Question

我正在尝试创建填充Admin和AdminHealthPlan表。

两者之间的关系是一对多（一个管理员可以拥有许多健康计划）。

Admin类的定义如下：

@Entity
@Table(name = "admin")
@NamedQueries(value = {
        @NamedQuery(name = "admin.getAdminById", query = "SELECT a "
            + "from  Admin a WHERE a.id=:id and deleted = 0"),
        @NamedQuery(name = "admin.getAdminByEmail", query = "SELECT a "
            + "from Admin a where a.email = :email and deleted = 0") })
public class Admin extends Persistent {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    @Column(name = "password")
    private String password;
    @Column(name = "email")
    private String email;
    @Column(name = "first_name")
    private String firstName;
    @Column(name = "last_name")
    private String lastName;
    @Column(name = "phone")
    private String phone;
    @OneToOne(mappedBy = "admin", fetch = FetchType.EAGER,
            cascade = CascadeType.ALL)
    private AdminRole role;
    @OneToOne(mappedBy = "admin", fetch = FetchType.EAGER,
            cascade = CascadeType.ALL)
    private AdminPasswordToken passwordToken;
    @OneToMany(mappedBy = "admin", fetch = FetchType.EAGER,
            cascade = CascadeType.ALL)
    private Set<AdminHealthPlan> healthPlans;
    @Column(name = "deleted")
    private Boolean deleted = false;
    @Column(name = "security_question")
    private Integer securityQuestion;
    @Column(name = "security_answer")
    private String securityAnswer;

    public Admin() {}

    // Getters/setters
}

AdminHealthPlan类定义如下。

@Entity
@Table(name = "admin_health_plan")
public class AdminHealthPlan implements Serializable {

    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    @Column(name = "health_plan")
    private String healthPlan;
    @ManyToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "admin_id", nullable = false)
    private Admin admin;
    @Column(name = "deleted")
    private Boolean deleted = false;

    public AdminHealthPlan() {}

    public AdminHealthPlan(String healthPlan) {
        this.healthPlan = healthPlan;
    }

    // Getters/setters
}

我通过从请求中检索值并将其设置为admin对象来创建管理对象，类似地创建管理员健康计划集对象并在admin对象中设置它，最后发送对象以保存到db。

它给出了如下例外

May 21, 2012 6:49:16 PM org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet appServlet threw exception
com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException: Column 'admin_id' cannot be null
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:39)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:27)
    at java.lang.reflect.Constructor.newInstance(Constructor.java:513)
    at com.mysql.jdbc.Util.handleNewInstance(Util.java:406)
    at com.mysql.jdbc.Util.getInstance(Util.java:381)
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1015)
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:956)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3558)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3490)
    at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:1959)
    at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2109)
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2648)
    at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2077)
    at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2362)
    at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2280)
    at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2265)
    at org.apache.tomcat.dbcp.dbcp.DelegatingPreparedStatement.executeUpdate(DelegatingPreparedStatement.java:105)
    at org.apache.tomcat.dbcp.dbcp.DelegatingPreparedStatement.executeUpdate(DelegatingPreparedStatement.java:105)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at org.hibernate.engine.jdbc.internal.proxy.AbstractStatementProxyHandler.continueInvocation(AbstractStatementProxyHandler.java:122)
    at org.hibernate.engine.jdbc.internal.proxy.AbstractProxyHandler.invoke(AbstractProxyHandler.java:81)
    at $Proxy70.executeUpdate(Unknown Source)

WS

用于创建Admin和AdminHealthPlan对象：

private Admin createAdminDummyObject() {
        Admin admin = new Admin();
        admin.setFirstName("Zuned");
        admin.setLastName("Ahmed");
        admin.setEmail("zuned@hcentive.com");
        admin.setPhone("123-1234454");
        admin.setHealthPlans(getHealthPlans(admin));
        return admin;
    }

    private Set<AdminHealthPlan> getHealthPlans(Admin admin) {
         Set<AdminHealthPlan> adminAHP = new HashSet<AdminHealthPlan>();
         AdminHealthPlan ahp = new AdminHealthPlan();
         ahp.setAdmin(admin);
         ahp.setHealthPlan("Plan1");
         adminAHP.add(ahp);
        return adminAHP;
    }

我用过

getJpaTemplate().persist(t);

坚持使用db。

任何人都可以告诉我我错过了什么吗？

Answer 1

我使用EclipseLink，所以我不确定它在hibernate中是否有所不同，但这是我看错了：

1-您没有将@Column注释添加到Admin实体的id属性。
2 - 我总是使用GenerationType.IDENTITY来表示pks。 3 - 在表AdminHealthPlan中，admin属性具有JoinColumn批注，但您未包含referencedColumnName属性。它应该是这样的：@JoinColumn(name = "admin_id", referencedColumnName = "admin_id")
4-为了向Admin实体添加新的AdminHealthPlan，我建议您执行以下操作：在Admin类中创建一个名为“addAdminHealthPlan”的方法：

public void addAdminHealthPlan(AdminHealthPlan ahp) {
    if(healthPlans == null) {
          healthPlans = new HashSet<AdminHealthPlan>();
    }
    ahp.setAdmin(this);
    healthPlans.add(ahp);
}

我不知道getJpaTemplate（）方法是什么，对象“t”是什么，但这是坚持管理员的逻辑：

Admin admin = createAdminDummyObject();
AdminHealthPlan ahp = new AdminHealthPlan();
ahp.setAdmin(admin);
ahp.setHealthPlan("Plan1");
admin.addAdminHealthPlan(ahp);
entityManager.persist(admin);

jpa一对多的关系无法给予例外

1 个答案: