Question

您好我正在尝试在Mysql中插入一条记录，并在Spring启动JPA中获取插入用户的ID。我得到了以下错误。我已经看过这样的几个问题，但是他们为这些输出提供了USE JDBC模板的答案。是不是可以这样做？或者这与其他一些问题有关。

控制器

@RequestMapping(value = "/signup", method = RequestMethod.POST)
 public String createsignup(@RequestParam String name,@RequestParam String email,@RequestParam String password, ModelMap model) {
     int userid = 0;
     User user  = new User(name,email);
     userrepository.save(user);
     userid = user.getId();
     Authentication auth = new Authentication(email,password,userid);
     authrepository.save(auth);
     model.put("remember_token", auth.getRemember_token());
     return "redirect:/profile";
 }

模型

  @Entity

公共类用户{

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
private String name;
private String email;
private String location;

public User() {

}
}

错误

      018-01-31 23:45:32.890  INFO 2512 --- [nio-8080-exec-1]  o.s.web.servlet.DispatcherServlet        : FrameworkServlet 'dispatcherServlet': initialization completed in 7 ms
      Hibernate: insert into user (email, location, name) values (?, ?, ?)
      Hibernate: select next_val as id_val from hibernate_sequence for update
2018-01-31 23:45:41.783 ERROR 2512 --- [nio-8080-exec-2]  o.hibernate.id.enhanced.TableStructure   : could not read a hi value

    com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table  'blog.hibernate_sequence' doesn't exist
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) ~[na:1.8.0_45]
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62) ~[na:1.8.0_45]

更新

应用程序属性

 spring.mvc.view.prefix=/WEB-INF/jsp/
 spring.mvc.view.suffix=.jsp
 logging.level.org.springframework.web=INFO

 spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.MySQL5Dialect
 spring.jpa.show-sql = true
 spring.jpa.hibernate.ddl-auto=none
 spring.datasource.url=jdbc:mysql://localhost:3306/blog
 spring.datasource.username=pranava
 spring.datasource.password=**********

来自服务器日志的消息

2018-02-01 22:21:29.691  INFO 5648 --- [  restartedMain] org.hibernate.Version                    : HHH000412: Hibernate Core {5.2.10.Final}
2018-02-01 22:21:29.693  INFO 5648 --- [  restartedMain] org.hibernate.cfg.Environment            : HHH000206: hibernate.properties not found
o.hibernate.annotations.common.Version   : HCANN000001: Hibernate Commons Annotations {5.0.1.Final}

 2018-02-01 22:21:29.928  INFO 5648 --- [  restartedMain]               org.hibernate.dialect.Dialect            : HHH000400: Using dialect: org.hibernate.dialect.MySQL5Dialect

2018-02

Answer 1

经过大量搜索并尝试我发现User Table是Postgresql中的默认表。这就是为什么不允许的原因。我已经更改了表名并且它有效。有没有其他方法可以重复使用它？。

Answer 2

确保为hibernate和MySQL指定了正确的方言。似乎无法正确设置标识列的值。

例如，如果您使用Spring Boot，请检查您的应用程序属性，例如：

spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.MySQL5Dialect

或者，如果您使用没有Spring Boot的Spring MVC，那么您有：

hibernateJpa.setDatabasePlatform("org.hibernate.dialect.MySQL5Dialect");

在LocalContainerEntityManagerFactoryBean配置中。

在spring boot JPA中插入数据并获取ID

2 个答案: