如何将 swagger.json 映射到 Swagger 对象？

Question

I am currently trying to map a swagger.json file to a io.swagger.models.Swagger.class. I've tried to use com.fasterxml.jackson.databind.ObjectMapper.class to solve the issue by :

new ObjectMapper().readValue(file.getInputStream(), Swagger.class)`

but i got this exception:

"InvalidDefinitionException: Cannot construct instance of io.swagger.models.parameters.Parameter (no Creators, like default construct, exist): abstract types either need to be mapped to concrete types, have custom deserializer, or contain additional type information at [Source: (FileInputStream); line: 28, column: 21] (through reference chain: io.swagger.models.Swagger["paths"]->java.util.LinkedHashMap["/myPath"]->io.swagger.models.Path["post"]->io.swagger.models.Operation["parameters"]->java.util.ArrayList[0])"

The problem is that the parameters field in Swagger class goes like this :

protected Map<String, Parameter> parameters;

Parameter here is an Interface, so the ObjectMapper doesn't know which class to instantiate (BodyParameter, PathParameter, FormParameter...)

I expect to load a swagger.json file into a Swagger Object.

Thank you for your help.

Answer 1

Similar to Rayen's answer, you have to register deserialisers. Swagger offers serializers for its objects (eg Parameter, Property, Model) in io.swagger.util within io.swagger:swagger-core .

Answer 2

I found it, I needed to implement Deserializers in order to declare explicitly which class constructor to use. I had problem with Parameter, Model and Property which were all interfaces.

Here is the Code :

ObjectMapper mapper = new ObjectMapper();
        SimpleModule module = new SimpleModule();
        module.addDeserializer(io.swagger.models.parameters.Parameter.class, new ParameterDeserializer());
        module.addDeserializer(io.swagger.models.Model.class, new ModelDeserializer());
        module.addDeserializer(io.swagger.models.properties.Property.class, new PropertyDeserializer());
        mapper.registerModule(module);

Just an example to clarify things, I can make it more flexible later :

import java.io.IOException;

import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.deser.std.StdDeserializer;

import io.swagger.models.parameters.BodyParameter;
import io.swagger.models.parameters.FormParameter;
import io.swagger.models.parameters.Parameter;
import io.swagger.models.parameters.PathParameter;
import io.swagger.models.parameters.QueryParameter;

public class ParameterDeserializer extends StdDeserializer<io.swagger.models.parameters.Parameter> {

    /**
     * 
     */
    private static final long serialVersionUID = 1L;

    public ParameterDeserializer() {
        this(null);
    }

    public ParameterDeserializer(Class<?> vc) {
        super(vc);
    }

    @Override
    public Parameter deserialize(JsonParser parser, DeserializationContext ctxt)
            throws IOException, JsonProcessingException {
        final JsonNode node = parser.getCodec().readTree(parser);
        final ObjectMapper mapper = (ObjectMapper) parser.getCodec();
        if (node.get("in").toString().toLowerCase().replaceAll("\"", "").equals("path")) {
            return mapper.treeToValue(node, PathParameter.class);
        } else if (node.get("in").toString().toLowerCase().replaceAll("\"", "").equals("body")) {
            return mapper.treeToValue(node, BodyParameter.class);
        } else if (node.get("in").toString().toLowerCase().replaceAll("\"", "").equals("query")) {
            return mapper.treeToValue(node, QueryParameter.class);
        } else {
            return mapper.treeToValue(node, FormParameter.class);
        }
    }

}

Same for the other deserializers. Hope it helps.