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Posted to users@cocoon.apache.org by beyaNet Consultancy <be...@ntlworld.com> on 2004/01/22 14:10:42 UTC

Exception in Hibernate: could not insert: [test.User#32]

Hi,
I have an issue with saving data to a postgresql database via 
hibernate, for which I get the above mentioned exception. Can anybody 
shine any light on this issue please?

1. Table schema.

CREATE TABLE "users" (
     "Name" character varying(40),
     "Password" character varying(20),
     "EmailAddress" character varying(40),
     "Lastlogon" date,
     "logonID" integer NOT NULL DEFAULT nextval('seq_id_mytable'::text),
     CONSTRAINT "users_pkey" PRIMARY KEY ("logonID")
) WITH OIDS;

2. hibernate.cfg.xml (snippet)

     <session-factory>
         	<property 
name="connection.driver_class">org.postgresql.Driver</property>
		<property 
name="connection.url">jdbc:postgresql://localhost:5432/test</property>
		<property name="connection.username">x</property>
		<property name="connection.password">x</property>
		<property name="jdbc.batch_size">0</property>
		<property name="jdbc.use_scrollable_resultsets">false</property>
		<property name="show_sql">true</property>
         <property name="use_outer_join">true</property>
         <property 
name="dialect">net.sf.hibernate.dialect.PostgreSQLDialect</property>

         <mapping resource="test/User2.hbm.xml"/>


     </session-factory>

3. User2.hbm.xml file (snippet)

	<class name="test.User" table="users">
	    <id name="ID" type="integer" column="LogonID">
				<generator class="sequence">
						<param name="sequence">seq_id_mytable</param>
				</generator>
		</id>
		
		<property name="userName" column="Name" type="string"/>
		<property name="password" column="Password" type="string"/>
		<property name="emailAddress" column="EmailAddress" type="string"/>
		<property name="lastLogon" column="Lastlogon" type="date"/>
	</class>

4. User.class (snippet)

private Integer userID;
	private String userName;
	private String password;
	private String emailAddress;
	private Date lastLogon;
	
	public User(){
	}
	
	public Integer getID() {
     		return userID;
    	}
	public void setID(Integer newUserID) {
     		userID = newUserID;
    	}

     public String getUserName() {
         return (userName == null ? "" : userName);
     }
     public void setUserName(String userName) {
         this.userName = userName;
     }

      public String getPassword() {
          return (password == null ? "" : password);
      }
      public void setPassword(String password) {
          this.password = password;
      }

     public String getEmailAddress() {
         return emailAddress;
     }
     public void setEmailAddress(String emailAddress) {
         this.emailAddress = emailAddress;
     }

	public Date getLastLogon() {
		return lastLogon;
	}
     public void setLastLogon(Date newLastLogon) {
		this.lastLogon = newLastLogon;
	}

5. The action class (snippet)

             Configuration cfg = new Configuration()
             .configure ();
             SessionFactory sf = cfg.buildSessionFactory();
             Session session = sf.openSession();
			Transaction transaction = session.beginTransaction();
			
			User usr = new User();
			usr.setUserName("postgres");
			usr.setPassword("postgres");
			usr.setEmailAddress("test@hotmail.coml");
			usr.setLastLogon(newDate);
			
             session.save(usr);
			//session.flush();
			transaction.commit();


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Re: Exception in Hibernate: could not insert: [test.User#32]

Posted by Ugo Cei <u....@cbim.it>.
beyaNet Consultancy wrote:
> Hi,
> I have an issue with saving data to a postgresql database via hibernate, 
> for which I get the above mentioned exception. Can anybody shine any 
> light on this issue please?

Look in the hibernate.log file. Look in log4j.properties to see where it 
is stored.

	Ugo


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