使用jQuery异步传递Model到控制器方法,并异步返回错误信息

需要通过jquery传递到控制器方法的Model为：

    public class Person

        public string Name { get; set; }

        public int Age { get; set; }

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使用下面这个类的Message属性装载成功或错误信息，并最终转换成json格式返回给视图：

    public class PersonVm

        public string Message { get; set; }

控制器通过遍历ModelState.Keys来显示错误信息。

      public ActionResult Demo()

            return View();

        [HttpPost]

        public ActionResult SavePerson(Person person)

            if (ModelState.IsValid)

                string message = string.Format("用户姓名:{0},年龄：{1}", person.Name, person.Age);

                return Json(new PersonVm() {Message = message});

            else

                string errorMessage = "<div class=\"validation-summary-errors\">发生以下错误：<ul>";

                foreach (var key in ModelState.Keys)

                    var error = ModelState[key].Errors.FirstOrDefault();

                    if (error != null)

                        errorMessage += "<li class=\"field-validation-error\">"+error.ErrorMessage+"</li>";

                errorMessage += "</ul>";

                return Json(new PersonVm() {Message = errorMessage});

视图使用 $.toJSON()方法，把Model转换成json格式传递给控制器方法。把对象转换成json格式的jQuery扩展，在这里。

@model MvcApplication1.Models.Person

@{

    ViewBag.Title = "Demo";

    Layout = "~/Views/Shared/_Layout.cshtml";

@Html.EditorForModel()

<input type="submit" value="保存" id="personCreate"/>

<div>

    <span id="resultMessage"></span>

</div>

@section scripts

    <script src="~/Scripts/json.js"></script>

    <script type="text/javascript">

        $(function() {

            $('#personCreate').click(function() {

                var person = getPerson();

                if (person == null) {

                    alert("用户名不能为空");

                    return;

                var json = $.toJSON(person);

                $.ajax({

                    url: '@Url.Action("SavePerson","Home")',

                    type: 'POST',

                    dataType: 'json',

                    data: json,

                    contentType: 'application/json; charset=utf-8',

                    success: function(data) {

                        var message = data.Message;

                        $("#resultMessage").html(message);

});

});

});

        function getPerson() {

            var name = $('#Name').val();

            var age = $('#Age').val();

            return (name == "") ? null : { Name: name, Age: age };

    </script>

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使用jQuery异步传递Model到控制器方法,并异步返回错误信息

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