当我从html表单传递值时，为什么在发布api时会出现“列不能为空”错误

Question

当我尝试通过运行应用程序进行测试时，请导航至admin.html页面。我在表格中填写名字，姓氏和电子邮件的值。 单击提交按钮后，“列电子邮件不能为空”出现错误。 为了简洁起见，我已经排除了诸如吸气剂，设置器，构造函数等代码。 这是我的admin.html页面，其中有一个表单，用于将值发布到我的api，在其中，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类中的POST方法，我在其中处理该帖子，并使用从表单传入的值创建一个对象，并尝试保留该新对象

@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模型类-我为一个雇员构造了一个只有firstName，lastName和email的构造函数，并且所有值都为一个。

@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的过程中，前端的值为空