Spring MVC的嵌套模型验证

Question

我有两个型号：User,Project

public class Project{ 
    private int id; 
    @NotEmpty(message="Project Name can not be empty") 
    private String name; 
    private User manager; 
    private User operator; 
    //getter/setter omitted 
} 

public class User{ 
    private int id; 
    private String name; 
    //omit other properties and getter/setter 
} 

现在，当我创建一个新的项目，我将提交下列参数ProjectController：

projects?name=jhon&manager.id=1&operator.id=2...

然后我将创建一个新的Project对象并将其插入到数据库中。

但是，我必须验证经理和运营商的ID是否有效，也就是说我会验证，如果在用户表中有匹配的ID。

所以我想知道如何实现这种验证？

---

UPDATE1：使用验证

这是创建新项目的形式：

<sf:form method="${project.id==0?'post':'put'}" commandName="project" action="${context}${action}"> 
    Manager:<sf:input path="manager.id" /> <sf:errors path="manager.id" /> <br /> 
    Operator:<sf:input path="operator.id" /> <sf:errors path="operator.id" /> <br /> 
    Name:<sf:input path="name" /> <sf:errors path="name" /> <br /> 
    <input type="submit" value="Submit" /> 
</sf:form> 


@Override 
public void validate(Object obj, Errors errors) { 
    User user = (User) obj; 
    int id=user.getId(); 
    User u=userDao.query(id); 
    if(u==null){ 
     errors.rejectValue("id", "user does not exist!"); 
    } 
} 

看来，这验证工作。

但是，错误消息无法显示在窗体中。

然后虽然调试我检查结果对象，我发现这一点：

org.springframework.validation.BeanPropertyBindingResult: 2 errors 
Field error in object 'project' on field 'id': rejected value [0]; codes [user does not exist!.project.id,user does not exist!.id,user does not exist!.int,user does not exist!]; arguments []; default message [null] 
Field error in object 'project' on field 'id': rejected value [0]; codes [user does not exist!.project.id,user does not exist!.id,user does not exist!.int,user does not exist!]; arguments []; default message [null] 

看来，结果确实有错误，但它的路径是project.id而在我的形式，它是project.manager.id

如何修理？

Answer 1

前的最后一段做 here

这是一个可能的解决方案。

创建类下面：

... 
import org.springframework.validation.Validator; 
... 

@Component 
public class ProjectValidator implements Validator { 

    @Override 
    public boolean supports(Class<?> clazz) { 
     return Project.class.equals(clazz); 
    } 

    @Override 
    public void validate(Object target, Errors errors) { 
     Project project = (Project) target; 

     /* Do your checks here */ 
     ... 

     if (managerIdDoesNotMatch) { 
      errors.rejectValue("manager.id", "your_error_code"); 
     } 

     ... 

     if (operatorIdDoesNotMatch) { 
      errors.rejectValue("operator.id", "your_error_code"); 
     } 

     ... 
    } 
} 

而在你的控制器做这样的事情：

... 
public class ProjectController { 

    @Autowired 
    ProjectValidator projectValidator; 

    ... 

    @RequestMapping(...) 
    public String yourCreateMethod(..., @ModelAttribute @Valid Project project, BindingResult result) { 
     projectValidator.validate(project, result);   

     if (result.hasErrors()){ 
      // do something 
     } 
     else { 
      // do something else 
     } 
    } 

} 

这应该让你开始。你可以以不同的方式实例化/设置验证器，拥有一个用户子验证器，但你可以得到一般想法。

参考文献：

• Latest Spring validation documentation
• This Stack Overflow post about how to perform validation

Answer 2

在你的控制器，你可以添加自定义的验证：

@InitBinder 
protected void initBinder(WebDataBinder binder) { 
    binder.setValidator(new ProjectValidator()); 
} 

在此验证，您可以检查User对象或委托给一个UserValidator，为第6.3

Answer 3

我做了什么杰罗姆Dalbert建议，另外增加了一个自定义的BeanValidator委派验证到JSR 303实施的实际工作。

前缀用于表示表单中属性的路径。

@Component 
public class BeanValidator implements org.springframework.validation.Validator, InitializingBean {  

private Validator validator; 
public void afterPropertiesSet() throws Exception { 
    ValidatorFactory validatorFactory = Validation.buildDefaultValidatorFactory(); 
    validator = validatorFactory.usingContext().getValidator(); 
} 

public boolean supports(Class clazz) { 
    return true; 
} 

public void validate(Object target, Errors errors, String prefix) { 
    Set<ConstraintViolation<Object>> constraintViolations = validator.validate(target); 
    for (ConstraintViolation<Object> constraintViolation : constraintViolations) { 
    String propertyPath = constraintViolation.getPropertyPath().toString(); 
    String message = constraintViolation.getMessage(); 
    errors.rejectValue(prefix + "." + propertyPath, "", message); 
    } 
} 

public void validate(Object target, Errors errors) { 
    validate(target, errors, ""); 
} 
} 

在这里我如何用它在UserValidator：

@Component 
public class UserValidator implements Validator { 

@Autowired 
BeanValidator beanValidator; 

@Override 
public boolean supports(Class<?> clazz) { 
    return User.class.equals(clazz); 
} 

@Override 
public void validate(Object target, Errors errors) { 
    User user = (User) target; 
    beanValidator.validate(user.getAddress(), errors, "address"); 
} 
} 

参考文献：

• http://blog.trifork.com/2009/08/04/bean-validation-integrating-jsr-303-with-spring/

Answer 4

其实你需要做的是

添加 @Valid

private User manager; 
private User operator; 

这样

@Valid 
private User manager; 
@Valid 
private User operator;