我正在尝试使用hibernate从java连接到mysql数据库。我试图了解是否真的值得使用ORM，例如hibernate用于关系。特别是注释与注释的关系似乎让我非常恼火。因此，对于友谊表，我希望通过使用来自users表的外键来查看特定用户的友谊请求。
在helper表的hibernate类中，还有一个表示主表的字段。因此，它不是保持与简单int的关系，而是将User类作为主要字段。如果您需要访问用户对象，这可能很有用，但在某些情况下，您只需要外键ID。
我把我为两个表写的示例代码，Users和UserFriendRequests。这是表示以下两个表的最佳方法吗？我觉得这对于代表这样的基本表来说太过分了。我更喜欢将Hibernate用于连接和单独表示表并手动处理关系。你会推荐这种方法吗？如果您能分享您关于最佳实践的想法和文档，我将不胜感激。
用户表包含以下列
UserFriendRequests表包含以下列：
用户表格的表示
@Entity
@Table(name = "Users", catalog = "example_schema", uniqueConstraints = {
@UniqueConstraint(columnNames = "userId"),
@UniqueConstraint(columnNames = "facebookId")})
public class User implements Serializable {
@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name = "userId", unique = true, nullable = false)
private int userId;
@Column(name = "facebookId", unique = true, nullable = false)
private String facebookId;
@Column(name = "userName", nullable = false)
private String userName;
@Column(name = "createdAt", nullable = false)
@Temporal(javax.persistence.TemporalType.DATE)
private Date createdAt;
@Column(name = "lastLogin")
@Temporal(javax.persistence.TemporalType.DATE)
private Date lastLogin;
@Column(name = "lastNotifiedAt")
@Temporal(javax.persistence.TemporalType.DATE)
private Date lastNotifiedAt;
@OneToMany(fetch = FetchType.LAZY, mappedBy = "id.firstUser") // I would prefer to fill these relations manually
private List<UserFriendRequest> userFriendRequests = Lists.newArrayList();
@Transient
private boolean newUser;
public User() {
this.createdAt = new Date();
this.newUser = false;
}
public User(int userId) {
this();
this.userId = userId;
}
public User(String facebookId, String userName) {
this();
this.facebookId = facebookId;
this.userName = userName;
}
public User(int userId, String facebookId, String userName) {
this(facebookId, userName);
this.userId = userId;
}
public int getUserId() {
return userId;
}
public void setUserId(int userId) {
this.userId = userId;
}
public String getFacebookId() {
return facebookId;
}
public void setFacebookId(String facebookId) {
this.facebookId = facebookId;
}
public boolean isNewUser() {
return newUser;
}
public void setNewUser(boolean newUser) {
this.newUser = newUser;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public Date getCreatedAt() {
return createdAt;
}
public void setCreatedAt(Date createdAt) {
this.createdAt = createdAt;
}
public Date getLastLogin() {
return lastLogin;
}
public void setLastLogin(Date lastLogin) {
this.lastLogin = lastLogin;
}
public Date getLastNotifiedAt() {
return lastNotifiedAt;
}
public void setLastNotifiedAt(Date lastNotifiedAt) {
this.lastNotifiedAt = lastNotifiedAt;
}
public List<UserFriendRequest> getUserFriendRequests() {
return userFriendRequests;
}
public void setUserFriendRequests(List<UserFriendRequest> userFriendRequests) {
this.userFriendRequests = userFriendRequests;
}
@Override
public int hashCode() {
int hash = 7;
hash = 53 * hash + this.userId;
return hash;
}
@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final User other = (User) obj;
if (this.userId != other.userId) {
return false;
}
return true;
}
@Override
public String toString() {
return ToStringBuilder.reflectionToString(this, ToStringStyle.MULTI_LINE_STYLE);
}
}
UserFriendRequests表的表示
@Entity
@Table(name = "UserFriendRequests", catalog = "example_schema")
public class UserFriendRequest implements Serializable {
@Embeddable
public static class UserFriendRequestKey implements Serializable {
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "firstUser", nullable = false)
//Instead of user class I would prefer to keep only the int field. If I would need the user object I would handle it myself.
private User firstUser;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "secondUser", nullable = false)
private User secondUser;
public UserFriendRequestKey() {
}
public UserFriendRequestKey(User firstUser, User secondUser) {
this.firstUser = firstUser;
this.secondUser = secondUser;
}
public User getFirstUser() {
return firstUser;
}
public void setFirstUser(User firstUser) {
this.firstUser = firstUser;
}
public User getSecondUser() {
return secondUser;
}
public void setSecondUser(User secondUser) {
this.secondUser = secondUser;
}
@Override
public int hashCode() {
int hash = 7;
hash = 89 * hash + Objects.hashCode(this.firstUser);
hash = 89 * hash + Objects.hashCode(this.secondUser);
return hash;
}
@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final UserFriendRequestKey other = (UserFriendRequestKey) obj;
if (!Objects.equals(this.firstUser, other.firstUser)) {
return false;
}
if (!Objects.equals(this.secondUser, other.secondUser)) {
return false;
}
return true;
}
@Override
public String toString() {
return ToStringBuilder.reflectionToString(this, ToStringStyle.MULTI_LINE_STYLE);
}
}
@Id
private UserFriendRequestKey id;
@Column(name = "requestedAt", nullable = false)
@Temporal(javax.persistence.TemporalType.DATE)
private Date requestedAt;
public UserFriendRequest() {
}
public UserFriendRequest(UserFriendRequestKey id) {
this.id = id;
this.requestedAt = new Date();
}
public UserFriendRequestKey getId() {
return id;
}
public void setId(UserFriendRequestKey id) {
this.id = id;
}
public Date getRequestedAt() {
return requestedAt;
}
public void setRequestedAt(Date requestedAt) {
this.requestedAt = requestedAt;
}
@Override
public String toString() {
return ToStringBuilder.reflectionToString(this, ToStringStyle.MULTI_LINE_STYLE);
}
}
答案 0 :(得分：1)
Hibernate具有陡峭的学习曲线，但它也具有以下优势：
所以它不是默认的＆＃39;持久性解决方案，因为有数百万个用PHP编写的Web应用程序，没有成功运行的ORM框架。
我认为Hibernate对于缓存，审计，并发可靠性是强制性非功能性需求的企业应用程序更有意义。
答案 1 :(得分：0)
Hibernate用于对象关系映射。你编写HQL查询，如果你正在使用hibernate。我们在不同的方法中使用了hibernate。我告诉你怎么做。
使用hibernate插件，您可以为DB中存在的每个表生成pojo类。因此，无需从您身边编写pojo类以及许多xml相关文件。您的项目中必须有一个persisten.xml文件。