自动递增在使用Hibernate向PostgreSQL添加项时不会发生。

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英文:

Auto incrementing does not occur when adding an item in PostgreSQL using Hibernate

问题

我在PostgreSQL中创建了一个包含个人数据的表，并将列 "id" 标记为类型 'SERIAL' 和 '主键'，'非空'。

以下是创建脚本：

  1. CREATE TABLE people (
  2.   id serial NOT NULL,
  3.   name varchar(15),
  4.   surname varchar(25),
  5.   department varchar(20),
  6.   salary int,
  7.   PRIMARY KEY (id)
  8. );

然后，我为我的表创建了一个实体类：

  1. package hibernate_test.entity;
  2. import jakarta.persistence.*;
  4. @Entity
  5. @Table(name = "people")
  6. public class Employee {
  7.     @Id
  8.     @Column(name = "id")
  9.     private int id;
  10.     @Column(name = "name")
  11.     private String name;
  12.     @Column(name = "surname")
  13.     private String surname;
  14.     @Column(name = "department")
  15.     private String department;
  16.     @Column(name = "salary")
  17.     private int salary;
  19.     public Employee() {
  20.     }
  22.     public Employee(String name, String surname, String department, int salary) {
  23.         this.name = name;
  24.         this.surname = surname;
  25.         this.department = department;
  26.         this.salary = salary;
  27.     }
  29.     @Override
  30.     public String toString() {
  31.         return "Employee{" +
  32.                 "id=" + id +
  33.                 ", name='" + name + '\'' +
  34.                 ", surname='" + surname + '\'' +
  35.                 ", department='" + department + '\'' +
  36.                 ", salary=" + salary +
  37.                 '}';
  38.     }
  40.     public int getId() {
  41.         return id;
  42.     }
  44.     public void setId(int id) {
  45.         this.id = id;
  46.     }
  48.     public String getName() {
  49.         return name;
  50.     }
  52.     public void setName(String name) {
  53.         this.name = name;
  54.     }
  56.     public String getSurname() {
  57.         return surname;
  58.     }
  60.     public void setSurname(String surname) {
  61.         this.surname = surname;
  62.     }
  64.     public String getDepartment() {
  65.         return department;
  66.     }
  68.     public void setDepartment(String department) {
  69.         this.department = department;
  70.     }
  72.     public int getSalary() {
  73.         return salary;
  74.     }
  76.     public void setSalary(int salary) {
  77.         this.salary = salary;
  78.     }
  79. }

接下来，我编写了一个程序来向我的表中添加记录：

  1. package hibernate_test;
  3. import hibernate_test.entity.Employee;
  4. import org.hibernate.Session;
  5. import org.hibernate.SessionFactory;
  6. import org.hibernate.cfg.Configuration;
  8. public class Test1 {
  9.     public static void main(String[] args) {
  10.         SessionFactory factory = new Configuration()
  11.                 .configure("hibernate.cfg.xml")
  12.                 .addAnnotatedClass(Employee.class)
  13.                 .buildSessionFactory();
  15.         Session session = factory.getCurrentSession();
  16.         Employee emp = new Employee("Jack", "Jack", "IT", 5000);
  18.         try {
  19.             session.beginTransaction();
  20.             session.save(emp);
  21.             session.getTransaction().commit();
  23.         } finally {
  24.             factory.close();
  25.         }
  26.     }
  27. }

当我运行此代码时，第一条记录被添加到表中，但id是0。我认为id应该从1开始，因为我在id中指定了类型 'SERIAL'。当我尝试添加另一项时，出现错误：

错误表明不能有两条id为0的记录，但我没有直接将id值传递给构造函数。Hibernate应该自己递增id，并且理想情况下从1开始。

如果可以的话，我会很高兴帮助你找出我哪里出错了。

英文:

I created a table in PostgeSQL with data for a person and marked the column "id" with the type 'SERIAL' as 'primary key', 'not null'

CREATE Script:

  1. CREATE TABLE people (
  2. id serial NOT NULL,
  3. name varchar(15),
  4. surname varchar(25),
  5. department varchar(20),
  6. salary int,
  7. PRIMARY KEY (id)
  8. );

