我想通过对父实体调用 save 将父实体和子实体一起保存到 MySQL 数据库中。 Parent 和 Child 实体之间存在一对一的映射。父 ID 是自动生成的，我们也需要在 child 中使用与 child 的 pk 相同的 ID。

我正在使用 Spring Data JPA 2.0(JPA 提供程序是 Hibernate)和 Spring MVC 框架。 当尝试插入实体时，出现以下错误。

根本原因

org.springframework.dao.DataIntegrityViolationException: not-null property references a null or transient value: com.serro.cbmapi.model.Child.parent; nested exception is org.hibernate.PropertyValueException: not-null property references a null or transient value: com.serro.cbmapi.model.Child.parent org.springframework.orm.jpa.vendor.HibernateJpaDialect.convertHibernateAccessException(HibernateJpaDialect.java:207)

这是我的数据库架构:

Parent Table:
   CREATE TABLE `parent` (
   `pid` int(11) NOT NULL AUTO_INCREMENT,
   `parent_name` varchar(256) CHARACTER SET utf8 COLLATE utf8_bin DEFAULT NULL,
    PRIMARY KEY (`pid`)    ) ENGINE=InnoDB AUTO_INCREMENT=13 DEFAULT CHARSET=latin1;

    ) ENGINE=InnoDB DEFAULT CHARSET=latin1;

Child Table:
    CREATE TABLE `child` (
    `cid` int(11) NOT NULL,
    `child_name` varchar(256) CHARACTER SET utf8 COLLATE utf8_bin DEFAULT NULL,
    PRIMARY KEY (`cid`),
    CONSTRAINT `child_f1` FOREIGN KEY (`cid`) REFERENCES `parent` (`pid`)

    ) ENGINE=InnoDB DEFAULT CHARSET=latin1;

这是我的 Java 实体 父实体:

@Entity(name="parent")
@NamedQuery(name="Parent.findAll", query="SELECT p FROM parent p")
public class Parent implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue
    private int pid;

    @Column(name="parent_name")
    private String parentName;

    //bi-directional one-to-one association to Child
    @OneToOne(mappedBy="parent",cascade=CascadeType.ALL)
    private Child child;

    //getter, setters
}

子实体:

@Entity(name="child")
@NamedQuery(name="Child.findAll", query="SELECT c FROM child c")
public class Child implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    private int cid;

    @Column(name="child_name")
    private String childName;

    //bi-directional one-to-one association to Parent
    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="cid")
    @MapsId("cid")
    private Parent parent;

    //getter, setters
}

这是我的主要方法

AbstractApplicationContext context = new ClassPathXmlApplicationContext("applicationContext.xml");  
ParentRepository parentResp = context.getBean(ParentRepository.class);  
Parent parent = new Parent();
parent.setParentName("Parent1");
Child child = new Child();
child.setChildName("Child1");
parent.setChild(child);
parentResp.save(parent);

最佳答案

您应该在两个对象中都使用级联。试试这个:

@Entity(name="parent")
public class Parent implements Serializable {
    //...
    @OneToOne(mappedBy="parent",cascade=CascadeType.ALL)
    private Child child;
    //...
}

@Entity(name="child")
public class Child implements Serializable {
    //...
    @OneToOne(optional = false, fetch = FetchType.LAZY, cascade = CascadeType.REMOVE)
    @JoinColumn(name="cid", referencedColumnName = "id")
    private Parent parent;
    //...
}

关于java - Spring Data JPA 通过从父实体获取 id 来插入子实体和父实体，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25490774/