JDBCExceptionReporter [ERROR] Field 'id' doesn't have a default value

Question

After trying different suggested solutions I haven't managed to figure out this one.

I'm using hibernate 3.6 and mysql 5.x and trying to persist my entity (I excluded the getters & setters):

@Entity
@Table(name = "brand_managers")
public class BrandManager implements Serializable {

    /** Serial version unique id */
    private static final long serialVersionUID = -7992146584570782015L;

    /*--- Members ---*/

    /** The unique, internal ID of the entity. */
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id;

    /**
     * The creation time of this user
     */
    @Temporal(TemporalType.DATE)
    @Column(name = "creation_time")
    protected Calendar creationTime;

    /**
     * The hashed password
     */
    @Column(name = "password")
    protected String password;

    @Column(name = "first_name")
    protected String firstName;

    @Column(name = "last_name")
    protected String lastName;

    @Column(name = "email")
    protected String eMail;

    @Column(name = "address1")
    protected String address1;

    @Column(name = "address2", nullable = true)
    protected String address2;

    @Column(name = "city")
    protected String city;

    @Column(name = "state")
    protected String state;

    @Column(name = "zip", nullable = true)
    protected String zip;

    @Column(name = "country")
    protected String country;

    @Column(name = "phone")
    protected String phone;

    @Column(name = "brand_id")
    protected int brandId;

    /*--- Constructors ---*/

    /**
     * default
     */
    public BrandManager() {
    setCreationTime(Calendar.getInstance());
    }

    /**
     * @param password
     *            The hashed user password
     * @param firstName
     *            The first name
     * @param lastName
     *            The last name
     * @param eMail
     *            User eMail
     * @param address1
     *            User address
     * @param address2
     *            Another user address
     * @param city
     *            City of residence
     * @param state
     *            User state
     * @param country
     *            Country of of residence
     * @param phone
     *            User phone number
     * @param The
     *            id of the brand, managed by this brand manager
     */
    public BrandManager(String password, String firstName, String lastName, String eMail, String address1, String address2, String city,
        String state, String zip, String country, String phone, int brandId) {
    this();
    this.password = password;
    this.firstName = firstName;
    this.lastName = lastName;
    this.eMail = eMail;
    this.address1 = address1;
    this.address2 = address2;
    this.city = city;
    this.state = state;
    this.zip = zip;
    this.country = country;
    this.phone = phone;
    this.brandId = brandId;
    }

    /*--- Overridden Methods ---*/

    /**
     * Equality is based on the e-mail of this brand manager
     */
    @Override
    public boolean equals(Object obj) {

    if ((obj == null) || !(obj instanceof BrandManager)) {
        return false;
    }

    // reference comparison
    if (obj == this) {
        return true;
    }

    final BrandManager other = (BrandManager) obj;

    return new EqualsBuilder().append(geteMail(), other.geteMail()).isEquals();
    }

    /**
     * The unique hash code based on the e-mail of this brand manager
     */
    @Override
    public int hashCode() {
    return new HashCodeBuilder().append(this.geteMail()).toHashCode();
    }

    /**
     * Returning first name, last name and email.
     */
    @Override
    public String toString() {
    return ("First name: " + getFirstName() + ", Last name: " + getLastName() + ", E-Mail: " + geteMail());
    }

I created a corresponding table in my DB:

CREATE TABLE `brand_managers` (
  `id` bigint(20) NOT NULL,
  `creation_time` datetime NOT NULL,
  `password` varchar(45) NOT NULL,
  `first_name` varchar(45) NOT NULL,
  `last_name` varchar(45) NOT NULL,
  `email` varchar(45) NOT NULL,
  `address1` varchar(45) NOT NULL,
  `address2` varchar(45) DEFAULT NULL,
  `city` varchar(45) NOT NULL,
  `state` varchar(45) NOT NULL,
  `zip` varchar(45) DEFAULT NULL,
  `country` varchar(45) NOT NULL,
  `phone` varchar(45) NOT NULL,
  `brand_id` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `id_UNIQUE` (`id`),
  UNIQUE KEY `email_UNIQUE` (`email`)

When trying to persist a new instance of this entity I get:

2012-07-21 12:24:03 JDBCExceptionReporter [WARN] SQL Error: 1364, SQLState: HY000
2012-07-21 12:24:03 JDBCExceptionReporter [ERROR] Field 'id' doesn't have a default value
2012-07-21 12:24:03 HibernateTask [ERROR] Hibernate exception caught in me.comocomo.server.dao.objectModel.club.register.BrandManager - could not insert: [me.comocomo.server.dao.objectModel.club.register.BrandManager]
2012-07-21 12:24:03 AssertionFailure [ERROR] an assertion failure occured (this may indicate a bug in Hibernate, but is more likely due to unsafe use of the session)
org.hibernate.AssertionFailure: null id in me.comocomo.server.dao.objectModel.club.register.BrandManager entry (don't flush the Session after an exception occurs)
    at org.hibernate.event.def.DefaultFlushEntityEventListener.checkId(DefaultFlushEntityEventListener.java:82)
    at org.hibernate.event.def.DefaultFlushEntityEventListener.getValues(DefaultFlushEntityEventListener.java:190)
    at org.hibernate.event.def.DefaultFlushEntityEventListener.onFlushEntity(DefaultFlushEntityEventListener.java:147)
    at org.hibernate.event.def.AbstractFlushingEventListener.flushEntities(AbstractFlushingEventListener.java:219)
    at org.hibernate.event.def.AbstractFlushingEventListener.flushEverythingToExecutions(AbstractFlushingEventListener.java:99)
    at org.hibernate.event.def.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:50)
    at org.hibernate.impl.SessionImpl.flush(SessionImpl.java:1216)
    at org.hibernate.impl.SessionImpl.managedFlush(SessionImpl.java:383)
    at org.hibernate.transaction.JDBCTransaction.commit(JDBCTransaction.java:133)

Now, I have managed to resolve it by using auto-increment, Meaning that when I altered the id column (which is the PK) as follows:

ALTER TABLE `MYDB`.`brand_managers` CHANGE COLUMN `id` `id` BIGINT(20) NOT NULL AUTO_INCREMENT  ;

It worked just fine!

I don't mind actually working this way (using auto-incremented id) but I simply don't understand why this is the only constellation that actually works. Why can't I generate a long type unique id using the "IDENTITY" generator?

Answer 1

That's what an IDENTITY generator is: a generator that relies on an auto-increment field to generate the unique ID, and asks the database for the generated ID once the record has been saved.

If you want the ID to be generated in memory you could use a UUID generator.

Most of the generators ask the database to generate the ID, because it's the only component in the architecture that is reliable and shared. If you have several applications each saving records in the same table, and each using their own generator, it will obviously lead to conflicts. Asking the database solves this problem: every application agrees to use an auto-increment field, or a sequence, or a table, and everything is fine.