所以我目前有一个包含用户信息的数据库,定义用户的是user_id。
然后我有一个名为 token 的表,它有一个 token_id 和 user_id 作为主键和其余信息,使它成为一个一对多的数据库。
@Entity
@Table(name = "user")
public class User implements Serializable {
@Id
@Column(name = "user_id")
private long userId;
//Other variables and getters and setters
@OneToMany(orphanRemoval = true, mappedBy = "user", fetch = FetchType.LAZY, cascade = CascadeType.ALL)
@Access(AccessType.PROPERTY) //I need this as is since I have other things in the setter
private List<Token> tokens = new ArrayList<>();
public List<Token> getTokens() {
return tokens;
}
public void setTokens(List<Token> tokens) {
this.tokens = tokens;
}
}
在这段代码片段之后,我得到了 token 的类
public class Token implements Serializable{
@Id
private long tokenId;
@Id
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "user_id")
private User user;
@Column(nullable = false)
private String token;
@Access(AccessType.PROPERTY)
private Instant lastUsed;
@Column(nullable = false)
private Instant dateCreated;
@Transient
private boolean expired;
//Getters and setters go here
//Static methods and generating the token
private static String generateToken(){
Random random = new Random();
byte[] randomString = new byte[256];
random.nextBytes(randomString);
return Base64.encodeBase64String(randomString);
}
public static Token generateUserToken(User user){
Token token = new Token();
token.setTokenId(new Random().nextLong());
token.setUser(user);
token.setDateCreated(Instant.now());
token.setToken(generateToken());
return token;
}
//Static methods and generating the token
}
现在出于某种原因,每当 User user
未标记为 @Id 时,它就可以工作(即使在数据库中它是主键)。
任何帮助;
应用程序属性:
spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.MySQL57InnoDBDialect
spring.jpa.show-sql=true
logging.level.org.hibernate.type=TRACE
SQL 输出:
Hibernate: insert into tokens (date_created, last_used, token, user_id, token_id) values (?, ?, ?, ?, ?)
binding parameter [1] as [TIMESTAMP] - [2018-05-14T08:29:00.719764Z]
binding parameter [2] as [TIMESTAMP] - [null] //This is okay to be null this is last_used
binding parameter [3] as [VARCHAR] - [<Token too long to write in question>] //Actual data type is LONGTEXT
binding parameter [4] as [BIGINT] - [null] //this is a problem (user_id should not be - should be a long numebr such as: 5531405900210671089)
binding parameter [5] as [BIGINT] - [0] //this is a problem (token_id should be a long number such as: -8824825685434914749)
SQL Error: 1048, SQLState: 23000
Column 'user_id' cannot be null
最佳答案
您不需要在 Token 中使用 @Id 注释 user_id:您看到这有效。 同样在数据库中,定义被 tokednId 的表 token 的主键就足够了。 当然user_id必须设置为不能为空的外键。
关于java - JPA 复合外主键,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50326307/