在MVC 4中的同一弹出对话框中显示模型错误 [英] Display model error on same popup dialog in MVC 4

问题描述

我正在开发具有Index.cshtml的应用程序（带搜索条件和搜索按钮的主要页面），_PartialView.cshtml（显示带有更新/删除链接的记录的部分视图），Edit.cshtml（用于更新的视图）。



点击更新链接后，打开弹出模式对话框，其中包含我应用的编辑视图（Edit.cshtml）必需& StringLength属性到某些字段。如果某些属性未满足，我想在同一个弹出窗口中显示它们。但是在控制器动作方法中，我正在返回View，因此模态弹出消失了&编辑视图在div中打开。



我的记录在partialView（_PartialView.cshtml）中显示为网格，更新链接和javascript函数为

I am developing application which has Index.cshtml (main pg with searching criteria & Search button), _PartialView.cshtml (partial view displaying records with update/delete links), Edit.cshtml (View for updating).

After clicking Update link, popup modal dialog is opened containing Edit view (Edit.cshtml) I applied Required & StringLength attribute to some fields. If some of attributes not fulfilled, i want to display them in same popup. But in controller action method, I am returning View so modal popup disappears & Edit view is opened in div.

My records are displayed as grid in partialView (_PartialView.cshtml) with Update link and javascript function as

<a href="javascript:void(0);" class="anchorDetail" onclick="editFunc(@item.SrNo)" ">Update</a>

function editFunc(id) {
    var TeamDetailPostBackURL = '/Home/Edit';
    var $buttonClicked = $(this);
    var options = { "backdrop": "static", keyboard: true };

    $.ajax({
        type: "GET",
        url: TeamDetailPostBackURL,
        cache: false,
        contentType: "application/json; charset=utf-8",
        data: { "SrNo": id },
        datatype: "json",
        success: function (data) {
            $('#myModalContent').html(data);
            $('#myModal').css('display', 'block');
            $("#myModal").dialog({
                title: 'Edit Data',
                modal: true,
                bgiframe: true,
                show: 'slide',
                hide: 'slide',
                width: 750
            });
            $('#myModal').dialog("show");
        },

        error: function () {
            alert("Dynamic content load failed.");
        }
    });
}

我的Edit.cshtml：

My Edit.cshtml :

@using (Ajax.BeginForm("Edit", "Home", new AjaxOptions { UpdateTargetId = "divPartial", HttpMethod = "POST", InsertionMode = InsertionMode.Replace, OnBegin = "CallEditBegin", OnSuccess = "CallUnblock" }))
 {
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)   
    
    <fieldset>
        
        <div class="editor-label" style="display:none" >
            @Html.LabelFor(model => model.SrNo)
        </div>
        <div class="editor-field" style="display:none">
            @Html.EditorFor(model => model.SrNo)
            @Html.ValidationMessageFor(model => model.SrNo)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.UserName)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.UserName)
            @Html.ValidationMessageFor(model => model.UserName)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.ReaderName)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.ReaderName)
            @Html.ValidationMessageFor(model => model.ReaderName)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.UploadDate)
        </div>
        <div class="editor-field">
            @Html.TextBoxFor(model => model.UploadDate, new { @readonly = "readonly" })
            @Html.ValidationMessageFor(model => model.UploadDate)
        </div>

        <p>
            <input type="submit" value="Save" />
        </p>
    </fieldset>
 }

我的HomeController.cs

My HomeController.cs

[HttpPost]
  public ActionResult Edit(UploadData uploadData)
  {
      if (!ModelState.IsValid)
      {

          return View(uploadData);
      }
       else 
      {
          // Processing Logic here

           string strResult = uploadDataBusinessLayer.InsertUpdateUploadData(uploadData, "Update");
            if (strResult == "Success")
            {   // Do return Partial View
            }
            else
            {   // How to display exception error in same popup ???
            }
      }
   }

现在需要提交一些文件，而ReaderName的最小长度为6个字符。如果任何条件未满足

Now some filed are required and ReaderName has restiction of min length of 6 characters. If any of the condition not fulfilled

if (!ModelState.IsValid)
{
 
          return View(uploadData);
}

此块将被执行。



但接下来发生了什么，我的弹出对话框消失了& div将在主pg上打开显示错误消息。

我想在控件前弹出窗口显示它们。



我很困惑如果出现模型错误该怎么办。

请给出一些建议，我缺少什么.....



我的尝试：



搜索谷歌但无法找到任何合适的信息.......

this block will be executed.

But what happened next, my popup dialog disappears & div will be open on main pg displaying error message.
I want to show them on popup itself in front of control.

I am confusing what to do in case of model errors.
Please give some advice, what I am missing.....

What I have tried:

Searched on google but cant find any suitable information.......

推荐答案

buttonClicked =

buttonClicked =

（此）;
var options = { backdrop： static，键盘： true };

(this); var options = { "backdrop": "static", keyboard: true };

.ajax（{
type： GET，
url：TeamDetailPostBackURL，
cache： false ，
contentType： application / json; charset = utf-8，
data：{ SrNo：id}，
数据类型： json，
成功： function （数据）{

.ajax({ type: "GET", url: TeamDetailPostBackURL, cache: false, contentType: "application/json; charset=utf-8", data: { "SrNo": id }, datatype: "json", success: function (data) {

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