如何使用 Spring MVC JSR-303 验证器将不同模型 class 验证为一种形式

Question

我在 jsp 页面中有如下表格：

<springForm:form action="${addAction}" name="frm" method="post" commandName="employee">
    <table>
        <tr>
            <td><label>First Name</label>   </td>
            <td><springForm:input path="firstName" /></td>
            <td><springForm:errors path="firstName" cssClass="error" /></td>
        </tr>
        <tr>
            <td><label>Last Name</label></td>
            <td><springForm:input path="lastName" /></td>
            <td><springForm:errors path="lastName" cssClass="error" /></td>
        </tr>
        <tr>
            <td><label>Email</label>    </td>
            <td><springForm:input path="email" /></td>
            <td><springForm:errors path="email" cssClass="error" /></td>
        </tr>
        <tr>
            <td><label>Street</label>   </td>
            <td><springForm:input path="employeeDetail.street" /></td>
            <td><springForm:errors path="employeeDetail.street" cssClass="error" /></td>
        </tr>
        <tr>
            <td><label>City</label> </td>
            <td><springForm:input path="employeeDetail.city" /></td>
            <td><springForm:errors path="employeeDetail.city" cssClass="error" /></td>
        </tr>
        <tr>
            <td><label>State</label>    </td>
            <td><springForm:input path="employeeDetail.state" /></td>
            <td><springForm:errors path="employeeDetail.state" cssClass="error" /></td>
        </tr>
        <tr>
            <td><label>Country</label>  </td>
            <td><springForm:input path="employeeDetail.country" /></td>
            <td><springForm:errors path="employeeDetail.country" cssClass="error" /></td>
        </tr>
        <tr>
            <td colspan="2" style="text-align: center;">
                <input type="submit" value="Add Info" />
            </td>
        </tr>
    </table>
</springForm:form>

在此 jsp 中，名字、姓氏和电子邮件属于模型 class，即员工。最后所有属性，如 Street、City 等都属于 EmployeeDetails 模型 class。

public class Employee {

private Long empId;

@Size(min=2, max=30)
private String firstName;

@Size(min=2, max=30)
private String lastName;

@NotEmpty @Email
private String email;

@DateTimeFormat(pattern="MM/dd/yyyy")
@NotNull @Past
private Date doj;

@Phone
private String phone;

private EmployeeDetail employeeDetail;

Getter/Setter
}

员工详情：

public class EmployeeDetail {

@Size(min=2, max=30)
private String street;

@Size(min=2, max=30)
private String city;

@Size(min=2, max=30)
private String state;

@Size(min=2, max=30)
private String country;
//Getter/Setter
}

我正在单个表单中对两个模型执行验证。我的 Cotroller 映射是：

@RequestMapping(value = "/saveEmpInfo", method = RequestMethod.POST)
public String saveEmployee(@Valid Employee employee,
        @Valid EmployeeDetail employeeDetail, BindingResult bindingResult) {
    employee.setEmployeeDetail(employeeDetail);
    if (bindingResult.hasErrors()) {
        System.out.println("Returning addInfo.jsp page");
        return "addInfo";
    }
    //future logic
}

但是收到 Bad Request 没有用。

Answer 1

您应该将属于 EmployeeDetail 的属性名称从例如name="street" 至 name="employeeDetail.street"

另外，你也应该在Employee参数后面添加一个BindingResult对象，这样框架可以存储与Employee相关的绑定错误，当前的BindingResult将只适用于EmployeeDetail 实例，所以

saveEmployee(@Valid Employee employee, BindingResult bindingResultEmployee,
        @Valid EmployeeDetail employeeDetail, BindingResult bindingResult)

没有它你会得到一个错误的请求，有了它，请求实际上会到达方法，如果设置为调试级别，日志将显示确切的绑定错误

Answer 2

要完成这项工作，您应该在 Employee.class 中使用 @Valid 注释 employeeDetails，如下所示：

@Valid
private EmployeeDetail employeeDetail;

现在在控制器中你应该像这样使用 Employee 对象：

public String saveEmployee(@Valid Employee employee, BindingResult bindingResult) {

如何使用 Spring MVC JSR-303 验证器将不同模型 class 验证为一种形式

How to Validate different model class into one form using Spring MVC JSR-303 Validator

java

spring

spring-mvc

bean-validation

hibernate-validator