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Home/ Questions/Q 3935024
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T23:46:58+00:00 2026-05-19T23:46:58+00:00

I have a class (Activity) that with a field (signatureSecret) that is defined as

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I have a class (Activity) that with a field (signatureSecret) that is defined as an interface (SignatureSecret), the implementation of which (SharedConsumerSecret) comes from the Spring Security OAuth package.

When persisting instances of my class with Hibernate, I want to use a specific property of SharedConsumerSecret as the value that should be saved/loaded from the database. I can’t annotate SharedConsumerSecret, as it’s part of the Spring framework.

Is there any way to do this?

@Entity
public class Activity implements ConsumerDetails
{
    @Transient
    private List<GrantedAuthority> authorities = new ArrayList<GrantedAuthority>();
    private String consumerKey;
    private String consumerName;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Integer id;
    @ManyToOne
    private ActivityOwner activityOwner;
    //THIS IS THE TRICKY ONE
    private SignatureSecret signatureSecret;
    @Size(min = 1, max = 36)
    private String uuid;
    ...

The only way around this I can think of would be to have a one-to-one relationship of SignatureSecrets to Activities, but that seems a bit silly.

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  1. Editorial Team
    Editorial Team
    2026-05-19T23:46:58+00:00Added an answer on May 19, 2026 at 11:46 pm

    I want to use a specific property of SharedConsumerSecret as the value that should be saved/loaded from the database

    You’ll need to implement a UserType for SignatureSecret. It will need to know how to convert the object into a value (probably a String), and how to convert a String into the object.

    See this example from Hibernate test suite on how to build a UserType: https://github.com/hibernate/hibernate-core/blob/master/hibernate-core/src/test/java/org/hibernate/test/annotations/entity/PhoneNumberType.java

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