MC1*_*123 5 forms spring user-registration thymeleaf spring-boot
我正在为网站注册页面。我了解要创建一个新用户,需要一个id,因此我们有以下字段:
<input type="hidden" th:field="{*id} />
Run Code Online (Sandbox Code Playgroud)
但是,当我转到页面时,我得到了本文标题中提到的错误。
这是有问题的表格:
<form th:action="@{/users/register}" th:object="${user}" class="form-signin" method="POST">
<h2 class="form-signin-heading">Register</h2>
<input type="hidden" th:field="*{id}" />
<label for="inputUsername" class="sr-only">Username*</label>
<input type="text" th:field="*{username}" name="username" id="inputUsername" class="form-control" placeholder="Username" required="required" autofocus="autofocus" />
<label for="inputEmail" class="sr-only">Email Address*</label>
<input type="text" th:field="*{email}" name="email" id="inputEmail" class="form-control" placeholder="Email address" required="required" autofocus="autofocus" />
<label for="inputPassword" class="sr-only">Password</label>
<input type="password" th:field="*{password}" name="password" id="inputPassword" class="form-control" placeholder="Password" required="required" />
<label for="inputConfirmPassword" class="sr-only">Confirm Password</label>
<input type="password" th:field="${confirmPassword}" name="confirmPassword" id="inputConfirmPassword" class="form-control" placeholder="Confirm password" required="required" />
<button class="btn btn-lg btn-primary btn-block" type="submit">Register</button>
</form>
Run Code Online (Sandbox Code Playgroud)
这是我的UserController:
@RequestMapping("/register")
public String registerAction(Model model) {
model.addAttribute("user", new User());
model.addAttribute("confirmPassword", "");
return "views/users/register";
}
@RequestMapping(value="/register", method = RequestMethod.POST)
public String doRegister(User user) {
User savedUser = userService.save(user);
return "redirect:/"; //redirect to homepage
}
Run Code Online (Sandbox Code Playgroud)
和用户实体的第一部分:
@Entity
@Table(name = "users")
public class User {
// Default constructor require by JPA
public User() {}
@Column(name = "id")
@Id @GeneratedValue
private Long id;
public void setId(long id) {
this.id = id;
}
public long getId() {
return id;
}
Run Code Online (Sandbox Code Playgroud)
据我所知,这里没有错,所以我被卡住了。
我正在关注以下示例:https : //github.com/cfaddict/spring-boot-intro
有任何想法吗?
问题在于您声明 id 属性的方式。该字段使用引用类型 Long,该类型为 null。getter 使用原始 long。当 Spring 访问 id 字段时,它会尝试取消装箱空值,从而导致错误。将您的域类更改为
@Entity
@Table(name = "users")
public class User {
// Default constructor required by JPA
public User() {}
@Id
@Column(name = "id")
@GeneratedValue
private Long id;
public void setId(Long id) {
this.id = id;
}
public Long getId() {
return id;
}
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
25197 次 |
| 最近记录: |