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问题:
I have a a map that looks like this:
public class VerbResult { @JsonProperty("similarVerbs") private Map> similarVerbs; }
My verb class looks like this:
public class Verb extends Word { @JsonCreator public Verb(@JsonProperty("start") int start, @JsonProperty("length") int length, @JsonProperty("type") String type, @JsonProperty("value") VerbInfo value) { super(length, length, type, value); } //... }
I want to serialize and deserialize instances of my VerbResult class, but when I do I get this error: Can not find a (Map) Key deserializer for type [simple type, class my.package.Verb]
I read online that you need to tell Jackson how to deserialize map keys, but I didn't find any information explaining how to go about doing this. The verb class needs to be serialized and deserialzed outside of the map as well, so any solution should preserve this functionality.
回答1:
Building on the answer given here that suggests to implement a Module with a deserializer I refer to the Jackson Module documentation. The JodaTime Module is an easy to understand full example of a module containing serializers and deserializers.
Please note that the Module feature was introduced in Jackson version 1.7 so you might need to upgrade.
So step by step:
- create a module containing a (de)serializer for your class based on the Joda example
- register that module with
mapper.registerModule(module);
and you'll be all set
回答2:
After a day of searching, I came across a simpler way of doing it based on this question. The solution was to add the @JsonDeserialize(keyUsing = YourCustomDeserializer.class)
annotation to the map. Then implement your custom deserializer by extending KeyDeserializer
and override the deserializeKey
method. The method will be called with the string key and you can use the string to build the real object, or even fetch an existing one from the database.
So first in the map declaration:
@JsonDeserialize(keyUsing = MyCustomDeserializer.class) private Map> similarVerbs;
Then create the deserializer that will be called with the string key.
public class MyCustomDeserializer extends KeyDeserializer { @Override public MyMapKey deserializeKey(String key, DeserializationContext ctxt) throws IOException, JsonProcessingException { //Use the string key here to return a real map key object return mapKey; } }
Works with Jersey and Jackson 2.x
回答3:
As mentioned above the trick is that you need a key deserializer (this caught me out as well). In my case a non-String map key was configured on my class but it wasn't in the JSON I was parsing so an extremely simple solution worked for me (simply returning null in the key deserializer).
public class ExampleClassKeyDeserializer extends KeyDeserializer { @Override public Object deserializeKey( final String key, final DeserializationContext ctxt ) throws IOException, JsonProcessingException { return null; } } public class ExampleJacksonModule extends SimpleModule { public ExampleJacksonModule() { addKeyDeserializer( ExampleClass.class, new ExampleClassKeyDeserializer() ); } } final ObjectMapper mapper = new ObjectMapper(); mapper.registerModule( new ExampleJacksonModule() );