3 回答

TA貢獻(xiàn)2041條經(jīng)驗(yàn) 獲得超4個(gè)贊
客戶端
使用該jQuery.validate庫應(yīng)該非常簡單。
在Web.config文件中指定以下設(shè)置:
<appSettings>
<add key="ClientValidationEnabled" value="true"/>
<add key="UnobtrusiveJavaScriptEnabled" value="true"/>
</appSettings>
建立視圖時(shí),您將定義以下內(nèi)容:
@Html.LabelFor(Model => Model.EditPostViewModel.Title, true)
@Html.TextBoxFor(Model => Model.EditPostViewModel.Title,
new { @class = "tb1", @Style = "width:400px;" })
@Html.ValidationMessageFor(Model => Model.EditPostViewModel.Title)
注意:這些需要在表單元素中定義
然后,您需要包括以下庫:
<script src='@Url.Content("~/Scripts/jquery.validate.js")' type='text/javascript'></script>
<script src='@Url.Content("~/Scripts/jquery.validate.unobtrusive.js")' type='text/javascript'></script>
這應(yīng)該能夠使您進(jìn)行客戶端驗(yàn)證
資源資源
http://msdn.microsoft.com/zh-cn/vs2010trainingcourse_aspnetmvccustomvalidation_topic5.aspx
服務(wù)器端
注意:這僅用于jQuery.validation庫頂部的其他服務(wù)器端驗(yàn)證
也許這樣的事情可能會(huì)有所幫助:
[ValidateAjax]
public JsonResult Edit(EditPostViewModel data)
{
//Save data
return Json(new { Success = true } );
}
當(dāng)ValidateAjax一個(gè)屬性定義為:
public class ValidateAjaxAttribute : ActionFilterAttribute
{
public override void OnActionExecuting(ActionExecutingContext filterContext)
{
if (!filterContext.HttpContext.Request.IsAjaxRequest())
return;
var modelState = filterContext.Controller.ViewData.ModelState;
if (!modelState.IsValid)
{
var errorModel =
from x in modelState.Keys
where modelState[x].Errors.Count > 0
select new
{
key = x,
errors = modelState[x].Errors.
Select(y => y.ErrorMessage).
ToArray()
};
filterContext.Result = new JsonResult()
{
Data = errorModel
};
filterContext.HttpContext.Response.StatusCode =
(int) HttpStatusCode.BadRequest;
}
}
}
這樣做是返回一個(gè)JSON對(duì)象,該對(duì)象指定所有模型錯(cuò)誤。
示例響應(yīng)將是
[{
"key":"Name",
"errors":["The Name field is required."]
},
{
"key":"Description",
"errors":["The Description field is required."]
}]
這將返回到您的錯(cuò)誤處理回調(diào)$.ajax調(diào)用
您可以遍歷返回的數(shù)據(jù)以根據(jù)返回的鍵根據(jù)需要設(shè)置錯(cuò)誤消息(我認(rèn)為類似的東西$('input[name="' + err.key + '"]')會(huì)找到您的輸入元素

TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超7個(gè)贊
這是一個(gè)相當(dāng)簡單的解決方案:
在控制器中,我們返回如下錯(cuò)誤:
if (!ModelState.IsValid)
{
return Json(new { success = false, errors = ModelState.Values.SelectMany(x => x.Errors).Select(x => x.ErrorMessage).ToList() }, JsonRequestBehavior.AllowGet);
}
以下是一些客戶端腳本:
function displayValidationErrors(errors)
{
var $ul = $('div.validation-summary-valid.text-danger > ul');
$ul.empty();
$.each(errors, function (idx, errorMessage) {
$ul.append('<li>' + errorMessage + '</li>');
});
}
這就是我們通過ajax處理它的方式:
$.ajax({
cache: false,
async: true,
type: "POST",
url: form.attr('action'),
data: form.serialize(),
success: function (data) {
var isSuccessful = (data['success']);
if (isSuccessful) {
$('#partial-container-steps').html(data['view']);
initializePage();
}
else {
var errors = data['errors'];
displayValidationErrors(errors);
}
}
});
另外,我通過以下方式通過ajax渲染部分視圖:
var view = this.RenderRazorViewToString(partialUrl, viewModel);
return Json(new { success = true, view }, JsonRequestBehavior.AllowGet);
RenderRazorViewToString方法:
public string RenderRazorViewToString(string viewName, object model)
{
ViewData.Model = model;
using (var sw = new StringWriter())
{
var viewResult = ViewEngines.Engines.FindPartialView(ControllerContext,
viewName);
var viewContext = new ViewContext(ControllerContext, viewResult.View,
ViewData, TempData, sw);
viewResult.View.Render(viewContext, sw);
viewResult.ViewEngine.ReleaseView(ControllerContext, viewResult.View);
return sw.GetStringBuilder().ToString();
}
}
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