我在这个网站上看过fimiliar问题。但是没有什么对我有用。我试着在没有xml的情况下配置Hibernate。我想创建一个表 * animal * 并持久化一条记录。
- 堆栈跟踪异常**
java.lang.IllegalArgumentException: Unable to locate persister: gameCenter.Animal
at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:735)
at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:716)
at gameCenter.Test.saveAnimal(Test.java:50)
at gameCenter.Test.main(Test.java:21)
- 测试类**
package gameCenter;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;
import org.hibernate.cfg.Configuration;
import org.hibernate.cfg.Environment;
import org.hibernate.service.ServiceRegistry;
import java.util.Properties;
public class Test {
public static void main(String[] args) {
Animal animal = new Animal("Dog");
saveAnimal(animal);
}
public static SessionFactory getSessionFactory() {
Properties settings = new Properties();
settings.put(Environment.DRIVER, "org.postgresql.Driver");
settings.put(Environment.URL, "jdbc:postgresql://localhost:5432/ineffable");
settings.put(Environment.USER, "postgres");
settings.put(Environment.PASS, "ewqe3121");
settings.put(Environment.DIALECT, "org.hibernate.dialect.PostgreSQLDialect");
settings.put(Environment.SHOW_SQL, "true");
settings.put(Environment.HBM2DDL_AUTO, "create");
Configuration configuration = new Configuration();
configuration.setProperties(settings);
configuration.addAnnotatedClass(Animal.class);
ServiceRegistry serviceRegistry = new StandardServiceRegistryBuilder()
.applySettings(configuration.getProperties()).build();
return configuration.buildSessionFactory(serviceRegistry);
}
public static void saveAnimal(Animal animal) {
Transaction transaction = null;
try(Session session = getSessionFactory().openSession()) {
transaction = session.beginTransaction();
session.persist(animal);
transaction.commit();
} catch (Exception e) {
System.out.println("Message: " + e.getMessage());
e.printStackTrace();
}
}
}
- Animal类(实体)**
package gameCenter;
import javax.persistence.*;
import java.io.Serializable;
@Entity
@Table(name = "animal", schema = "public")
class Animal implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@Column(name = "one")
private String one;
public Animal(String one) {
this.one = one;
}
public int getId() {
return id;
}
public String getOne() {
return one;
}
public void setOne(String one) {
this.one = one;
}
}
- 依赖关系**
<dependencies>
<dependency>
<groupId>org.postgresql</groupId>
<artifactId>postgresql</artifactId>
<version>42.3.3</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-core</artifactId>
<version>6.0.0.Final</version>
</dependency>
<dependency>
<groupId>javax.persistence</groupId>
<artifactId>persistence-api</artifactId>
<version>1.0.2</version>
</dependency>
</dependencies>
Postgres版本为14....................................................................................................................................................................................................................................................
2条答案
按热度按时间njthzxwz1#
我和你有同样的问题...
我不知道为什么,但我需要使用实体的默认构造函数,然后用它的访问器赋值。
试试这个,告诉我它是否有效:
fumotvh32#
您必须将软件包从
javax.persistance.*
更改为jakarta.persitance.*
。