Right way to write JSON deserializer in Spring or extend it

Question

I am trying to write a custom JSON deserializer in Spring. I want to use default serializer for most part of fields and use a custom deserializer for few properties. Is it possible? I am trying this way because, most part of properties are values, so for these I can let Jackson use default deserializer; but few properties are references, so in the custom deserializer I have to query a database for reference name and get reference value from database.

I'll show some code if needed.

Answer 1

I've searched a lot and the best way I've found so far is on this article:

Class to serialize

package net.sghill.example;  import net.sghill.example.UserDeserializer import net.sghill.example.UserSerializer import org.codehaus.jackson.map.annotate.JsonDeserialize; import org.codehaus.jackson.map.annotate.JsonSerialize;  @JsonDeserialize(using = UserDeserializer.class) public class User {     private ObjectId id;     private String   username;     private String   password;      public User(ObjectId id, String username, String password) {         this.id = id;         this.username = username;         this.password = password;     }      public ObjectId getId()       { return id; }     public String   getUsername() { return username; }     public String   getPassword() { return password; } } 

Deserializer class

package net.sghill.example;  import net.sghill.example.User; import org.codehaus.jackson.JsonNode; import org.codehaus.jackson.JsonParser; import org.codehaus.jackson.ObjectCodec; import org.codehaus.jackson.map.DeserializationContext; import org.codehaus.jackson.map.JsonDeserializer;  import java.io.IOException;  public class UserDeserializer extends JsonDeserializer<User> {      @Override     public User deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException {         ObjectCodec oc = jsonParser.getCodec();         JsonNode node = oc.readTree(jsonParser);         return new User(null, node.get("username").getTextValue(), node.get("password").getTextValue());     } } 

Edit: Alternatively you can look at this article which uses new versions of com.fasterxml.jackson.databind.JsonDeserializer.

Answer 2

I was trying to @Autowire a Spring-managed service into my Deserializer. Somebody tipped me off to Jackson using the new operator when invoking the serializers/deserializers. This meant no auto-wiring of Jackson's instance of my Deserializer. Here's how I was able to @Autowire my service class into my Deserializer:

context.xml

<mvc:annotation-driven>   <mvc:message-converters>     <bean class="org.springframework.http.converter.json.MappingJackson2HttpMessageConverter">       <property name="objectMapper" ref="objectMapper" />     </bean>   </mvc:message-converters> </mvc> <bean id="objectMapper" class="org.springframework.http.converter.json.Jackson2ObjectMapperFactoryBean">     <!-- Add deserializers that require autowiring -->     <property name="deserializersByType">         <map key-type="java.lang.Class">             <entry key="com.acme.Anchor">                 <bean class="com.acme.AnchorDeserializer" />             </entry>         </map>     </property> </bean> 

Now that my Deserializer is a Spring-managed bean, auto-wiring works!

AnchorDeserializer.java

public class AnchorDeserializer extends JsonDeserializer<Anchor> {     @Autowired     private AnchorService anchorService;     public Anchor deserialize(JsonParser parser, DeserializationContext context)              throws IOException, JsonProcessingException {         // Do stuff     } } 

AnchorService.java

@Service public class AnchorService {} 

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Update: While my original answer worked for me back when I wrote this, @xi.lin's response is exactly what is needed. Nice find!