Play Framework [1.2.4]: unique constraints for an entity - jpa

Question

I am using playframwork with mySQL for my web application. I have a table which requires an unique_constraint on multiple columns.
I have the entity defined as follows...
package models;
import java.util.Date;
import javax.persistence.Entity;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
import javax.persistence.UniqueConstraint;
#Entity
#Table(name = "table",
uniqueConstraints = { #UniqueConstraint(columnNames =
{ "col_1", "col_2" }) })
public class Table extends Model{
#ManyToOne
#JoinColumn(name = "col_1")
public Column1 col1;
#ManyToOne
#JoinColumn(name = "col_2")
public Column2 col2;
#ManyToOne
#JoinColumn(name = "col_3")
public Column3 col_3;
}
Column1 and Column2 are different entities with relationship with
Table entity.
When i try to insert data with duplicate "col_1 and col_2" values as
shown below i am getting no error. The data is inserted fine in the
table (MySQL).
Table table1 = new Table();
table1.col1 = new Column1("1");
table1.col2= new Column2("2);
table1.col3= new Column3("3");
table1.save();
Table table2 = new Table();
table2 .col1 = new Column1("1");
table2 .col2= new Column2("2);
table2 .col3= new Column3("3");
table2 .save();
Do i need to create unique_constraint manually on the table?
Please help me understand, if i have anything missing in the above
implementation
Thanks,
Karthik

Answer 1

If the table is not created by JPA, yes, you'll have to manually create the unique constraints for them to be taking affect.
If the table does not pre-exist, JPA should create the constraint for you.
JPA does not create a constraint on an already existing table.