当我从 html 表单传递值时，为什么我在 post api 期间收到 'Column cannot be null' 错误

Question

当我尝试通过运行我的应用程序对此进行测试时，导航到 admin.html 页面。我在表单中填写了名字、姓氏和电子邮件的值。单击提交按钮时，出现 'Column email cannot be null' 的错误。为了简洁起见，我排除了诸如 getter、setter、contructor 等代码。这是我的 admin.html 页面，我在其中有一个表单，我使用该表单将 post 值转换为 api，其中这些值用于创建员工对象

<form role="form" action="api/employees/create" method="post">
    <div class="form-group">
        <label for="firstName">First Name</label>
        <input type="text" class="form-control" id="firstName" placeholder="Enter first name">
    </div>
    <div class="form-group">
        <label for="lastName">Last Name</label>
        <input type="text" class="form-control" id="lastName" placeholder="Enter last name">
    </div>
    <div class="form-group">
        <label for="email">Email</label>
        <input type="text" class="form-control" id="email" placeholder="Enter email">
    </div>
    <button type="submit" class="btn btn-success btn-block">Create</button>
</form>

这是我在 EmployeeAPI.java class 中的 POST 方法，我在其中处理 post 并使用从表单传入的值创建一个对象并尝试保留这个新对象

@POST
@Path("create")
@Consumes(MediaType.APPLICATION_FORM_URLENCODED)
@Produces(MediaType.TEXT_HTML)
public Response createEmployee(@FormParam(value = "firstName") String firstName,
                               @FormParam(value = "lastName") String lastName,
                               @FormParam(value = "email") String email) {
    SessionFactory factory = HibernateUtil.getSessionFactory();
    Session session = factory.getCurrentSession();
    URI location;
    try{
        session.getTransaction().begin();

        Employee newEmployee = new Employee();
        newEmployee.setFirstName(firstName);
        newEmployee.setLastName(lastName);
        newEmployee.setEmail(email);

        session.persist(newEmployee);
        session.getTransaction().commit();
        session.close();

        location = new URI("http://localhost:8080/index.html");
        return Response.temporaryRedirect(location).build();
    } catch (Exception e) {
        session.getTransaction().rollback();
        e.printStackTrace();
    }
    return null;
}

这是我的 Employee.java 模型 class - 我有一个只包含 firstName、lastName 和 email 的员工构造函数，以及一个用于所有值的构造函数。

@XmlRootElement
@Entity
public class Employee {

@Id
@GeneratedValue
private int id;

@Expose
@Column(nullable = false)
private String firstName;

@Expose
@Column(nullable = false)
private String lastName;

@Expose
@Column(nullable = false, unique = true)
private String email;

这是我在服务器端看到的错误

Hibernate: insert into Employee (availability_id, email, firstName, isAdmin, isManager, isMentee, isMentor, lastName, mentorDuration, topic_name, id) values (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
2019-03-04 16:07:41 WARN  SqlExceptionHelper:129 - SQL Error: 1048, SQLState: 23000
2019-03-04 16:07:41 ERROR SqlExceptionHelper:131 - Column 'email' cannot be null
2019-03-04 16:07:41 INFO  AbstractBatchImpl:193 - HHH000010: On release of batch it still contained JDBC statements
2019-03-04 16:07:41 ERROR ExceptionMapperStandardImpl:39 - HHH000346: Error during managed flush [org.hibernate.exception.ConstraintViolationException: could not execute statement]

Answer 1

在调试 EmployeeAPI 期间，前端的值变为 null

当我从 html 表单传递值时，为什么我在 post api 期间收到 'Column cannot be null' 错误

Why am I getting an error for 'Column cannot be null' during my post api when I am passing a value from a html form

java

hibernate

jax-rs

jakarta-ee