我想我误解了@多对一关系上下文中级联的含义。
案例:
public class User {
@OneToMany(fetch = FetchType.EAGER)
protected Set<Address> userAddresses;
}
public class Address {
@ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
protected User addressOwner;
}
cascade = CascadeType.ALL是什么意思?例如，如果我从数据库中删除了某个地址，我添加了cascade = CascadeType这一事实如何。所有影响我的数据(我猜是用户)?
我是Hibernate的新手，我不确定是否使用Hibernate SessionFactory或JPA EntityManagerFactory来创建Hibernate会话。
这两者有什么不同?使用这些工具的优缺点是什么?
这个问题几乎说明了一切。使用JPARepository我如何更新一个实体?
JPARepository只有一个save方法，它并没有告诉我它是创建还是更新。例如，我插入一个简单的对象到数据库User，它有三个字段:姓，名和年龄:
@Entity
public class User {
private String firstname;
private String lastname;
//Setters and getters for age omitted, but they are the same as with firstname and lastname.
private int age;
@Column
public String getFirstname() {
return firstname;
}
public void setFirstname(String firstname) {
this.firstname = firstname;
}
@Column
public String getLastname() {
return lastname;
}
public void setLastname(String lastname) {
this.lastname = lastname;
}
private long userId;
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
public long getUserId(){
return this.userId;
}
public void setUserId(long userId){
this.userId = userId;
}
}
然后我简单地调用save()，这在这一点上实际上是插入到数据库:
User user1 = new User();
user1.setFirstname("john"); user1.setLastname("dew");
user1.setAge(16);
userService.saveUser(user1);// This call is actually using the JPARepository: userRepository.save(user);
到目前为止一切顺利。现在我要更新这个用户，比如改变他的年龄。为此，我可以使用QueryDSL或NamedQuery之类的查询。但是，考虑到我只想使用spring-data-jpa和JPARepository，我如何告诉它我想做更新而不是插入呢?
具体来说，我如何告诉spring-data-jpa具有相同用户名和名字的用户实际上是EQUAL，并且现有实体应该被更新?重写等号并不能解决这个问题。
我有一个包含多对一关系的jpa持久化对象模型:一个Account有多个transaction。一个事务有一个帐户。
下面是一段代码:
@Entity
public class Transaction {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@ManyToOne(cascade = {CascadeType.ALL},fetch= FetchType.EAGER)
private Account fromAccount;
....
@Entity
public class Account {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@OneToMany(cascade = {CascadeType.ALL},fetch= FetchType.EAGER, mappedBy = "fromAccount")
private Set<Transaction> transactions;
我能够创建Account对象，向其添加事务，并正确地持久化Account对象。但是，当我创建一个事务，使用现有的已经持久化的帐户，并持久化的事务，我得到一个异常:
导致:org.hibernate.PersistentObjectException:传递给persist: com.paulsanwald.Account的分离实体 org.hibernate.event.internal.DefaultPersistEventListener.onPersist (DefaultPersistEventListener.java: 141)
因此，我能够持久化一个包含事务的Account，但不能持久化一个具有Account的Transaction。我认为这是因为帐户可能没有附加，但这段代码仍然给了我相同的异常:
if (account.getId()!=null) {
account = entityManager.merge(account);
}
Transaction transaction = new Transaction(account,"other stuff");
// the below fails with a "detached entity" message. why?
entityManager.persist(transaction);
如何正确地保存与已经持久化的帐户对象相关联的事务?
Hibernate和Spring Data JPA之间的主要区别是什么?
什么时候不应该使用Hibernate或Spring Data JPA?
另外，Spring JDBC模板什么时候比Hibernate和Spring Data JPA表现更好?
