Issue
I came across a blog of using UUID with Hibernate and MySql. Now the problem is, whenever I take a look at the database the ID's will be non-readable format (binary-16). How can I store UUID as a readable format like 7feb24af-fc38-44de-bc38-04defc3804fe
instead of ¡7ôáßEN¹º}ÅÑs
I was using this code
@Id
@GeneratedValue( generator = "uuid2" )
@GenericGenerator( name = "uuid2", strategy = "uuid2" )
@Column( name = "id", columnDefinition = "BINARY(16)" )
private UUID id;
And the result is ¡7ôáßEN¹º}ÅÑs. But I want it as readable UUID so I used the following code which didn't help me
@Id
@GeneratedValue( generator = "uuid2" )
@GenericGenerator( name = "uuid2", strategy = "uuid2" )
@Column( name = "id", columnDefinition = "CHAR(32)" )
private UUID id;
How to save the UUID as a string instead of binary(16) without changing the java type UUID
Solution
just use
@org.hibernate.annotations.Type(type="uuid-char")
There is three levels of data types:
- Java types
- Hibernate's types
- Database Specific types.
Hibernate data type presentation is a bridge between Java data type and Database types to be independent from database.
You can check this mappings.
As you can find there java.util.UUID
can be mapped to diferent types (binary or char/varchar). uuid-binary
is key to hibernate's UUIDBinaryType, you get this type by default and it will be mapped to BINARY
of your database.
If you want to get CHAR type under your UUID, you should explain to hibernate that you want his UUIDCharType
. To do that you use uuid-char
key and as you can check in JavaDoc of @Type annotation: Defines a Hibernate type mapping.
. So, you use annotation to explain hibernate which bridge it should use.
Answered By - Ivan Osipov
Answer Checked By - Mildred Charles (JavaFixing Admin)