Hibernate Criteria 创建一个“？”在生成的 SQL

Question

我创建了一个示例 Web 应用程序。我使用 MS SQL Server 2008 作为数据库并使用 hibernate + 注释来访问数据库。我的休眠配置xml如下。 问题是，Criteria.list() 返回一个空列表，而且我看到一个“？”在生成的 HSQL 中，而不是我在 Criteria 中传递的参数。

<session-factory name="">
    <property name="hibernate.connection.driver_class">sun.jdbc.odbc.JdbcOdbcDriver</property>
    <property name="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</property>
    <property name="hibernate.connection.url">jdbc:odbc:dbname</property>

    <property name="connection.pool_size">10</property>
    <property name="current_session_context_class">thread</property>

    <!-- Disable the second-level cache -->
    <property name="cache.provider_class">org.hibernate.cache.NoCacheProvider</property>       

     <!-- Echo all executed SQL to stdout -->
    <property name="show_sql">true</property>       
    <property name="hibernate.connection.username"></property>
    <property name="hibernate.connection.password"></property>

<mapping class="com.demo.Person" />
</session-factory>

这是我的注释豆

    @Entity
    @Table(name = "person")
    public class Person implements Serializable {

        public Person(){

        }
        private static final long serialVersionUID = 1L;
        @Id
        @GeneratedValue(strategy=GenerationType.AUTO) 
        @Basic(optional = false)
        @Column(name = "personid")
        private Integer personid;

        @Basic(optional = false)
        @Column(name = "firstname")
        private String firstname;

        @Column(name = "lastname")
        private String lastname;

        @Basic(optional = false)
        @Column(name = "phone")
        private String phone;

        @Column(name = "mobile")
        private String mobile;

        @Column(name = "street")
        private String street;

        @Basic(optional = false)
        @Column(name = "city")
        private String city;

        @Basic(optional = false)
        @Column(name = "country")
        private String country;

        @Basic(optional = false)
        @Column(name = "bussinessowner")
        private int bussinessowner;

        @OneToMany(cascade = CascadeType.ALL, mappedBy = "resultid1")
        private Collection<Recent> recentCollection;

//setters & getters
}

而我正在运行的代码是

Session session;
        List list = new ArrayList();
        try{
        session = HibernateUtil.getSessionFactory().openSession();
        Criteria criteria = session.createCriteria(Person.class);
        criteria.add(Restrictions.like("firstname", name, MatchMode.START));
        list= criteria.list();
        System.out.println(list.size());

        }catch (Exception e) {
            e.printStackTrace();
        }

生成的 HSQL：

Hibernate: select this_.personid as personid2_0_,  this_.city as city2_0_, this_.country as country2_0_, this_.firstname as firstname2_0_, this_.lastname as lastname2_0_ from person this_ where this_.firstname like ?

除此之外，我也没有遇到任何异常。你能帮我解决这个问题吗？ 谢谢！

Answer 1

将其更改为criteria.add(Restrictions.like("firstname", name + "%")); 有帮助吗？ 顺便说一下，? 是 jbdc 语句的正确参数占位符。

Answer 2

'?'是 JDBC 语句中参数的占位符。你看不到它的实际价值，仅此而已。没有问题，只是使用hibernate的跟踪仍然不完整。

如果您想查看 '?' 的实际值，则必须使用额外的产品，例如 log4jdbc。

Answer 3

@Rakesh 试试下面的代码：

criteria.add(Restrictions.ilike("firstname", name +"%"));

Answer 4

代替这一行

criteria.add(Restrictions.like("firstname", name, MatchMode.START)); 

试试

criteria.add(Restrictions.like("firstname", name, MatchMode.ANYWHERE)); 

或

criteria.add(Restrictions.ilike("firstname", name, MatchMode.ANYWHERE)); 

如果您将 firstname 作为 "abcd" 传递，它将生成 SQL 作为 firstname like '%abcd%'

你可以试试这个吗

session = HibernateUtil.getSessionFactory().openSession(); 
List<Person> personList = session.createQuery("Select * from Person p where p.firstname like 'rake%'").list();

System.out.println("Person result :" + personList .size());