FindByUUID() using Spring Data's JPA Repository

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面向向阳花
面向向阳花 2020-12-21 15:39

for some reason I have not being able to find a suitable answer for this. I have the following simple entity:

@Entity
@Table(name = \"simple_entity\")
@Acce         


        
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  •  醉梦人生
    2020-12-21 16:21

    Change binary column to String. Default is binary you must add this addnotation

    @Type(type="org.hibernate.type.UUIDCharType")
    

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