并发访问SQLite Da(concurrent access SQLite Da)

 我正在开发一个包含3个部分的应用程序： - JavaFX桌面应用程序。 - Java Server WebApp - AndroidApp 
 
 我使用Hibernate映射SQLite数据库。 
 
 但是当桌面应用程序打开并尝试通过服务器从AndroidApp插入新的ibject时，它会给我一个错误： java.sql.SQLException: database is locked 
 
 我的hibernate.cfg.xml文件： 
 
<property name="show_sql">true</property>
<property name="format_sql">true</property>
    <property name="dialect">dialect.SQLiteDialect</property>
    <property name="connection.driver_class">org.sqlite.JDBC</property>
    <property name="connection.url">jdbc:sqlite:grainsa_provisional.sqlite</property>
    <property name="connection.username"></property>
    <property name="connection.password"></property>
 
 和我的“对象管理器”，在服务器和桌面上的方式相同： 
 
private Session mSession;
private Transaction mTransaction;

private void initQuery() throws HibernateException {
    mSession = HibernateUtil.getSessionFactory().openSession();
    mTransaction = mSession.beginTransaction();
}

private void manejaExcepcion(HibernateException hibernateException) {
    mTransaction.rollback();
    throw new HibernateException("ha ocurrido un error con la Base de Datos!!!", hibernateException);
}


public Conductor selectConductorByID(Integer id) {
    Conductor conductor = new Conductor();
    try{
        initQuery();
        conductor = (Conductor) mSession.get(Conductor.class, id);
    } catch (HibernateException e){
        manejaExcepcion(e);
        throw e;
    } finally {
        mSession.close();
    }
    return conductor;
}
 
 如果您需要更多信息，请询问！ 
 
 我做错了什么？ 
 
 谢谢大家，对不起我的英语！ 
 
 编辑：思考改变mi桌面JavaFX应用程序的访问模式以通过服务器进行查询，但这需要我很多时间，而且我认为这不是最好的方法。 
 
 Edit2：这是打开，查询和关闭conatabion到databasa锁定/查询/解锁的正确方法吗？ 
 
private void initQuery() throws HibernateException {
    mSession = HibernateUtil.getSessionFactory().openSession();
    mTransaction = mSession.beginTransaction();
}

private void manejaExcepcion(HibernateException hibernateException) {
    mTransaction.rollback();
    throw new HibernateException("ha ocurrido un error con la Base de Datos!!!", hibernateException);
}


public Conductor selectConductorByID(Integer id) {
    Conductor conductor = new Conductor();
    try{
        initQuery();
        conductor = (Conductor) mSession.get(Conductor.class, id);
    } catch (HibernateException e){
        manejaExcepcion(e);
        throw e;
    } finally {
        mSession.close();
    }
    return conductor;
}
 
 请帮忙！ 再次感谢！ 我有点沮丧...... 

I´m developing an app with 3 parts: - JavaFX Desktop app. - Java Server WebApp - AndroidApp
 
I´m using Hibernate for mapping a SQLite Database.
 
But when the desktop app is open and try to insert a new ibject from the AndroidApp throug the Server it gives me an error: java.sql.SQLException: database is locked
 
My hibernate.cfg.xml file:
 
<property name="show_sql">true</property>
<property name="format_sql">true</property>
    <property name="dialect">dialect.SQLiteDialect</property>
    <property name="connection.driver_class">org.sqlite.JDBC</property>
    <property name="connection.url">jdbc:sqlite:grainsa_provisional.sqlite</property>
    <property name="connection.username"></property>
    <property name="connection.password"></property>
 
And my "Objects Manager",the same way in the Server and in the Desktop by example:
 
private Session mSession;
private Transaction mTransaction;

private void initQuery() throws HibernateException {
    mSession = HibernateUtil.getSessionFactory().openSession();
    mTransaction = mSession.beginTransaction();
}

private void manejaExcepcion(HibernateException hibernateException) {
    mTransaction.rollback();
    throw new HibernateException("ha ocurrido un error con la Base de Datos!!!", hibernateException);
}


public Conductor selectConductorByID(Integer id) {
    Conductor conductor = new Conductor();
    try{
        initQuery();
        conductor = (Conductor) mSession.get(Conductor.class, id);
    } catch (HibernateException e){
        manejaExcepcion(e);
        throw e;
    } finally {
        mSession.close();
    }
    return conductor;
}
 
If you need more information please ask!
 
What i´m doing wrong?
 
Thanks everyone and sorry about my english!
 
Edit: ím thinking to change the acces mode of mi desktop JavaFX app to make the query through the server, but it will take me alot of time, and i do not think that is the best way to do it..
 
Edit2: This is the right way to open, make query and close the conexion to the databasa to lock/query/unlock?
 
private void initQuery() throws HibernateException {
    mSession = HibernateUtil.getSessionFactory().openSession();
    mTransaction = mSession.beginTransaction();
}

private void manejaExcepcion(HibernateException hibernateException) {
    mTransaction.rollback();
    throw new HibernateException("ha ocurrido un error con la Base de Datos!!!", hibernateException);
}


public Conductor selectConductorByID(Integer id) {
    Conductor conductor = new Conductor();
    try{
        initQuery();
        conductor = (Conductor) mSession.get(Conductor.class, id);
    } catch (HibernateException e){
        manejaExcepcion(e);
        throw e;
    } finally {
        mSession.close();
    }
    return conductor;
}
 
Please help! and thanks again! I´m a little bit deseperated...

原文：https://stackoverflow.com/questions/22741382