CODEWITHSUNDEEP
javahibernatehbm2ddlhbm
Maarten Blokker
Maarten Blokker

Reputation: 604

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.

Upvotes: 7

Views: 38581

Answers (6)

Maarten Blokker
Maarten Blokker

Reputation: 604

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 :)

Upvotes: 4

EricParis16
EricParis16

Reputation: 809

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 :)

Upvotes: 4

bsimic
bsimic

Reputation: 926

This is a tricky one because you have no access to the database. But it can be done with a bit of work

What you will need to do is create a custom type class. This class will basically retrieve the value from the database (1 or 0) and it will contain logic that returns either a true or false Boolean object.

Here is an example:

http://alenovarini.wikidot.com/mapping-a-custom-type-in-hibernate

your mapping for your new type will look like:

    <typedef class="com.path.to.my.package.CustomBooleanType" name="myBoolType" />

That classes nullSafeGet method will return your Boolean object that will contain true or false.

so your new mapping will contain:

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

Upvotes: 0

axtavt
axtavt

Reputation: 242786

You can try to use numeric_boolean as a type:

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

Upvotes: 1

EricParis16
EricParis16

Reputation: 809

try this :

<property column="opener" name="opener" access="field" />

assuming that you got a getter 

 boolean isOpener() ;

and a setter

void setOpener(boolean b);

Upvotes: 1

mut1na
mut1na

Reputation: 795

You can define your DB column as a char(1) and in your Hibernate mapping file define the property as type="yes_no", which is a Java Boolean type. These will appear as Y and N values in the DB.

If you want to use tiny_int then the size will have to be 1, but I'm not 100% sure how this is mapped in the HBM file.

Upvotes: 3

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