如何手动验证具有属性的模型？

Question

我有一个叫做User属性的类Name

public class User
{
    [Required]
    public string Name { get; set; }
}

我想验证它,如果有任何错误添加到控制器ModelState或实例化另一个模型状态...

[HttpPost]
public ActionResult NewUser(UserViewModel userVM)
{
    User u = new User();
    u.Name = null;

    /* something */

    // assume userVM is valid
    // I want the following to be false because `user.Name` is null
    if (ModelState.IsValid)
    {
        TempData["NewUserCreated"] = "New user created sucessfully";

        return RedirectToAction("Index");
    }

    return View();
}

这些属性适用于UserViewModel,但我想知道如何在不将其发布到操作的情况下验证类.

我怎么能做到这一点？

Answer 1

您可以使用Validator来完成此任务.

var context = new ValidationContext(u, serviceProvider: null, items: null);
var validationResults = new List<ValidationResult>();

bool isValid = Validator.TryValidateObject(u, context, validationResults, true);

Answer 2

我在Stack Overflow Documentation中做了一个条目,解释了如何执行此操作:

验证上下文

任何验证都需要一个上下文来提供有关正在验证的内容的一些信息.这可以包括各种信息,例如要验证的对象,一些属性,错误消息中显示的名称等.

ValidationContext vc = new ValidationContext(objectToValidate); // The simplest form of validation context. It contains only a reference to the object being validated.

创建上下文后,有多种方法可以进行验证.

验证对象及其所有属性

ICollection<ValidationResult> results = new List<ValidationResult>(); // Will contain the results of the validation
bool isValid = Validator.TryValidateObject(objectToValidate, vc, results, true); // Validates the object and its properties using the previously created context.
// The variable isValid will be true if everything is valid
// The results variable contains the results of the validation

验证对象的属性

ICollection<ValidationResult> results = new List<ValidationResult>(); // Will contain the results of the validation
bool isValid = Validator.TryValidatePropery(objectToValidate.PropertyToValidate, vc, results, true); // Validates the property using the previously created context.
// The variable isValid will be true if everything is valid
// The results variable contains the results of the validation

和更多

要了解有关手动验证的更多信息,请

• ValidationContext类文档
• 验证器类文档

Answer 3

我编写了一个包装器，以使其使用起来不那么笨重。

用法：

var response = SimpleValidator.Validate(model);

var isValid = response.IsValid;
var messages = response.Results; 

或者，如果您只关心检查有效性，那就更严格了：

var isValid = SimpleValidator.IsModelValid(model);

完整来源：

using System;
using System.Collections.Generic;
using System.ComponentModel.DataAnnotations;

namespace Ether.Validation
{
    public static class SimpleValidator
    {
        /// <summary>
        /// Validate the model and return a response, which includes any validation messages and an IsValid bit.
        /// </summary>
        public static ValidationResponse Validate(object model)
        {
            var results = new List<ValidationResult>();
            var context = new ValidationContext(model);

            var isValid = Validator.TryValidateObject(model, context, results, true);
         
            return new ValidationResponse()
            {
                IsValid = isValid,
                Results = results
            };
        }

        /// <summary>
        /// Validate the model and return a bit indicating whether the model is valid or not.
        /// </summary>
        public static bool IsModelValid(object model)
        {
            var response = Validate(model);

            return response.IsValid;
        }
    }

    public class ValidationResponse
    {
        public List<ValidationResult> Results { get; set; }
        public bool IsValid { get; set; }

        public ValidationResponse()
        {
            Results = new List<ValidationResult>();
            IsValid = false;
        }
    }
}

或者在这个要点：https://gist.github.com/kinetiq/faed1e3b2da4cca922896d1f7cdcc79b

Answer 4

由于问题是专门询问 ASP.NET MVC，因此您可以使用TryValidateObjectinsideController操作。

您想要的方法重载是 TryValidateModel(Object)

验证指定的模型实例。

如果模型验证成功，则返回 true；否则为假。

您修改后的源代码

[HttpPost]
public ActionResult NewUser(UserViewModel userVM)
{
    User u = new User();
    u.Name = null;

    if (this.TryValidateObject(u))
    {
        TempData["NewUserCreated"] = "New user created sucessfully";
        return RedirectToAction("Index");
    }

    return View();
}