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Mapping a boolean with hibernate

I'm running into trouble with hibernate. I recently set my hbm2ddl to validate, and it has been complaining a lot about wrong datatypes. I have fixed every problem except for booleans.

I have a field opener in my class, which is mapped as:

<property column="opener" name="opener" type="boolean"/>

The column opener is a tinyint (4) and has a value of 1 or 0. So far I've tried changing the types, but to no avail. I have also tried using the following setting in my hibernate.cfg:

<property name="hibernate.query.substitutions">true 1, false 0</property>

But I am still getting the same error. What am I doing wrong?

org.hibernate.HibernateException: Wrong column type: opener, expected: bit
    at org.hibernate.mapping.Table.validateColumns(Table.java:261)
    at org.hibernate.cfg.Configuration.validateSchema(Configuration.java:1083)
    at org.hibernate.tool.hbm2ddl.SchemaValidator.validate(SchemaValidator.java:116)
    at org.hibernate.impl.SessionFactoryImpl.<init>(SessionFactoryImpl.java:317)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1294)
    at org.hibernate.cfg.AnnotationConfiguration.buildSessionFactory(AnnotationConfiguration.java:859)

note: I have no access to the database.

like image 888
Terraego Avatar asked May 31 '11 13:05

Terraego


2 Answers

If you can't change your SQL type in your table, i recommend you to do this :

<property name="opener" column="opener" type="path.to.your.package.YourClassUserType"/>

and create your class :

import org.hibernate.usertype.UserType;

public class YourClassUserType implements UserType{
 ...
}

you have to implement methods from the interface UserType. The implementation will transform byte to boolean (because a TINYINT is mapped in byte in Java)

see examples here

good luck :)

like image 107
EricParis16 Avatar answered Oct 16 '22 10:10

EricParis16


For anyone who ran into the same trouble as me, I used a combination of two answers posted here.

I implemented a custom usertype to handle my tinyint field:

public class TinyIntegerToBoolean implements UserType {

    public int[] sqlTypes() {
        return new int[]{Types.TINYINT};
    }

    public Class returnedClass() {
        return Boolean.class;
    }

    public Object nullSafeGet(ResultSet rs, String[] names, SessionImplementor si, Object owner) throws HibernateException, SQLException {
        return (rs.getByte(names[0]) != 0);
    }

    public void nullSafeSet(PreparedStatement st, Object value, int index, SessionImplementor si) throws HibernateException, SQLException {
        st.setByte(index, Boolean.TRUE.equals(value) ? (byte) 1 : (byte) 0);
    }

    /* boilerplate... */
    public boolean isMutable() {
        return false;
    }

    public boolean equals(Object x, Object y) throws HibernateException {
        if (x == null || y == null) {
            return false;
        } else {
            return x.equals(y);
        }
    }

    public int hashCode(Object x) throws HibernateException {
        assert (x != null);
        return x.hashCode();
    }

    public Object deepCopy(Object value) throws HibernateException {
        return value;
    }

    public Object replace(Object original, Object target, Object owner)
            throws HibernateException {
        return original;
    }

    public Serializable disassemble(Object value) throws HibernateException {
        return (Serializable) value;
    }

    public Object assemble(Serializable cached, Object owner)
            throws HibernateException {
        return cached;
    }
}

Then I added the following to my mappings:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <typedef class="com.test.model.TinyIntegerToBoolean" name="tinyint_boolean"/>
</hibernate-mapping>

Then in my opener field I use type=tinyint_boolean and it works like a charm :)

like image 39
Terraego Avatar answered Oct 16 '22 10:10

Terraego