Then I created an Entity class for my table

  1. package hibernate_test.entity;
  2. import jakarta.persistence.*;
  3. @Entity
  4. @Table(name = "people")
  5. public class Employee {
  6. @Id
  7. @Column(name = "id")
  8. private int id;
  9. @Column(name = "name")
  10. private String name;
  11. @Column(name = "surname")
  12. private String surname;
  13. @Column(name = "department")
  14. private String department;
  15. @Column(name = "salary")
  16. private int salary;
  17. public Employee() {
  18. }
  19. public Employee(String name, String surname, String department, int salary) {
  20. this.name = name;
  21. this.surname = surname;
  22. this.department = department;
  23. this.salary = salary;
  24. }
  25. @Override
  26. public String toString() {
  27. return "Employee{" +
  28. "id=" + id +
  29. ", name='" + name + '\'' +
  30. ", surname='" + surname + '\'' +
  31. ", department='" + department + '\'' +
  32. ", salary=" + salary +
  33. '}';
  34. }
  35. public int getId() {
  36. return id;
  37. }
  38. public void setId(int id) {
  39. this.id = id;
  40. }
  41. public String getName() {
  42. return name;
  43. }
  44. public void setName(String name) {
  45. this.name = name;
  46. }
  47. public String getSurname() {
  48. return surname;
  49. }
  50. public void setSurname(String surname) {
  51. this.surname = surname;
  52. }
  53. public String getDepartment() {
  54. return department;
  55. }
  56. public void setDepartment(String department) {
  57. this.department = department;
  58. }
  59. public int getSalary() {
  60. return salary;
  61. }
  62. public void setSalary(int salary) {
  63. this.salary = salary;
  64. }
  65. }

Next, I wrote a program that adds a record to my table

  1. package hibernate_test;
  2. import hibernate_test.entity.Employee;
  3. import org.hibernate.Session;
  4. import org.hibernate.SessionFactory;
  5. import org.hibernate.cfg.Configuration;
  6. public class Test1 {
  7. public static void main(String[] args) {
  8. SessionFactory factory = new Configuration()
  9. .configure("hibernate.cfg.xml")
  10. .addAnnotatedClass(Employee.class)
  11. .buildSessionFactory();
  12. Session session = factory.getCurrentSession();
  13. Employee emp = new Employee("Jack", "Jack", "IT", 5000);
  14. try {
  15. session.beginTransaction();
  16. session.save(emp);
  17. session.getTransaction().commit();
  18. }finally {
  19. factory.close();
  20. }
  21. }
  22. }

When I run this code, the first record is added to the table, but the id is at index 0. I think the index should start at 1 because I specified the type 'SERIAL' in the id.
When I try to add another item, an error occurs:

