JPA OneToMany雙向

我知道有很多關於它的問題，但我找不到一個好的回答我的問題。JPA OneToMany雙向

我使用JBoss AS 7，Spring和Hibernate（4）JPA 2.0提供商所以我有簡單@OneToMany雙向關係：

我有超類人那樣：

@MappedSuperclass 
@Inheritance(strategy=InheritanceType.JOINED) 
public abstract class Person { 

@Id 
@GeneratedValue(strategy=GenerationType.IDENTITY) 
private Long id; 

@NotNull 
@Size(min = 1, max = 25) 
@Pattern(regexp = "[A-Za-z ]*", message = "must contain only letters and spaces") 
private String name; 


public Person(String name) { 
    super(); 
    this.name = name; 
}

和類會員：

@Entity 
@Table(uniqueConstraints = @UniqueConstraint(columnNames = "email")) 
public class Member extends Person implements Serializable 
{ 
    /** Default value included to remove warning. Remove or modify at will. **/ 
    private static final long serialVersionUID = 1L; 


    @NotNull 
    @NotEmpty 
    @Email 
    private String email; 

    @NotNull 
    @Size(min = 10, max = 12) 
    @Digits(fraction = 0, integer = 12) 
    @Column(name = "phone_number") 
    private String phoneNumber; 

    @OneToMany(cascade=CascadeType.ALL , mappedBy="member" , fetch=FetchType.EAGER) 
    private List<Order> orders;

而且還Order類：

@Entity 
public class Order { 

@Id 
@GeneratedValue(strategy=GenerationType.IDENTITY) 
private Integer id; 

private float price; 

@ManyToOne(optional=false) 
private Member member; 

private String name;

所以我認爲這是一個很好的配置，但我測試的HSQL在內存中這個應用程序，我有錯誤：

Hibernate: alter table Order drop constraint FK48E972E548C740B 
2012-09-20 16:25:37 org.hibernate.tool.hbm2ddl.SchemaExport perform 
ERROR: HHH000389: Unsuccessful: alter table Order drop constraint FK48E972E548C740B 
2012-09-20 16:25:37 org.hibernate.tool.hbm2ddl.SchemaExport perform 
ERROR: Blad skladniowy w wyrazeniu SQL "ALTER TABLE ORDER[*] DROP CONSTRAINT FK48E972E548C740B "; oczekiwano "identifier" 
Syntax error in SQL statement "ALTER TABLE ORDER[*] DROP CONSTRAINT FK48E972E548C740B "; expected "identifier"; SQL statement: 
alter table Order drop constraint FK48E972E548C740B [42001-165]

而且也：

Syntax error in SQL statement "CREATE TABLE ORDER[*] (ID INTEGER GENERATED BY DEFAULT AS IDENTITY, NAME VARCHAR(255), PRICE FLOAT NOT NULL, MEMBER_ID BIGINT NOT NULL, PRIMARY KEY (ID)) "; expected "identifier"; SQL statement:

我的JUnit測試失敗我不知道什麼是錯的這個配置...

這是我簡單的JUnit：

@Test 
public void testInsertWithOrder(){ 
    Order order = new Order(20.0f, "first stuff"); 
    Order order2 = new Order(40.0f, "secondary stuff"); 
    List<Order> orders = new ArrayList<Order>(); 
    orders.add(order2); 
    orders.add(order); 

    Member member = new Member("Member name", "[email protected]", "2125552141", orders); 

    memberDao.register(member); 

    List<Member> members = memberDao.findAllOrderedByName(); 

    Assert.assertNotNull(members); 
    Assert.assertEquals(1, members.size());  

}

來源

2012-09-20 Łukasz Woźniczka

變化表名從「秩序」，以不同的東西，像PersonOrder

來源

2012-09-20 14:58:54 user1516873

是，'user1516873'是對的，如果你使用'MySqL'，'Order'字是MySQL的保留字。 – CycDemo

+1你說得對。但我仍然有錯誤：NULL不允許列「MEMBER_ID」; SQL語句： insert into UOrder（id，member_id，name，price）values（null，？，？，？）[23502-165]' –

感謝user1516873，同樣的關鍵字在PostgreSQL上保留， – dsuess

在你member在Order類中，有缺失@JoinColumn註解。嘗試如下。

@ManyToOne(optional=false) 
@JoinColumn(name = "memberId", referencedColumnName = "id") 
private Member member;

來源

2012-09-20 15:00:18 CycDemo

我嘗試了，但我也有錯誤，我的junit失敗： –

在H2保留的Keywork中，'Order'就是其中之一;請參閱http://www.scribd.com/doc/88837291/224/Keywords-Reserved-Words'CROSS，CURRENT_DATE，CURRENT_TIME，CURRENT_TIMESTAMP，DISTINCT，EXCEPT，EXISTS，FALSE，FOR，FROM，FULL，GROUP，HAVING，INNER ，INTERSECT，IS，JOIN，LIKE，LIMIT，MINUS，NATURAL，NOT，NULL，ON，ORDER，PRIMARY，ROWNUM，SELECT，SYSDATE，SYSTIME，SYSTIMESTAMP，TODAY，TRUE，UNION，UNIQUE，Where' – CycDemo

是的，我是如此愚蠢的:)它是如此容易...我改變我的班級名稱爲UOrder –

@CycDemo

我只是弄清楚，在我constuctor我現在已經得到了：

@OneToMany(cascade=CascadeType.ALL , mappedBy="member" , fetch=FetchType.EAGER) 
    private List<UOrder> orders = new ArrayList<UOrder>(); 



     public Member(String name, String email, String phoneNumber ,List<UOrder> orders) { 
     super(name); 
     this.orders = orders; 
     this.email = email; 
     for(UOrder o : orders){ 
      o.setMember(this);   
     } 
     this.orders = orders; 

    }

螞蟻這是我所需要的:)））

來源

2012-09-20 17:06:03

JPA OneToMany雙向

回答

相關問題