如何在Hibernate中定义未生成的主键字段

Question

Consider class Person :

package model;
import java.io.Serializable;
import openAPI.modelInterfaces.PersonInterface;


/**
 * 
 * @author X3
 *
 */
public class Person implements PersonInterface , Serializable
{

    private static final long serialVersionUID = 1L;

    private String m_id = "Missing ID number";
    private String m_firstName = "Missing first name";;
    private String m_lastName = "Missing last name";;
    private String m_userName = "Missing user name";;
    private String m_password = "Missing password";
    private String m_address = "Missing address";
    private String m_status = "Missing status";

    /**
     * 
     * @param _id
     * @param _firstName
     * @param _lastName
     * @param _userName
     * @param _password
     * @param _address
     */
    public Person(String _id, String _firstName , String _lastName , 
            String _userName , String _password , String _address , String _status)
    {
        this.m_id = _id;
        this.m_firstName = _firstName;
        this.m_lastName = _lastName;
        this.m_userName = _userName;
        this.m_password = _password;
        this.m_address = _address;
        this.m_status = _status;
    }

    public Person() {}


    //////////////////////////  getters /////////////////////////


    @Override
    public String getLastName() {

        return this.m_lastName;
    }


    @Override
    public String getIdnumber() {
        return this.m_id;
    }



    @Override
    public String getFirstName() {
        return this.m_firstName;
    }



    @Override
    public String getUserName() {
        return this.m_userName;
    }




    @Override
    public String getPassword() {
        return this.m_password;
    }


    @Override
    public String getAddress() {
        return this.m_address;
    }


    ///////////////////// setters ///////////////////////

    @Override
    public void setAddress(String _address) {

        if (!isMissing(_address)) 
            this.m_address = _address;
    }

    @Override
    public void setLastName(String _lastName) {

        if (!isMissing(_lastName)) 
            this.m_lastName = _lastName;
    }

    @Override
    public void setPassword(String _password) 
    {
        if (!isMissing(_password)) 
            this.m_password = _password;
    }


    @Override
    public void setUsername(String _userName) {

            this.m_userName = _userName;
    }


    @Override
    public void setFirstName(String _newName) {

        if (!isMissing(_newName)) 
            this.m_firstName = _newName;
    }


    @Override
    public void setidNumber(String _newID) {

        if (!isMissing(_newID)) 
            this.m_id = _newID;
    }


    // others 

    private boolean isMissing(String value) 
    {
        return((value == null) || (value.trim().equals("")));
    }

    @Override
    public void setStatus(String _status) {
        if (!isMissing(_status)) 
            this.m_status = _status;

    }

    @Override
    public String getStatus() {
        return this.m_status;
    }
}

And its matching table in MYSQL :

/**
 * creating a table for each type person in the bank 
 * @param tableType
 * @throws SQLException
 */
public void createTable(String tableType) throws SQLException
{
    m_statement.executeUpdate (

            "CREATE TABLE IF NOT EXISTS "+ tableType +" ("
            + "FirstName CHAR(20), LastName CHAR(20),"
            + "Address CHAR(50), IdNumber CHAR(20),"
            + "UserName CHAR(20), Password CHAR(20) , Status CHAR(20) , PRIMARY KEY (IdNumber))");
}

I want to use Hibernate to map a Person object to a record in the MYSQL DB .

Here is the hbm.xml file :

<?xml version='1.0'?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">

<!-- name of the package -->
<hibernate-mapping package="model">

<!-- name of the class & name of the table , that we want to MAP the object to  -->
<class name="Person" table="PersonnelTable">

<id name="m_id" type="String" column="IdNumber" >
<generator class="identity"/>
</id>

<property name="m_firstName" column="FirstName" type="string"/>
<property name="m_lastName" column="LastName" type="string"/>
<property name="m_userName" column="UserName" type="string"/>
<property name="m_passWord" column="Password" type="string"/>
<property name="m_address" column="Address" type="string"/>
<property name="m_status" column="PersonType" type="string"/>

</class>
</hibernate-mapping>

How can I set the primary key privateId to a non auto increment primary key ?

I don't want MYSQL to auto generate a unique key for each person that I map , since the ID is given by the user .

Thanks

Answer 1

The freely available, easily searchable documentation says:

assigned

lets the application assign an identifier to the object before save() is called. This is the default strategy if no element is specified.

Also, you should definitely not use char as the type of your columns, especially for IDs. char columns are padded with spaces. Use varchar instead.

And also: annotations exist since Java 5, 10 years ago. Why are you still using proprietary XML files instead of standard, much easier and safer JPA annotations to define your mappings?