87

Anyone know of any step by step tutorials on how to upload/display images from a database using Entity Framework? I've checked out code snippets, but I'm still not clear on how it works. I have no code, because aside from writing an upload form, I'm lost. Any (and I mean any) help is greatly appreciated.

On a sidenote, why don't any books cover this topic? I have both Pro ASP.NET MVC 4 and Professional MVC4, and they make no mention of it.

4

3 回答 3

144

看看以下

@using (Html.BeginForm("FileUpload", "Home", FormMethod.Post, 
                            new { enctype = "multipart/form-data" }))
{  
    <label for="file">Upload Image:</label> 
    <input type="file" name="file" id="file" style="width: 100%;" /> 
    <input type="submit" value="Upload" class="submit" /> 
}

您的控制器应该具有可以接受的操作方法HttpPostedFileBase；

 public ActionResult FileUpload(HttpPostedFileBase file)
    {
        if (file != null)
        {
            string pic = System.IO.Path.GetFileName(file.FileName);
            string path = System.IO.Path.Combine(
                                   Server.MapPath("~/images/profile"), pic); 
            // file is uploaded
            file.SaveAs(path);

            // save the image path path to the database or you can send image 
            // directly to database
            // in-case if you want to store byte[] ie. for DB
            using (MemoryStream ms = new MemoryStream()) 
            {
                 file.InputStream.CopyTo(ms);
                 byte[] array = ms.GetBuffer();
            }

        }
        // after successfully uploading redirect the user
        return RedirectToAction("actionname", "controller name");
    }

更新 1

如果你想使用 jQuery 异步上传文件，那么试试这篇文章。

处理服务器端（用于多次上传）的代码是；

 try
    {
        HttpFileCollection hfc = HttpContext.Current.Request.Files;
        string path = "/content/files/contact/";

        for (int i = 0; i < hfc.Count; i++)
        {
            HttpPostedFile hpf = hfc[i];
            if (hpf.ContentLength > 0)
            {
                string fileName = "";
                if (Request.Browser.Browser == "IE")
                {
                    fileName = Path.GetFileName(hpf.FileName);
                }
                else
                {
                    fileName = hpf.FileName;
                }
                string fullPathWithFileName = path + fileName;
                hpf.SaveAs(Server.MapPath(fullPathWithFileName));
            }
        }

    }
    catch (Exception ex)
    {
        throw ex;
    }

此控件还返回图像名称（在 javascript 回调中），然后您可以使用它在 DOM 中显示图像。

更新 2

或者，您可以尝试MVC 4 中的异步文件上传。

于 2013-04-27T19:35:32.473 回答
46

这是一个简短的教程：

模型：

namespace ImageUploadApp.Models
{
    using System;
    using System.Collections.Generic;

    public partial class Image
    {
        public int ID { get; set; }
        public string ImagePath { get; set; }
    }
}

看法：

  1. 创造：

    @model ImageUploadApp.Models.Image
@{
    ViewBag.Title = "Create";
}
<h2>Create</h2>
@using (Html.BeginForm("Create", "Image", null, FormMethod.Post, 
                              new { enctype = "multipart/form-data" })) {
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)
    <fieldset>
        <legend>Image</legend>
        <div class="editor-label">
            @Html.LabelFor(model => model.ImagePath)
        </div>
        <div class="editor-field">
            <input id="ImagePath" title="Upload a product image" 
                                  type="file" name="file" />
        </div>
        <p><input type="submit" value="Create" /></p>
    </fieldset>
}
<div>
    @Html.ActionLink("Back to List", "Index")
</div>
@section Scripts {
    @Scripts.Render("~/bundles/jqueryval")
}

  2. 索引（用于显示）：

    @model IEnumerable<ImageUploadApp.Models.Image>

@{
    ViewBag.Title = "Index";
}

<h2>Index</h2>

<p>
    @Html.ActionLink("Create New", "Create")
</p>
<table>
    <tr>
        <th>
            @Html.DisplayNameFor(model => model.ImagePath)
        </th>
    </tr>

@foreach (var item in Model) {
    <tr>
        <td>
            @Html.DisplayFor(modelItem => item.ImagePath)
        </td>
        <td>
            @Html.ActionLink("Edit", "Edit", new { id=item.ID }) |
            @Html.ActionLink("Details", "Details", new { id=item.ID }) |
            @Ajax.ActionLink("Delete", "Delete", new {id = item.ID} })
        </td>
    </tr>
}

</table>

  3. 控制器（创建）

    public ActionResult Create(Image img, HttpPostedFileBase file)
{
    if (ModelState.IsValid)
    {
        if (file != null)
        {
            file.SaveAs(HttpContext.Server.MapPath("~/Images/") 
                                                  + file.FileName);
            img.ImagePath = file.FileName;
        }  
        db.Image.Add(img);
        db.SaveChanges();
        return RedirectToAction("Index");
    }
    return View(img);
}

希望这会有所帮助:)

于 2013-06-17T08:33:11.233 回答
-16 
        <input type="file" id="picfile" name="picf" />
       <input type="text" id="txtName" style="width: 144px;" />
 $("#btncatsave").click(function () {
var Name = $("#txtName").val();
var formData = new FormData();
var totalFiles = document.getElementById("picfile").files.length;

                    var file = document.getElementById("picfile").files[0];
                    formData.append("FileUpload", file);
                    formData.append("Name", Name);

$.ajax({
                    type: "POST",
                    url: '/Category_Subcategory/Save_Category',
                    data: formData,
                    dataType: 'json',
                    contentType: false,
                    processData: false,
                    success: function (msg) {

                                 alert(msg);

                    },
                    error: function (error) {
                        alert("errror");
                    }
                });

});

 [HttpPost]
    public ActionResult Save_Category()
    {
      string Name=Request.Form[1]; 
      if (Request.Files.Count > 0)
        {
            HttpPostedFileBase file = Request.Files[0];
         }


    }
于 2014-07-28T05:44:26.197 回答