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Question

我在使用 spring 和 thymeleaf 绑定集合时遇到问题。每次我发送表单时，我的对象集合都设置为 null (User.postions)，我的示例如下：

我的控制器：

@RequestMapping(value = urlFragment + "/add", method = RequestMethod.GET)
public String addPosition(Model model) {

    HashSet<Position> positions = new HashSet<Position>(positionRepository.findByEnabledTrueOrderByNameAsc());

    User employee = new User();

    for (Position position : positions) {
        employee.addPosition(position);
    }

    model.addAttribute("employee", employee);

    return "crud/employee/add";
}

@RequestMapping(value = urlFragment + "/add", method = RequestMethod.POST)
public String processNewEmployee(Model model, @Valid @ModelAttribute("employee") User employee, BindingResult result) {
    String templatePath = "crud/employee/add";

    if (!result.hasErrors()) {
        userRepository.save(employee);
        model.addAttribute("success", true);
    }

    return templatePath;
}

还有我的员工表格：

<form action="#" th:action="@{/panel/employee/add}" th:object="${employee}" method="post">

    <div class="row">
        <div class="col-md-6">
            <label th:text="#{first_name}">First name</label>
            <input class="form-control" type="text" th:field="*{userProfile.firstName}"/>
        </div>
    </div>

    <div class="row">
        <div class="col-md-6">
            <label th:text="#{last_name}">Last name</label>
            <input class="form-control" type="text" th:field="*{userProfile.lastName}"/>
        </div>
    </div>

    <div class="row">
        <div class="col-md-6">
            <label th:text="#{email}">Email</label>
            <input class="form-control" type="text" th:field="*{email}"/>
        </div>
    </div>

    <div class="row">
        <div class="col-md-6">
            <label th:text="#{position}">Position</label>
            <select th:field="*{positions}" class="form-control">
                <option th:each="position : *{positions}"
                        th:value="${position.id}"
                        th:text="${position.name}">Wireframe
                </option>
            </select>
        </div>
    </div>


    <div class="row">
        <div class="col-md-5">
            <div class="checkbox">
                <button type="submit" class="btn btn-success" th:text="#{add_employee}">
                    Add employee
                </button>
            </div>
        </div>
    </div>
</form>

用户实体：

@Entity
@Table(name="`user`")
public class User extends BaseModel {

    @Column(unique = true, nullable = false, length = 45)
    private String email;

    @Column(nullable = false, length = 60)
    private String password;

    @Column
    private String name;

    @Column
    private boolean enabled;

    @ManyToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinTable(name = "user_role",
            joinColumns = {@JoinColumn(name = "user_id", nullable = false)},
            inverseJoinColumns = {@JoinColumn(name = "role_id", nullable = false)}
    )
    private Collection<Role> roles = new HashSet<Role>();

    @ManyToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinTable(name = "user_position",
            joinColumns = {@JoinColumn(name = "user_id", nullable = false)},
            inverseJoinColumns = {@JoinColumn(name = "position_id", nullable = false)}
    )
    private Collection<Position> positions = new HashSet<Position>();

    public User() {
    }

    public User(String email, String password, boolean enabled) {
        this.email = email;
        this.password = password;
        this.enabled = enabled;
    }

    public User(String email, String password, boolean enabled, Set<Role> roles) {
        this.email = email;
        this.password = password;
        this.enabled = enabled;
        this.roles = roles;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public boolean isEnabled() {
        return enabled;
    }

    public void setEnabled(boolean enabled) {
        this.enabled = enabled;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public Collection<Position> getPositions() {
        return positions;
    }

    private void setPositions(Collection<Position> positions) {
        this.positions = positions;
    }

    public boolean addPosition(Position position) {
        return positions.add(position);
    }

    public boolean removePosition(Position position) {
        return positions.remove(position);
    }

    public Collection<Role> getRoles() {
        return roles;
    }

    private void setRoles(Collection<Role> roles) {
        this.roles = roles;
    }

    public boolean addRole(Role role) {
        return roles.add(role);
    }

    public boolean removeRole(Role role) {
        return roles.remove(role);
    }

    @Override
    public String toString() {
        return User.class + " - id: " + getId().toString() + ", email: " + getEmail();
    }
}

我在某处读到我必须创建 equals() 和 hashCode()，所以我在我的 Position Entity 中做了。

public boolean equals(Position position) {
    return this.getId() == position.getId();
}

public int hashCode(){
    return this.getId().hashCode() ;
}

以下是通过 post 方法发送的数据：

这是我的结果：

我的春季版：4.1.6.RELEASE thymeleaf-spring4 版本：2.1.4.RELEASE thymeleaf-layout-dialect 版本：1.2.8

当然，我希望位置是 HashCode，其中一个元素为 id = 2 的对象位置。 你可以帮帮我吗？我做错了什么？

Answer 1

这是因为您使用${position.id} 作为您的期权价值。这意味着 spring 无法计算出 value 中使用的 id 和实际 Position 对象之间的关系。试试 ${position} 来获得你的价值，它应该可以工作：

