Serialize property of a field instead of entire field

Question

I've searched on internet about my problem.. but I not found a good solution.

I need to get the json showed at the end:

I've 2 classes

class Order {
    Long id;

    Client client;
}

class Client {
    Long id;
}

When I serialize Order I get:

[{"id":1,"client":{"id":1}]

But I want instead to obtain:

[{"id":1,"client":1}]

How can i reach this?

Thank you for any solution!!! Marco

score 1 · Accepted Answer · answered Oct 04 '12 at 14:36

1

- When a class implements Serializable, then its object or the object of its Sub-class are made to be Serialized.

- Now its the Fields of the Object, Not the Object itself that is being Serialized.

- And the entire object graph needs to be serialized, if not the Serialization fails.

- So Client being an Object Reference Variable in Order Class will get Serialize, and so does its Field id.

- You canNot Serialize property of a field instead of entire field, but if you want you can prevent Field from being serialized using transient keyword.

answered Oct 04 '12 at 14:36

Kumar Vivek Mitra

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I can hardly read what you are writing. Seriously you should not overuse `code` and **bold** words. – maba Oct 04 '12 at 14:39
This is also my JPA domain class.. I cannot make it transient. One solution is to use @JSonIgnore on the field and use a transient getter like getClientId().. – gipinani Oct 04 '12 at 14:41
@mserioli...you dont need to make it transient.... that will lead to the prevention of id field to be serialized...i think you don't want that... i just mentioned it, to tell you that what you want is not possible – Kumar Vivek Mitra Oct 04 '12 at 14:42
Frankly i have read it online before, but never tried implementing it..... Moreover i am bit confused...whats wrong in this.. `"client":{"id":1}` This says that when the object graph was Serialized, then `an Object Reference Variable in Order class, named client of type Client which have a Field name id whose value is 1` – Kumar Vivek Mitra Oct 04 '12 at 14:50

score 1 · Answer 2 · answered Oct 04 '12 at 21:12

1

Use @JsonValue annotation:

class Client {
    @JsonValue Long id;
}

answered Oct 04 '12 at 21:12

StaxMan

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score 0 · Answer 3 · edited May 23 '17 at 12:27

0

You need to write your own serialization method for this class.

For example check this Thread: Implementing my own serialization in java

edited May 23 '17 at 12:27

Community

1
1

answered Oct 04 '12 at 14:35

Robin

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score 0 · Answer 4 · answered Oct 04 '12 at 15:04

You could also try genson library http://code.google.com/p/genson/. It has most of Jackson features and some other nice ones that jackson does not provide.

To solve your problem with genson and few code you can do it that way:

public class Order {
    Long id;
    @JsonIgnore Client client;
    @JsonProperty("client") public Long getClientId() {
        return client.id;
    }
}

public static class Client {
    Long id;
}

System.out.println(new Genson().serialize(order));

If you prefer you can also write a custom serializer with genson see here. But in your case it is not necessary.

Serialize property of a field instead of entire field

4 Answers4