Hibernate Regexp MySQL

Question

I ask this question to show how MySQL and Hibernate work each other with Regular Expressions.

The problem:

SELECT * FROM table WHERE regexp column \'         


        

Answer 1

REGEXP is treated as a keyword in MySQL. User can use REGEXP in hibernate filter by registering the keyword. Create a class 'CustomMySQL5InnoDBDialect' extending MySQL5InnoDBDialect and register the keyword as follows :

public class CustomMySQL5InnoDBDialect extends MySQL5InnoDBDialect {

public CustomMySQL5InnoDBDialect() {
    super();
   /* register regexp keyword */
    registerKeyword("regexp");
}
}

Then change the hibernate-dialect property in persistence.xml as

<property name="hibernate.dialect" value="com.CustomMySQL5InnoDBDialect"/>

User can use the regexp in hibernate filter as follows

@Filters(value = {  @Filter(name="applyStudentFilter",condition="id in (select s.id from student s WHERE s.address REGEXP :addressValue)"),
})

Answer 2

String range = "ABCD";

List<HRTrainee> hrTrainees =
      (List<HRTrainee>)sessionFactory.getCurrentSession().createCriteria(HRTrainee.class)

.add(Restrictions.sqlRestriction("name REGEXP '^["+range+"]'")).list();

return hrTrainees;

Answer 3

Basically, to use MySQL regexp function in Hibernate we need to create a "SQLFunctionTemplate".

Now, how to do it:

First: Create a class called "AppMySQLDialect" and extends from MySQLDialect then override the empty constructor and finally register the regexp function:

public class AppMySQLDialect extends MySQLDialect {
    public AppMySQLDialect() {
        super();
        /**
         * Function to evaluate regexp in MySQL
         */
        registerFunction("regexp", new SQLFunctionTemplate(Hibernate.INTEGER, "?1 REGEXP ?2"));
    }
}

Ok, now lets use it as follow:

FROM Entity E WHERE regexp(E.string2evaluate, '\d') = 1

Create your HibernateQuery and execute.