Jackson - How to specify a single implementation for interface-referenced deserialization?

Question

I want to deserialize a JSON-Object with Jackson. Because the target is an interface I need to specify which implementation should be used.

This information could be stored in the JSON-Object, using @JsonTypeInfo-Annotation. But I want to specify the implementation in source code because it's always the same.

Is this possible?

The question here is about a single implementation, but one could research about inheritance in general, and we should then see the @JsonTypeInfo annotation. example https://stackoverflow.com/questions/28089484/deserialization-with-jsonsubtypes-for-no-value-missing-property-error/31016173#31016173 — pdem, Dec 21 '17 at 15:34

score 48 · Accepted Answer · edited May 26 '20 at 12:55

48

Use a SimpleAbstractTypeResolver:

ObjectMapper mapper = new ObjectMapper();

SimpleModule module = new SimpleModule("CustomModel", Version.unknownVersion());

SimpleAbstractTypeResolver resolver = new SimpleAbstractTypeResolver();
resolver.addMapping(Interface.class, Implementation.class);

module.setAbstractTypes(resolver);

mapper.registerModule(module);

edited May 26 '20 at 12:55

Mr00Anderson

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answered Oct 02 '12 at 10:24

David Grant

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Thanks, I just found an alternative if you want to specify the concrete implementation as an annotation http://wiki.fasterxml.com/JacksonFAQ#Deserializing_Abstract_types – Max Schmidt Oct 02 '12 at 10:32
Sounds like a nice compromise! – David Grant Oct 02 '12 at 10:33
1

@David Grant Where should we place SimpleAbstractTypeResolver in our code, i am using spring boot. – Harshit Saklecha Sep 21 '16 at 06:33
@har_shit take a look here for Spring: https://stackoverflow.com/a/30725492/6400036 – Frighi Jul 10 '20 at 12:26

score 35 · Answer 2 · answered Jun 21 '16 at 02:04

35

There is another approach that will work if you have just single interface implementation.

public class ClassYouWantToDeserialize {
    @JsonDeserialize(as = ImplementationClass.class)
    private InterfaceClass property;
...
}

This question was answered a while ago but I want to give you another option that doesn't require to tune ObjectMapper and also much simpler then @JsonTypeInfo annotation.

answered Jun 21 '16 at 02:04

Ilya Ovesnov

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This is by far the best answer to the OP question. – Adrian Hope-Bailie Nov 22 '16 at 13:50
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What if I have 2 Implementation and I want to deserialize based on some type field – Gaurav Oct 13 '17 at 08:31
@Gaurav this answer should help you: https://stackoverflow.com/questions/27506241/parse-to-a-subclass-by-default-with-jackson/27521288#27521288 – Ilya Ovesnov Oct 14 '17 at 21:05
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Note that if target interface is nested into collection then `@JsonDeserialize(contentAs=NestedType.class)` should be used – cybersoft Oct 01 '20 at 20:42

Jason Smiley · Answer 3 · 2017-09-11T21:58:05.613

9

You can use @JsonDeserialize(as = ImplementationClass.class) on the interface as well and all references will be deserialized the same way.

Note, if one of your Implementation classes is an enum, you might need @JsonFormat(shape = JsonFormat.Shape.OBJECT) on the enum as well.

edited Sep 11 '17 at 21:58

answered Jul 10 '17 at 19:49

Jason Smiley

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Jackson - How to specify a single implementation for interface-referenced deserialization?

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