<Location> t) {
          super(t);
      }

      @Override
     public void serialize(Location location, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException {
         jsonGenerator.writeStartObject();
         jsonGenerator.writeStringField("address", location.getAddress());
         jsonGenerator.writeStringField("city", location.getCity());
         jsonGenerator.writeStringField("zip", location.getZipcode());
         jsonGenerator.writeStringField("country", location.getCountry());
         jsonGenerator.writeStringField("state", location.getState());
         jsonGenerator.writeEndObject();
         String serialized = mapper.writeValueAsString(location);
     } 
}

我认为我的问题可能出在序列化上，但我无法弄清楚。也许有人注释了嵌套 obj。并可以告诉我我的问题是什么。

Answer 1

只需跳过jackson-jsonld部分并手动执行

1. 创建 JSON - 只需在 Java 类中引入 type 和 id 字段即可。
2. 创建 JSON-LD 上下文 - 将您的 id 和 type 字段映射到另一个 @context 对象中
3. 结合上下文和数据 - 例如只需使用标准 jackson API 在“正常”json 序列化之后添加您的 @context 对象即可。

示例

@Test
public void createJsonFromPojo() throws Exception {
    ObjectMapper mapper=new ObjectMapper();
    // Create object structure
    Venue venue = new Venue();
    venue.location = new Location();
    venue.id="12";
    venue.uri="http://localhost:8080/kangarooEvents/venue/12";
    venue.name="Joondalup Library - Ground Floor Meeting Room";
    venue.location.address="102 Boas Avenue";
    venue.location.city="Joondalup";
    venue.location.state="WA";
    venue.location.country="Australia";
    venue.location.zipcode="6027";

    //1. Create JSON 
    ObjectNode myData = mapper.valueToTree(venue);

    //2. Create a JSON-LD context
    ArrayNode context = mapper.createArrayNode();
    context.add("http://schema.org/");
    ObjectNode myContext=mapper.createObjectNode();
    myContext.put("id", "@id");
    myContext.put("type", "@type");
    context.add(myContext);

    //3. Combine context and data
    myData.set("@context",context);

    //4. Print
    StringWriter w = new StringWriter();
    mapper.configure(SerializationFeature.INDENT_OUTPUT, true).writeValue(w, myData);
    String result= w.toString();
    System.out.println(result);
}

public class Venue {
    public final String type = "http://schema.org/Organization";
    public String id;
    public String uri;
    public String name;
    public Location location;
}

public class Location {
    public final String type = "http://schema.org/PostalAddress";
    public String address;
    public String city;
    public String state;
    public String country;
    public String zipcode;
}

给你

{ 
    "@context": [
        "http://schema.org/",
        {
          "id": "@id",
          "type":"@type"
        }
     ],
     "uri":"http://localhost:8080/kangarooEvents/venue/12",
     "name":"Joondalup Library - Ground Floor Meeting Room",
     "location":{
         "address":"102 Boas Avenue",
         "city":"Joondalup",
         "zip":"6027",
         "country":"Australia",
         "state":"WA",
         "type":"http://schema.org/PostalAddress"
      },
      "id":"12",
      "type":"http://schema.org/Organization"
}

View Example in Playground