Best way to decode JSON string to Java object?

Question

I have a ServerEndpoint that will be recieving different JSON formats. Since we are only allowed one message handler per message type, my one decoder will have to convert the message to the corresponding Java objects.

In my decoder, I am trying to convert the message recieved to either SubClassA or SubClassB (which implemets the same interface) using the ObjectMapper class. The ObjectMapper class has a readValue method that requires the type I am trying to map the json to and will throw an exception when it cannot convert to the specified type.

I am currently decoding it like the following, but it is not very elegant.

@Override
public Message decode(String message) throws DecodeException {
    ObjectMapper mapper = new ObjectMapper();

    try {
        SubclassA obj = mapper.readValue(message, SubclassA .class);
        return obj;
    } catch (IOException e) {
    }

    try {
        SubclassB obj = mapper.readValue(message, SubclassB .class);
        return obj;
    } catch (IOException e) {
    }

    throw new DecodeException(message, "Failed to decode message.");
}

What is the best way to decode the JSON string into the corresponding Java object using ObjectMapper?

https://stackoverflow.com/questions/15736654/how-to-handle-deserializing-with-polymorphism — Dinesh Shekhawat, Feb 01 '19 at 09:08

score 3 · Answer 1 · edited May 23 '17 at 10:33

3

Use Jackson, you can declare subtypes for the parent object. Your Json will contain @type with with the name of the subtype, look at this post Deserialize JSON with Jackson into Polymorphic Types - A Complete Example is giving me a compile error

edited May 23 '17 at 10:33

Community

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answered Aug 31 '16 at 14:34

Vitalii Kravchenko

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Is it possible to not include the type within the JSON and still have it work? – dalawh Aug 31 '16 at 15:41
it's not possible, you can also have hack by implementing retry strategy for deserializing object with different classes. The second option is to add field with class name and during deserialization obtain class name and read Json – Vitalii Kravchenko Aug 31 '16 at 16:36
The hack is what I am currently using, which I don't like at all. – dalawh Aug 31 '16 at 16:40
Use Jackson approach then – Vitalii Kravchenko Aug 31 '16 at 16:49
I cannot modify the JSON as I have no control over the incoming messages to some extent – dalawh Aug 31 '16 at 16:55
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/122338/discussion-between-vitalii-kravchenko-and-dalawh). – Vitalii Kravchenko Aug 31 '16 at 19:00

score 0 · Answer 2 · edited May 23 '17 at 12:03

0

Jackson is one of the best options to work with JSON in Java.

For polymorphic deserialization, consider the following:

Using the @JsonTypeInfo annotation.
For situations where the @JsonTypeInfo annotation is not suitable, consider using a StdDeserializer. This answer will help you.

edited May 23 '17 at 12:03

Community

1
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answered Aug 31 '16 at 15:12

cassiomolin

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The StdDeserializer looks pretty interesting and looks like it would of worked. Unfortunately, SubclassA doesn't have a unique field. – dalawh Aug 31 '16 at 15:57
@dalawh What `SubclassA` and `SubclassB` look like? – cassiomolin Sep 01 '16 at 07:53
SubClassA and SubClass B extend different classes but implement the same interface. – dalawh Sep 06 '16 at 18:33

Best way to decode JSON string to Java object?

2 Answers2