それがで仕事ができるように、私はあなたのコードに非常にいくつかの変更をしましたけれども、私は、あなたが持っている問題を得ることはありません私のセットアップ。問題のあるユースケースのためにいくつかのデータ、またはfindMembersを初期化するcreateFriendshipsのいずれかに注意を払う：
ここ
/META-INF/persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="hibernate-entitymanager-demo" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/hibernate4?zeroDateTimeBehavior=convertToNull"/>
<property name="javax.persistence.jdbc.password" value="root"/>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
<property name="javax.persistence.jdbc.user" value="root"/>
<property name="hibernate.cache.provider_class" value="org.hibernate.cache.NoCacheProvider"/>
<!-- property name="javax.persistence.schema-generation.database.action" value="drop-and-create"/ -->
</properties>
</persistence-unit>
</persistence>
Main.java（2ユースケース。あなたがではなく、後者では）、最初にデータベースを作成したいと思うようにpersistence.xmlでjavax.persistence.schema-generation.database.actionプロパティは
public class Main {
public static void main(String[] args) {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("hibernate-entitymanager-demo");
EntityManager em = emf.createEntityManager();
EntityTransaction transaction = em.getTransaction();
transaction.begin();
// createFriendships(em);
findMembers(em);
transaction.commit();
em.close();
emf.close();
}
private static void findMembers(EntityManager em) {
List<Member> list = em.createQuery("from Member").getResultList();
for (Member m : list) {
System.out.println(m);
}
}
private static void createFriendships(EntityManager em) {
List<Member> members = createMembers(em);
for (int i = 0; i < members.size(); i++) {
for (int j = 0; j < members.size(); j++) {
if (i != j) {
createFriendship(em, members.get(i), members.get(j));
}
}
}
}
private static List<Member> createMembers(EntityManager em) {
List<Member> members = new ArrayList<>();
members.add(createMember(em, "Roberta", "Williams", "rwilliams"));
members.add(createMember(em, "Ken", "Williams", "kwilliams"));
members.add(createMember(em, "Dave", "Grossman", "dgrossman"));
members.add(createMember(em, "Tim", "Schafer", "tschafer"));
members.add(createMember(em, "Ron", "Gilbert", "rgilbert"));
return members;
}
private static Member createMember(EntityManager em, String fname, String lname, String username) {
Member m = new Member();
m.setFname(fname);
m.setLname(lname);
m.setUsername(username);
em.persist(m);
return m;
}
private static void createFriendship(EntityManager em, Member requester, Member friend) {
Friendship f = new Friendship();
f.setActive(true);
f.setDate(new Date());
f.setRequester(requester);
f.setFriend(friend);
em.persist(f);
}
}
主な生産：
Member [fname = Roberta, lname = Williams, requests = 4, friends = 4]
Member [fname = Ken, lname = Williams, requests = 4, friends = 4]
Member [fname = Dave, lname = Grossman, requests = 4, friends = 4]
Member [fname = Tim, lname = Schafer, requests = 4, friends = 4]
Member [fname = Ron, lname = Gilbert, requests = 4, friends = 4]
Friendship.java本当の実際の変更は、私は例外のcould not get a field value by reflection種類を得たように私はidにusernameから変更され参照される列の名前でした。横に私はそれがデータベースの正規化の面でより良いことだと思う：
@Entity
@Table(name = "FRIENDSHIPS")
public class Friendship implements Serializable {
@Id
@ManyToOne
@JoinColumn(referencedColumnName = "id")
Member requester;
@Id
@ManyToOne
@JoinColumn(referencedColumnName = "id")
Member friend;
@Temporal(javax.persistence.TemporalType.DATE)
Date date;
@Column(nullable = false)
boolean active;
// getters & setters
}
Member.java（toString()を除いて、ここで変更なし、）
@Entity
@Table(name = "MEMBERS")
public class Member implements Serializable {
@Id
@GeneratedValue
private long id;
@Column(name = "username", nullable = false, unique = true)
private String username;
@Column(nullable = false)
private String fname;
@Column(nullable = false)
private String lname;
@OneToMany(fetch = FetchType.EAGER, mappedBy = "requester")
private Set<Friendship> friendRequests = new HashSet<>();
@OneToMany(fetch = FetchType.EAGER, mappedBy = "friend")
private Set<Friendship> friends = new HashSet<>();
// getters & setters
@Override
public String toString() {
return "Member [fname = " + fname + ", lname = " + lname
+ ", requests = " + friendRequests.size()
+ ", friends = " + friends.size() + "]";
}
}
あなたが友人の要求のために別のテーブルを維持する必要があり、友情のためには、senderIdとreceiverIdだけのテーブルがあります。 –
ここにテストケースコードを掲載できますか？ –
@NiravChhatrola私はまだテストケースを持っていません。 –