ASP.NET - C＃MVC5

Question

基本上我有一個網站，用戶輸入的圖像標題，我想要在網站上顯示它。該圖像的標題。

Atm 我保存到一個文件夾並顯示所有圖像但我不能顯示標題。它只是當我沒有出現<h2>@Html.DisplayFor(m => m.title)</h2>

這樣的事情：但我想要的職位的標題。 https://i.gyazo.com/17828889116a77983f70fd8c8a4c2ebf.png

也許我需要將圖像保存到數據庫？請幫助我。

查看：

@foreach (var image in Model.Images) 
{ 
    <h2>@Html.DisplayFor(m => m.title)</h2> 
    <div class="row">  
     <div class="col-md-8 portfolio-item"> 
      <img class="portrait" src="@Url.Content(image)" alt="Hejsan" /> 
     </div> 
    </div> 

} 

型號

[Table("MemeImages")] 
public class UploadFileModel 
{ 
    [Key] 
    [DatabaseGenerated(System.ComponentModel.DataAnnotations.Schema.DatabaseGeneratedOption.Identity)] 
    public int id { get; set; } 

    public string location { get; set; } 

    public IEnumerable<string> Images { get; set; } 

    public int contentLength { get; set; } 
    public string contentType { get; set; } 

    public string userID { get; set; } 

    [Required] 
    [Display(Name = "Describe your post...")] 
    public string title { get; set; } 

    public void SavetoDatabase(UploadFileModel file) 
    { 
     ApplicationDbContext db = new ApplicationDbContext();      
     db.uploadedFiles.Add(file); 
     db.SaveChanges(); 
    } 
} 

控制器：

public class MemesController : Controller 
{ 

    private string memesDirectory = "~/Content/Images/Memes/"; 

    // GET: Memes/Upload 
    public ActionResult Upload() 
    { 
     var uploadFile = new UploadFileModel(); 
     return View(uploadFile); 
    } 

    // GET: Memes/Hot 
    public ActionResult Hot(UploadFileModel uploadFileModel) 
    { 
     //Select every image on the server memesdirectory and posts it 
     uploadFileModel.Images = Directory.EnumerateFiles(Server.MapPath(memesDirectory)).Select(fn => memesDirectory + Path.GetFileName(fn)); 
     return View(); 
    } 

    // GET: Memes/Trending 
    public ActionResult Trending() 
    { 
     return View(); 
    } 

    // GET: Memes/Fresh 
    public ActionResult Fresh() 
    { 
     return View(); 
    } 

    [HttpPost] 
    public ActionResult Upload(UploadFileModel uploadModel) 
    {   
     if(Request.Files.Count > 0) 
     { 
      var file = Request.Files[0]; 

      if(file != null && file.ContentLength > 0) 
      { 
       //saves image to the server 
       var fileName = Path.GetFileName(file.FileName); 
       var path = Path.Combine(Server.MapPath(memesDirectory), fileName); 
       file.SaveAs(path); 

       //saves image-related data to the database 
       uploadModel.userID = User.Identity.GetUserId(); 
       uploadModel.location = path; 
       uploadModel.contentType = file.ContentType; 
       uploadModel.contentLength = file.ContentLength; 


       //saves to the database 
       uploadModel.SavetoDatabase(uploadModel); 
      } 
     } 
     return RedirectToAction("Index", "Home"); 
    } 
} 

}

Answer 1

uploadFileModel.Images = Directory.EnumerateFiles(Server.MapPath(memesDirectory)).Select(fn => memesDirectory + Path.GetFileName(fn)); 
    return View(); 

可能需要更改爲：

// Add code here to get uploadFileModel from the databases 
    uploadFileModel.Images = Directory.EnumerateFiles(Server.MapPath(memesDirectory)).Select(fn => memesDirectory + Path.GetFileName(fn)); 
    uploadFileModel.title = "something you want"; // this line may not be necessary if getting it from the database did this for you already 

    return View(uploadFileModel);