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How to save image in database and display it into Views in MVC 4?

I have a table like followin:

CREATE TABLE [dbo].[tblA]
(
    [Id] INT NOT NULL PRIMARY KEY IDENTITY, 
    [fname] NVARCHAR(50) NULL, 
    [lname] NVARCHAR(50) NULL, 
    [imageUrl] NVARCHAR(50) NULL
)

I want use file upload to upload file. How to save image in database and display it into Views?

like image 317
Samiey Mehdi Avatar asked Dec 29 '13 12:12

Samiey Mehdi


1 Answers

  1. Creat an "Images" folder in Solution explorer.
  2. Create an ADO.NET Entity Data Model (in this example is "Database1Entities")
  3. Home Controller:

    using System;
    using System.Collections.Generic;
    using System.Linq;
    using System.Web;
    using System.Web.Mvc;
    
    namespace test2.Controllers
    {
    public class HomeController : Controller
    {
    public ActionResult Index()
    {
        return View();
    }
    
    public ActionResult FileUpload(HttpPostedFileBase file)
    {
    
        if (file != null)
        {
            Database1Entities db = new Database1Entities();
            string ImageName = System.IO.Path.GetFileName(file.FileName);
            string physicalPath =Server.MapPath("~/images/"+ ImageName);
    
            // save image in folder
            file.SaveAs(physicalPath);
    
            //save new record in database
            tblA newRecord = new tblA();
            newRecord.fname = Request.Form["fname"];
            newRecord.lname = Request.Form["lname"];
            newRecord.imageUrl = ImageName;
            db.tblAs.Add(newRecord);
            db.SaveChanges();
    
        }
        //Display records
        return RedirectToAction("../home/Display/");
    }
    
    public ActionResult Display()
    {
        return View();
    }
    }
    }
    
  4. Index View

    @{
       ViewBag.Title = "Index";
    }
    
    @using (Html.BeginForm("FileUpload", "Home", FormMethod.Post,
                        new { enctype = "multipart/form-data" }))
    {
    <div>
    First name<br />
    @Html.TextBox("fname") <br />
    Last name<br />
    @Html.TextBox("lname") <br />
    Image<br />
    <input type="file" name="file" id="file" style="width: 100%;" /> <br />
    <input type="submit" value="Upload" class="submit" />
    </div>    
    }
    
  5. Display View

    @{
        ViewBag.Title = "Display";
    }
    
    @{
       test2.Database1Entities db = new test2.Database1Entities();
    }
    @using (Html.BeginForm())
    {
    <table border="1">
    <thead>
        <tr>
            <th>Image</th>
            <th>First name</th>
            <th>Last name</th>
        </tr>
    </thead>
    <tbody>
        @foreach (var item in db.tblAs)
        {
            <tr>
                <td><img src="~/images/@item.imageUrl" width="100" height="100" /></td>
                <td>@item.fname</td>
                <td>@item.lname</td>
            </tr>
        }
    </tbody>
    </table>
    }
    

Output:

/Home/

enter image description here

/home/Display/

enter image description here

like image 85
Samiey Mehdi Avatar answered Oct 16 '22 09:10

Samiey Mehdi