我试图通过spring-jpa运行一个使用hibernate的spring-boot应用程序，但我得到这个错误:
Caused by: org.hibernate.HibernateException: Access to DialectResolutionInfo cannot be null when 'hibernate.dialect' not set
at org.hibernate.engine.jdbc.dialect.internal.DialectFactoryImpl.determineDialect(DialectFactoryImpl.java:104)
at org.hibernate.engine.jdbc.dialect.internal.DialectFactoryImpl.buildDialect(DialectFactoryImpl.java:71)
at org.hibernate.engine.jdbc.internal.JdbcServicesImpl.configure(JdbcServicesImpl.java:205)
at org.hibernate.boot.registry.internal.StandardServiceRegistryImpl.configureService(StandardServiceRegistryImpl.java:111)
at org.hibernate.service.internal.AbstractServiceRegistryImpl.initializeService(AbstractServiceRegistryImpl.java:234)
at org.hibernate.service.internal.AbstractServiceRegistryImpl.getService(AbstractServiceRegistryImpl.java:206)
at org.hibernate.cfg.Configuration.buildTypeRegistrations(Configuration.java:1885)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1843)
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:850)
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:843)
at org.hibernate.boot.registry.classloading.internal.ClassLoaderServiceImpl.withTccl(ClassLoaderServiceImpl.java:398)
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:842)
at org.hibernate.jpa.HibernatePersistenceProvider.createContainerEntityManagerFactory(HibernatePersistenceProvider.java:152)
at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:336)
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:318)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1613)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1550)
... 21 more
我的pom.xml文件是这样的:
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>1.1.8.RELEASE</version>
</parent>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-web</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-actuator</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-web</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-config</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-taglibs</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-jpa</artifactId>
</dependency>
<dependency>
<groupId>commons-dbcp</groupId>
<artifactId>commons-dbcp</artifactId>
</dependency>
</dependencies>
我的hibernate配置是这样的(方言配置在该类的最后一个方法中):
@Configuration
@EnableTransactionManagement
@ComponentScan({ "com.spring.app" })
public class HibernateConfig {
@Bean
public LocalSessionFactoryBean sessionFactory() {
LocalSessionFactoryBean sessionFactory = new LocalSessionFactoryBean();
sessionFactory.setDataSource(restDataSource());
sessionFactory.setPackagesToScan(new String[] { "com.spring.app.model" });
sessionFactory.setHibernateProperties(hibernateProperties());
return sessionFactory;
}
@Bean
public DataSource restDataSource() {
BasicDataSource dataSource = new BasicDataSource();
dataSource.setDriverClassName("org.postgresql.Driver");
dataSource.setUrl("jdbc:postgresql://localhost:5432/teste?charSet=LATIN1");
dataSource.setUsername("klebermo");
dataSource.setPassword("123");
return dataSource;
}
@Bean
@Autowired
public HibernateTransactionManager transactionManager(SessionFactory sessionFactory) {
HibernateTransactionManager txManager = new HibernateTransactionManager();
txManager.setSessionFactory(sessionFactory);
return txManager;
}
@Bean
public PersistenceExceptionTranslationPostProcessor exceptionTranslation() {
return new PersistenceExceptionTranslationPostProcessor();
}
Properties hibernateProperties() {
return new Properties() {
/**
*
*/
private static final long serialVersionUID = 1L;
{
setProperty("hibernate.hbm2ddl.auto", "create");
setProperty("hibernate.show_sql", "false");
setProperty("hibernate.dialect", "org.hibernate.dialect.PostgreSQLDialect");
}
};
}
}
我哪里做错了?
我实际上是在寻找一个“@Ignore”类型的注释，我可以用它来停止某个特定字段的持久化。如何实现这一目标?
Java有transient关键字。为什么JPA有@Transient而不是简单地使用已经存在的java关键字?
这里有一些关于JPA实体的讨论，以及应该为JPA实体类使用哪些hashCode()/equals()实现。它们中的大多数(如果不是全部)依赖于Hibernate，但是我想中立地讨论它们的jpa实现(顺便说一下，我使用的是EclipseLink)。
所有可能的实现都有其自身的优点和缺点:
hashCode()/equals()契约一致性(不可变性)用于列表/集操作 是否可以检测到相同的对象(例如来自不同会话的对象，来自惰性加载数据结构的动态代理) 实体在分离(或非持久化)状态下是否正确运行
在我看来，有三种选择:
Do not override them; rely on Object.equals() and Object.hashCode() hashCode()/equals() work cannot identify identical objects, problems with dynamic proxies no problems with detached entities Override them, based on the primary key hashCode()/equals() are broken correct identity (for all managed entities) problems with detached entities Override them, based on the Business-Id (non-primary key fields; what about foreign keys?) hashCode()/equals() are broken correct identity (for all managed entities) no problems with detached entities
我的问题是:
我是否错过了一个选择和/或赞成/反对的观点? 你选择了什么，为什么?
