无法读取上传的文件。 [英] Not able to read uploaded file.

查看:61
本文介绍了无法读取上传的文件。的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

点击任意图片,我需要上传所选的新文件。但是我的脚本出错了



 var file1 = $(this).file [0]; 





我的观点:



 @using(@ Html.BeginForm(uploadProductImage ,SellereLogin,FormMethod.Post,new {enctype =multipart / form-data}))
{
< table id =listOfProduct>
< tr>
< th>
@ Html.DisplayNameFor(model => model.ProductId)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image1Id)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image2Id)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image3Id)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image4Id)
< / th>
< th>< / th>
< / tr>

@foreach(模型中的var项目)
{
< tr>
< td>

@ Html.DisplayFor(modelItem => item.ProductId)

< / td>
< td>

@ Html.HiddenFor(m => item.Image1Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />

< / td>
< td>

@ Html.HiddenFor(m => item.Image2Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />

< / td>
< td>

@ Html.HiddenFor(m => item.Image3Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />

< / td>
< td>

@ Html.HiddenFor(m => item.Image4Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />
@ * @ Html.ActionLink(更改,uploadProductImage,新{productId = item.ProductId,imageId = item.Image4Id})* @

< / td>
< / tr>
}

< / table>
< div>
< input type =fileid =file1style =display:none/>
< / div>
}





我的剧本:



<脚本> 
$(function(){
$('。click')。on('click',function(){
$('#file1')。trigger('click') ;
});

$('#file1')。on('change',function(){
debugger;
var file1 = $(this ).file [0];
var formData = new FormData($('form')[0]);

$ .ajax({
type:POST ,
url:SellereLogin / uploadProductImage,
enctype:'multipart / form-data',
data:{
file:file1,
productId:123,
imageId:12345
},
成功:function(){
alert(数据上传:);
}
});
});
});
< / script>

解决方案

(this).file [0];





我的观点:



 @using(@ Html.BeginForm(uploadProductImage,SellereLogin) ,FormMethod.Post,new {enctype =multipart / form-data}))
{
< table id =listOfProduct>
< tr>
< th>
@ Html.DisplayNameFor(model => model.ProductId)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image1Id)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image2Id)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image3Id)
< / th>
< th>
@ Html.DisplayNameFor(model => model.Image4Id)
< / th>
< th>< / th>
< / tr>

@foreach(模型中的var项目)
{
< tr>
< td>

@ Html.DisplayFor(modelItem => item.ProductId)

< / td>
< td>

@ Html.HiddenFor(m => item.Image1Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />

< / td>
< td>

@ Html.HiddenFor(m => item.Image2Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />

< / td>
< td>

@ Html.HiddenFor(m => item.Image3Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />

< / td>
< td>

@ Html.HiddenFor(m => item.Image4Id)
< img src =@ System.Configuration.ConfigurationManager.AppSettings [prodImage]class =click />
@ * @ Html.ActionLink(更改,uploadProductImage,新{productId = item.ProductId,imageId = item.Image4Id})* @

< / td>
< / tr>
}

< / table>
< div>
< input type =fileid =file1style =display:none/>
< / div>
}





我的剧本:



< script> 


(function(){


('。click')。on('click',function() {

On click of any Image I need to upload the new file selected. But I get error in my script

var file1 = $(this).file[0];



My View :

@using (@Html.BeginForm("uploadProductImage", "SellereLogin", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    <table id="listOfProduct">
        <tr>
            <th>
                @Html.DisplayNameFor(model => model.ProductId)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image1Id)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image2Id)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image3Id)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image4Id)
            </th>
            <th></th>
        </tr>

        @foreach (var item in Model)
        {
            <tr>
                <td>

                    @Html.DisplayFor(modelItem => item.ProductId)

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image1Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image2Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image3Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image4Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>
                    @*@Html.ActionLink("Change", "uploadProductImage", new { productId = item.ProductId, imageId = item.Image4Id })*@

                </td>
            </tr>
        }

    </table>
    <div>
        <input type="file" id="file1" style="display: none" />
    </div>
}



My Script :

<script>
       $(function () {
           $('.click').on('click', function () {
               $('#file1').trigger('click');
           });

           $('#file1').on('change', function () {
               debugger;
               var file1 = $(this).file[0];
               var formData = new FormData($('form')[0]);

               $.ajax({
                   type: "POST",
                   url: "SellereLogin/uploadProductImage",
                   enctype: 'multipart/form-data',
                   data: {
                       file: file1,
                       productId: 123,
                       imageId: 12345
                   },
                   success: function () {
                       alert("Data Uploaded: ");
                   }
               });
           });
       });
   </script>

解决方案

(this).file[0];



My View :

@using (@Html.BeginForm("uploadProductImage", "SellereLogin", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    <table id="listOfProduct">
        <tr>
            <th>
                @Html.DisplayNameFor(model => model.ProductId)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image1Id)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image2Id)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image3Id)
            </th>
            <th>
                @Html.DisplayNameFor(model => model.Image4Id)
            </th>
            <th></th>
        </tr>

        @foreach (var item in Model)
        {
            <tr>
                <td>

                    @Html.DisplayFor(modelItem => item.ProductId)

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image1Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image2Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image3Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>

                </td>
                <td>

                    @Html.HiddenFor(m => item.Image4Id)
                    <img src ="@System.Configuration.ConfigurationManager.AppSettings["prodImage"]" class="click"/>
                    @*@Html.ActionLink("Change", "uploadProductImage", new { productId = item.ProductId, imageId = item.Image4Id })*@

                </td>
            </tr>
        }

    </table>
    <div>
        <input type="file" id="file1" style="display: none" />
    </div>
}



My Script :

<script>


(function () {


('.click').on('click', function () {


这篇关于无法读取上传的文件。的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