在Spring-boot和Thymeleaf中發布具有嵌套List物件的物件

Spring/Thymeleaf 初學者提前抱歉，但我有 2 個物體 Employee 和 MeetingInfo。Employee 與 MeetingInfo 具有 oneToMany 關系，因此基本上一個 Employee 可以有許多 MessageInfo。使用 psvm，我可以使用以下方法將具有多個 MessageInfo 的新員工添加到我的資料庫中：

Employee employee1 = new Employee("Employee 1");
MeetingInfo mInfo1 = new MeetingInfo(LocalDate.of(2021, 1, 1), "First Random message");
MeetingInfo mInfo2 = new MeetingInfo(LocalDate.of(2021, 2, 2), "Second Random message");
MeetingInfo mInfo3 = new MeetingInfo(LocalDate.of(2021, 3, 3), "Third Random message");

employee1.getMeetingInfo().add(mInfo1);
employee1.getMeetingInfo().add(mInfo2);
employee1.getMeetingInfo().add(mInfo3);

employeeRepository.save(employee1);

但是我怎么能用百里香中的表格做到這一點呢？我可以只添加一個員工，但不能添加一個新的 MeetingInfo 物件。當我這樣做時，我得到一個 passException 錯誤。

我的 new_employee.html

<form action="#" th:action="@{/ines/saveEmployee}" th:object="${employee}"
            method="POST">
            <input type="text" th:field="*{name}"
                placeholder="Employee Name" class="form-control mb-4 col-4">
            *** so if I remove between here***
            <input type="date" th:field="*{meetingInfo.meetingDate}"
               placeholder="Message Date" class="form-control mb-4 col-4">

            <input type="text" th:field="*{meetingInfo.message}"
               placeholder="Message" class="form-control mb-4 col-4">
            *** and here***
            *** how can I include a MessageInfo object with a new Employee?***
            <button type="submit" class="btn btn-info col-2">Save Meeting</button>
        </form>

我的控制器

@GetMapping("/showNewEmployeeForm")
    public String showNewEmployeeForm(Model model) {
        Employee employee = new Employee();
        model.addAttribute("employee", employee);
        return "meeting/new_employee.html";
    }

    @PostMapping("/saveEmployee")
    public String saveEmployee(@ModelAttribute("employee") Employee employee) {
        employeeService.saveMessage(employee);
        return "redirect:/ines/employees";
    }

員工

@Entity
@Table(name = "employee")
public class Employee {

    @Id
    @Column(name = "employee_id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long employeeId;

    @Column(nullable = false)
    private String name;

    @OneToMany(cascade = CascadeType.ALL)
    @JoinColumn(name = "employee_id", referencedColumnName = "employee_id")
    private List<MeetingInfo> meetingInfo = new ArrayList<>();
//Constructors, getters and setters

會議資訊

@Entity
@Table(name = "meeting_info")
public class MeetingInfo {

    @Id
    @Column(name = "meeting_id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long meetingId;

    private String message;


    @Column(name = "meeting_date")
    private LocalDate meetingDate;
//Constructors, getters and setters

uj5u.com熱心網友回復：

使用單個請求保存多個物體并不是您通常想要對 Spring Boot 應用程式執行的操作，但是，由于我知道這僅用于練習，您可以通過使用單個 DTO 物件來執行此操作，該物件將保存兩個物體的資訊：

public class EmployeeMeetingDTO {

private String employeeName;

private String meetingMessage;

private LocalDate meetingDate;

}

然后，您的控制器可以從請求中只接受一個 DTO 物體：

@PostMapping("/saveEmployee")
public String saveEmployee(@ModelAttribute("employeeDto") EmployeeMeetingDTO employeeDto) {
    employeeService.saveMessage(employeeDto);
    return "redirect:/ines/employees";
}

您可以在 EmployeeService 類中分別創建這兩個物體。您的 Thymeleaf 表單將如下所示：

<form action="#" th:action="@{/ines/saveEmployee}" th:object="${employeeDto}"
        method="POST">
        <input type="text" th:field="*{employeeName}"
            placeholder="Employee Name" class="form-control mb-4 col-4">
       
        <input type="date" th:field="*{meetingDate}"
           placeholder="Message Date" class="form-control mb-4 col-4">

        <input type="text" th:field="*{meetingMessage}"
           placeholder="Message" class="form-control mb-4 col-4">
        
        <button type="submit" class="btn btn-info col-2">Save Meeting</button>
    </form>

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