                    <select th:field="*{positions}" class="form-control">
                        <option th:each="position : *{positions}"
                                th:value="${position}"
                                th:text="${position.name}">Wireframe
                        </option>
                    </select>

（确保您已在 Position 类中实现 hashCode 和 equals）

如果这仍然不起作用，您可能必须为 Position 实现 Formatter，以使转换显式。请参阅此示例thymeleafexamples-selectmultiple。

Answer 2

我遇到了类似的问题，我通过添加 Formatter 类并将 Formatter 添加到 MVC 的配置中：

@Override
protected void addFormatters(FormatterRegistry registry){
    registry.addFormatter(new PositionFormater());
    ...
}

Position 类格式化程序应如下所示：

PositionFormatter：

public class PositionFormatter implements Formatter<Position>{

/** String representing null. */
private static final String NULL_REPRESENTATION = "null";

@Resource
private PositionRepository positionRepository;

public PositionFormatter() {
    super();
}

@Override
public String print(Position position, Locale locale) {
    if(position.equals(NULL_REPRESENTATION)){
        return null;
    }
    try {
        Position newPosition = new Position();
        newPosition.setId(position.getId());
        return newPosition.getId().toString();
    } catch (NumberFormatException e) {
        throw new RuntimeException("Failed to convert `" + position + "` to a valid id");
    }

}

@Override
public Position parse(String text, Locale locale) throws ParseException {
    if (text.equals(NULL_REPRESENTATION)) {
        return null;
    }
    try {
        Long id = Long.parseLong(text);
        Position position = new Position();
        position.setId(id);
        return position;
    } catch (NumberFormatException e) {
        throw new RuntimeException("Failed to convert `" + text + "` to valid Position");
    }
  }
}

就我而言，这两个解决了所有问题。我有几个格式化程序，我所做的只是制作一个并将其添加到配置文件（在我的例子中是 WebMVCConfig）

Check my original post where I resolved this problem

Answer 3

感谢大家回答我的问题。你帮了我很多。不幸的是，我不得不在一件事上不同意你的观点。您向我展示了以下示例：

newPosition.setId(position.getId());

Andrew github 存储库中也有相同的示例。我认为使用 setId() 方法是不好的做法。因此，我将提出我的解决方案，我将等待一些 cmets，然后将其标记为答案。

WebMvcConfig 类

@Configuration
@EnableWebMvc
@ComponentScan(basePackages = "com.smartintranet")
public class WebMvcConfig extends WebMvcConfigurerAdapter {

    @PersistenceContext
    private EntityManager entityManager;

    // (....rest of the methods.......)

    @Override
    public void addFormatters(FormatterRegistry formatterRegistry) {
        formatterRegistry.addFormatter(new PositionFormatter(entityManager));
    }
}

PositionFormatter 类

public class PositionFormatter implements Formatter<Position> {

    private EntityManager entityManager;

    public PositionFormatter(EntityManager entityManager) {
        this.entityManager = entityManager;
    }

    public String print(Position position, Locale locale) {
        if(position.getId() == null){
            return "";
        }

        return position.getId().toString();
    }

    public Position parse(String id, Locale locale) throws ParseException {
        return entityManager.getReference(Position.class, Long.parseLong(id));
    }
}

employeeForm.html

                <div class="col-md-6">
                    <label th:text="#{position}">Position</label>
                    <select th:field="*{position}" class="form-control">
                        <option th:each="position : ${allPositions}"
                                th:value="${position.id}"
                                th:text="${position.name}">Wireframe
                        </option>
                    </select>
                </div>

最后一个，EmployeeController 类

@Controller
public class EmployeeController extends AbstractCrudController {    
    // (...rest of dependency and methods....)

    @Transactional
    @RequestMapping(value = urlFragment + "/create", method = RequestMethod.GET)
    public String createNewEmployee(Model model) {
        prepareEmployeeForm(model);
        return "crud/employee/create";
    }

    @Transactional
    @RequestMapping(value = urlFragment + "/create", method = RequestMethod.POST)
    public String processNewEmployee(Model model, @ModelAttribute("employee") Employee employee, BindingResult result) {
        if (!result.hasErrors()) {
            // Look here it is important line!
            entityManager.merge(employee.getUser());
        }

        prepareEmployeeForm(model);

        return "crud/employee/create";
    }
}

这是我的解决方案。这里有什么不好？我认为那行：

entityManager.merge(employee.getUser());

我不能在这里使用：

userRepository.save(employee.getUser());

因为位置实体是分离的，当我使用保存方法时，它在这种情况下运行em.persist()，所以我手动运行em.merge()。我知道这段代码并不完美，但我认为这个解决方案比使用 setId() 更好。我会感谢建设性的批评者。

再次感谢Andrew 和Blejzer 的帮助，如果没有你，我不会这样做。我已将您的答案标记为有用。