java 错误：对于 JPA 中的 GeneratedValue，给定的 id 不能为 null
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Error: The given id must not be null for GeneratedValue in JPA
提问by Anil Bhaskar
My object:
我的对象：
@Entity
@Table(name="user")
public class User {
@Id
@Column(name="uid")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Long id;
//more code
}
When I POSTuserJSONwithout uid, I am getting error as the given id must not be null. Which should not be the case while uidshould be generated by the database. Please point what am I missing.
当我POSTuserJSON没有时uid，我收到错误，因为给定的 id 不能为 null。这不应该是这种情况，而uid应该由数据库生成。请指出我遗漏了什么。
JSON:
JSON：
{
"email": "[email protected]",
"name": "John Doe",
"phone": "98-765-4445"
}
Error:
错误：
{
"timestamp": 1501058952038,
"status": 500,
"error": "Internal Server Error",
"exception": "org.springframework.dao.InvalidDataAccessApiUsageException",
"message": "The given id must not be null!; nested exception is java.lang.IllegalArgumentException: The given id must not be null!",
"path": "/api/user/"
}
回答by Anil Bhaskar
It was my bad, I was calling foo(user.getId())before persisting the Userobject.
Anyways, take aways from this; @GeneratedValue(strategy=GenerationType.IDENTITY)is a correct code and it generates identical ids while persisting. And Longis not a problem. Thanks.
这是我的错，我foo(user.getId())在坚持User对象之前打电话。无论如何，请远离此；@GeneratedValue(strategy=GenerationType.IDENTITY)是一个正确的代码，它在持久化时生成相同的 ID。而且Long不是问题。谢谢。
回答by Mustafa
To generate string uuid's for primary keys (as I assume you are trying to do) you can try the following code:
要为主键生成字符串 uuid（我假设您正在尝试这样做），您可以尝试以下代码：
@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "uuid2")
private String id;