  1. 09:09:05.911 [main] INFO org.hibernate.Version -- HHH000412: Hibernate ORM core version 6.2.6.Final
  2. 09:09:05.916 [main] INFO org.hibernate.cfg.Environment -- HHH000406: Using bytecode reflection optimizer
  3. 09:09:06.366 [main] WARN org.hibernate.orm.connections.pooling -- HHH10001002: Using built-in connection pool (not intended for production use)
  4. 09:09:06.369 [main] INFO org.hibernate.orm.connections.pooling -- HHH10001005: Loaded JDBC driver class: org.postgresql.Driver
  5. 09:09:06.369 [main] INFO org.hibernate.orm.connections.pooling -- HHH10001012: Connecting with JDBC URL [jdbc:postgresql://127.0.0.1:5432/learnhbn]
  6. 09:09:06.369 [main] INFO org.hibernate.orm.connections.pooling -- HHH10001001: Connection properties: {password=****, user=postgres}
  7. 09:09:06.369 [main] INFO org.hibernate.orm.connections.pooling -- HHH10001003: Autocommit mode: false
  8. 09:09:06.373 [main] INFO org.hibernate.orm.connections.pooling -- HHH10001115: Connection pool size: 20 (min=1)
  9. 09:09:06.796 [main] INFO org.hibernate.bytecode.internal.BytecodeProviderInitiator -- HHH000021: Bytecode provider name : bytebuddy
  10. Hibernate: insert into people (department,name,salary,surname,id) values (?,?,?,?,?)
  11. 09:09:07.648 [main] WARN org.hibernate.engine.jdbc.spi.SqlExceptionHelper -- SQL Error: 0, SQLState: 23505
  12. 09:09:07.648 [main] ERROR org.hibernate.engine.jdbc.spi.SqlExceptionHelper -- ПОМИЛКА: повторювані значення ключа порушують обмеження унікальності "people_pkey"
  13. Подробности: Ключ (id)=(0) вже існує.
  14. 09:09:07.656 [main] INFO org.hibernate.orm.connections.pooling -- HHH10001008: Cleaning up connection pool [jdbc:postgresql://127.0.0.1:5432/learnhbn]
  15. Exception in thread "main" org.hibernate.exception.ConstraintViolationException: could not execute statement [ПОМИЛКА: повторювані значення ключа порушують обмеження унікальності "people_pkey"
  16. Подробности: Ключ (id)=(0) вже існує.] [insert into people (department,name,salary,surname,id) values (?,?,?,?,?)]
  17. at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:95)
  18. at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:56)
  19. at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:108)
  20. at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:278)
  21. at org.hibernate.engine.jdbc.mutation.internal.AbstractMutationExecutor.performNonBatchedMutation(AbstractMutationExecutor.java:108)
  22. at org.hibernate.engine.jdbc.mutation.internal.MutationExecutorSingleNonBatched.performNonBatchedOperations(MutationExecutorSingleNonBatched.java:40)
  23. at org.hibernate.engine.jdbc.mutation.internal.AbstractMutationExecutor.execute(AbstractMutationExecutor.java:53)
  24. at org.hibernate.persister.entity.mutation.InsertCoordinator.doStaticInserts(InsertCoordinator.java:170)
  25. at org.hibernate.persister.entity.mutation.InsertCoordinator.coordinateInsert(InsertCoordinator.java:112)
  26. at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2761)
  27. at org.hibernate.action.internal.EntityInsertAction.execute(EntityInsertAction.java:102)
  28. at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:633)
  29. at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:502)
  30. at org.hibernate.event.internal.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:358)
  31. at org.hibernate.event.internal.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:39)
  32. at org.hibernate.event.service.internal.EventListenerGroupImpl.fireEventOnEachListener(EventListenerGroupImpl.java:127)
  33. at org.hibernate.internal.SessionImpl.doFlush(SessionImpl.java:1412)
  34. at org.hibernate.internal.SessionImpl.managedFlush(SessionImpl.java:485)
  35. at org.hibernate.internal.SessionImpl.flushBeforeTransactionCompletion(SessionImpl.java:2301)
  36. at org.hibernate.internal.SessionImpl.beforeTransactionCompletion(SessionImpl.java:1966)
  37. at org.hibernate.engine.jdbc.internal.JdbcCoordinatorImpl.beforeTransactionCompletion(JdbcCoordinatorImpl.java:439)
  38. at org.hibernate.resource.transaction.backend.jdbc.internal.JdbcResourceLocalTransactionCoordinatorImpl.beforeCompletionCallback(JdbcResourceLocalTransactionCoordinatorImpl.java:169)
  39. at org.hibernate.resource.transaction.backend.jdbc.internal.JdbcResourceLocalTransactionCoordinatorImpl$TransactionDriverControlImpl.commit(JdbcResourceLocalTransactionCoordinatorImpl.java:267)
  40. at org.hibernate.engine.transaction.internal.TransactionImpl.commit(TransactionImpl.java:101)
  41. at hibernate_test.Test1.main(Test1.java:21)
  42. Caused by: org.postgresql.util.PSQLException: ПОМИЛКА: повторювані значення ключа порушують обмеження унікальності "people_pkey"
  43. Подробности: Ключ (id)=(0) вже існує.
  44. at org.postgresql.core.v3.QueryExecutorImpl.receiveErrorResponse(QueryExecutorImpl.java:2713)
  45. at org.postgresql.core.v3.QueryExecutorImpl.processResults(QueryExecutorImpl.java:2401)
  46. at org.postgresql.core.v3.QueryExecutorImpl.execute(QueryExecutorImpl.java:368)
  47. at org.postgresql.jdbc.PgStatement.executeInternal(PgStatement.java:498)
  48. at org.postgresql.jdbc.PgStatement.execute(PgStatement.java:415)
  49. at org.postgresql.jdbc.PgPreparedStatement.executeWithFlags(PgPreparedStatement.java:190)
  50. at org.postgresql.jdbc.PgPreparedStatement.executeUpdate(PgPreparedStatement.java:152)
  51. at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:275)
  52. ... 21 more
  53. Process finished with exit code 1

The error indicates that there can't be two records with index 0, but I don't pass any id value directly to the constructor. Hibernate should increment the id to one itself, and ideally start with one

I will be glad if you can help me figure out where I am making a mistake

答案1

得分: 1

答复来自 M. Deinum：

您没有在 @Id 字段上指定策略。应该在那里使用 @GeneratedValue 来告诉Hibernate数据库会生成标识符。

英文:

Answer from M. Deinum:

You didn't specify a strategy on the @Id field. There should be an @GeneratedValue on there to indicate to Hibernate that the database will generate the identifier.

huangapple
  • 本文由 发表于 2023年7月18日 14:14:54
  • 转载请务必保留本文链接：https://go.coder-hub.com/76709961.html
  • auto-increment
  • hibernate
  • java
  • postgresql
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