Jackson - 如何处理（反序列化）嵌套的 JSON？

Question

{
  vendors: [
    {
      vendor: {
        id: 367,
        name: "Kuhn-Pollich",
        company_id: 1,
      }
    },
    {
      vendor: {
        id: 374,
        name: "Sawayn-Hermann",
        company_id: 1,
      }
  }]
}

I have a Vendor object that can properly be deserialized from a single "vendor" json, but I want to deserialize this into a Vendor[] , I just can't figure out how to make Jackson cooperate. Any tips?

Answer 1

Here is a rough but more declarative solution. I haven't been able to get it down to a single annotation, but this seems to work well. Also not sure about performance on large data sets.

Given this JSON:

{
    "list": [
        {
            "wrapper": {
                "name": "Jack"
            }
        },
        {
            "wrapper": {
                "name": "Jane"
            }
        }
    ]
}

And these model objects:

public class RootObject {
    @JsonProperty("list")
    @JsonDeserialize(contentUsing = SkipWrapperObjectDeserializer.class)
    @SkipWrapperObject("wrapper")
    public InnerObject[] innerObjects;
}

and

public class InnerObject {
    @JsonProperty("name")
    public String name;
}

Where the Jackson voodoo is implemented like:

@Retention(RetentionPolicy.RUNTIME)
@JacksonAnnotation
public @interface SkipWrapperObject {
    String value();
}

and

public class SkipWrapperObjectDeserializer extends JsonDeserializer<Object> implements
        ContextualDeserializer {
    private Class<?> wrappedType;
    private String wrapperKey;

    public JsonDeserializer<?> createContextual(DeserializationContext ctxt,
            BeanProperty property) throws JsonMappingException {
        SkipWrapperObject skipWrapperObject = property
                .getAnnotation(SkipWrapperObject.class);
        wrapperKey = skipWrapperObject.value();
        JavaType collectionType = property.getType();
        JavaType collectedType = collectionType.containedType(0);
        wrappedType = collectedType.getRawClass();
        return this;
    }

    @Override
    public Object deserialize(JsonParser parser, DeserializationContext ctxt)
            throws IOException, JsonProcessingException {
        ObjectMapper mapper = new ObjectMapper();
        ObjectNode objectNode = mapper.readTree(parser);
        JsonNode wrapped = objectNode.get(wrapperKey);
        Object mapped = mapIntoObject(wrapped);
        return mapped;
    }

    private Object mapIntoObject(JsonNode node) throws IOException,
            JsonProcessingException {
        JsonParser parser = node.traverse();
        ObjectMapper mapper = new ObjectMapper();
        return mapper.readValue(parser, wrappedType);
    }
}

Hope this is useful to someone!

Answer 2

Your data is problematic in that you have inner wrapper objects in your array. Presumably your Vendor object is designed to handle id , name , company_id , but each of those multiple objects are also wrapped in an object with a single property vendor .

I'm assuming that you're using the Jackson Data Binding model.

If so then there are two things to consider:

The first is using a special Jackson config property. Jackson - since 1.9 I believe, this may not be available if you're using an old version of Jackson - provides UNWRAP_ROOT_VALUE . It's designed for cases where your results are wrapped in a top-level single-property object that you want to discard.

So, play around with:

objectMapper.configure(SerializationConfig.Feature.UNWRAP_ROOT_VALUE, true);

The second is using wrapper objects. Even after discarding the outer wrapper object you still have the problem of your Vendor objects being wrapped in a single-property object. Use a wrapper to get around this:

class VendorWrapper
{
    Vendor vendor;

    // gettors, settors for vendor if you need them
}

Similarly, instead of using UNWRAP_ROOT_VALUES , you could also define a wrapper class to handle the outer object. Assuming that you have correct Vendor , VendorWrapper object, you can define:

class VendorsWrapper
{
    List<VendorWrapper> vendors = new ArrayList<VendorWrapper>();

    // gettors, settors for vendors if you need them
}

// in your deserialization code:
ObjectMapper mapper = new ObjectMapper();
JsonNode rootNode = mapper.readValue(jsonInput, VendorsWrapper.class); 

The object tree for VendorsWrapper is analogous to your JSON:

VendorsWrapper:
    vendors:
    [
        VendorWrapper
            vendor: Vendor,
        VendorWrapper:
            vendor: Vendor,
        ...
    ]

Finally, you might use the Jackson Tree Model to parse this into JsonNodes , discarding the outer node, and for each JsonNode in the ArrayNode , calling:

mapper.readValue(node.get("vendor").getTextValue(), Vendor.class);

That might result in less code, but it seems no less clumsy than using two wrappers.

Answer 3

@Patrick I would improve your solution a bit

@Override
public Object deserialize(JsonParser jp, DeserializationContext ctxt)
        throws IOException, JsonProcessingException {        
    ObjectNode objectNode = jp.readValueAsTree();
    JsonNode wrapped = objectNode.get(wrapperKey);
    JsonParser parser = node.traverse();
    parser.setCodec(jp.getCodec());
    Vendor mapped = parser.readValueAs(Vendor.class);
    return mapped;
}

It works faster :)

Answer 4

I'm quite late to the party, but one approach is to use a static inner class to unwrap values:

import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;

class Scratch {
    private final String aString;
    private final String bString;
    private final String cString;
    private final static String jsonString;

    static {
        jsonString = "{\n" +
                "  \"wrap\" : {\n" +
                "    \"A\": \"foo\",\n" +
                "    \"B\": \"bar\",\n" +
                "    \"C\": \"baz\"\n" +
                "  }\n" +
                "}";
    }

    @JsonCreator
    Scratch(@JsonProperty("A") String aString,
            @JsonProperty("B") String bString,
            @JsonProperty("C") String cString) {
        this.aString = aString;
        this.bString = bString;
        this.cString = cString;
    }

    @Override
    public String toString() {
        return "Scratch{" +
                "aString='" + aString + '\'' +
                ", bString='" + bString + '\'' +
                ", cString='" + cString + '\'' +
                '}';
    }

    public static class JsonDeserializer {
        private final Scratch scratch;

        @JsonCreator
        public JsonDeserializer(@JsonProperty("wrap") Scratch scratch) {
            this.scratch = scratch;
        }

        public Scratch getScratch() {
            return scratch;
        }
    }

    public static void main(String[] args) throws JsonProcessingException {
        ObjectMapper objectMapper = new ObjectMapper();
        Scratch scratch = objectMapper.readValue(jsonString, Scratch.JsonDeserializer.class).getScratch();
        System.out.println(scratch.toString());
    }
}

However, it's probably easier to use objectMapper.configure(SerializationConfig.Feature.UNWRAP_ROOT_VALUE, true); in conjunction with @JsonRootName("aName") , as pointed out by pb2q