UUID as entity id

Question

I am using hibernate + mysql in my java application and I'd like to generate a unique string id (UUID) when I am inserting an entry into a table.

How do I create a table so that I can automatically generate uuid and use it as entity key?

+------------+-------------------------------------+
| uuid       | name                                |
+------------+-------------------------------------+
| asdf1221   | name 1                              |
| sksdkfk1   | name 2                              |
+------------+-------------------------------------+

with a `create table` SQL query? What exactly is the problem? — JB Nizet, Feb 06 '15 at 07:17
What do ypu mean with uuid? is a uuid a auto_increment or [that](http://en.wikipedia.org/wiki/Universally_unique_identifier) — Jens, Feb 06 '15 at 08:07

score 1 · Answer 1 · answered Jul 31 '18 at 23:46

MySQL doesn't allow a function for a default value, but see Can I use a function for a default value in MySql? for how to create a trigger to do this.

Note that MySQL's UUID() returns a CHAR(36), and storing UUIDs as text (as shown in those answers) is obviously inefficient. Instead, the column should be BINARY(16), and you can use UUID_TO_BIN() when inserting data and BIN_TO_UUID() when reading it back.

score 0 · Answer 2 · answered Aug 01 '18 at 01:18

I will suggest you to generate the uuid in Java.

Java UUID.randomUuid() is of type 4, which is cryptographically strong; while MySQL uuid() is of type 1 which is time based and relatively easy to be predicted.

If security is your concern, do it in Java.

As mentioned by Stephen, binary(16) is more efficient when storing uuid in database.

Hibernate has the native support from concerting between database binary (16) and java UUID. Just set the field type to UUID in your Java entity with column definition as Binary (16).

UUID as entity id

2 Answers2