更新1:
通过“hashCode()/equals()是坏的”，我的意思是连续的hashCode()调用可能返回不同的值，这(当正确实现时)在对象API文档的意义上不是坏的，但是当试图从Map、Set或其他基于哈希的集合中检索更改的实体时，会导致问题。因此，JPA实现(至少是EclipseLink)在某些情况下不能正确工作。
更新2:
谢谢你的回答——大部分问题都很有质量。 不幸的是，我仍然不确定哪种方法最适合实际应用程序，或者如何确定最适合我的应用程序的方法。所以，我将保持这个问题的开放性，希望有更多的讨论和/或意见。
我已经使用JPA(实现Hibernate)一段时间了，每次我需要创建实体时，我发现自己在AccessType、不可变属性、等于/hashCode、... .等问题上苦苦挣扎 所以我决定试着找出每个问题的一般最佳实践，并把它写下来供个人使用。 我不介意任何人对此发表评论，或者告诉我哪里错了。
实体类
实现序列化 原因:规范要求您必须这样做，但是一些JPA提供者并没有强制执行这一点。Hibernate作为JPA提供者并没有强制执行这一点，但是如果没有实现Serializable，它可能会在ClassCastException中失败。
构造函数
创建一个包含实体所有必需字段的构造函数 原因:构造函数应该始终让创建的实例处于正常状态。 除了这个构造函数:有一个包的私有默认构造函数 原因:默认构造函数需要Hibernate初始化实体;允许私有，但是需要包私有(或公共)可见性来生成运行时代理和高效的数据检索，而不需要字节码插装。
字段/属性
Use field access in general and property access when needed Reason: this is probably the most debatable issue since there are no clear and convincing arguments for one or the other (property access vs field access); however, field access seems to be general favourite because of clearer code, better encapsulation and no need to create setters for immutable fields Omit setters for immutable fields (not required for access type field) properties may be private Reason: I once heard that protected is better for (Hibernate) performance but all I can find on the web is: Hibernate can access public, private, and protected accessor methods, as well as public, private and protected fields directly. The choice is up to you and you can match it to fit your application design.
= / hashCode
Never use a generated id if this id is only set when persisting the entity By preference: use immutable values to form a unique Business Key and use this to test equality if a unique Business Key is not available use a non-transient UUID which is created when the entity is initialised; See this great article for more information. never refer to related entities (ManyToOne); if this entity (like a parent entity) needs to be part of the Business Key then compare the ID's only. Calling getId() on a proxy will not trigger the loading of the entity, as long as you're using property access type.
实体例子
@Entity
@Table(name = "ROOM")
public class Room implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue
@Column(name = "room_id")
private Integer id;
@Column(name = "number")
private String number; //immutable
@Column(name = "capacity")
private Integer capacity;
@ManyToOne(fetch = FetchType.LAZY, optional = false)
@JoinColumn(name = "building_id")
private Building building; //immutable
Room() {
// default constructor
}
public Room(Building building, String number) {
// constructor with required field
notNull(building, "Method called with null parameter (application)");
notNull(number, "Method called with null parameter (name)");
this.building = building;
this.number = number;
}
@Override
public boolean equals(final Object otherObj) {
if ((otherObj == null) || !(otherObj instanceof Room)) {
return false;
}
// a room can be uniquely identified by it's number and the building it belongs to; normally I would use a UUID in any case but this is just to illustrate the usage of getId()
final Room other = (Room) otherObj;
return new EqualsBuilder().append(getNumber(), other.getNumber())
.append(getBuilding().getId(), other.getBuilding().getId())
.isEquals();
//this assumes that Building.id is annotated with @Access(value = AccessType.PROPERTY)
}
public Building getBuilding() {
return building;
}
public Integer getId() {
return id;
}
public String getNumber() {
return number;
}
@Override
public int hashCode() {
return new HashCodeBuilder().append(getNumber()).append(getBuilding().getId()).toHashCode();
}
public void setCapacity(Integer capacity) {
this.capacity = capacity;
}
//no setters for number, building nor id
}
欢迎其他建议加入到这个列表中……
更新
读了这篇文章后，我调整了eq/hC的实现方式:
如果一个不可变的简单业务密钥可用:使用它 在所有其他情况下:使用uuid