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Home/ Questions/Q 8626971
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T08:09:30+00:00 2026-06-12T08:09:30+00:00

I want to deserialize a JSON-Object with Jackson. Because the target is an interface

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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?

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  1. Editorial Team
    Editorial Team
    2026-06-12T08:09:31+00:00Added an answer on June 12, 2026 at 8:09 am

    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);
    
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