你可以使用文件返回一个文件，如视图，内容等

 public ActionResult PrintDocInfo(string Attachment)
            {
                string test = Attachment;
                if (test != string.Empty || test != "" || test != null)
                {
                    string filename = Attachment.Split('\\').Last();
                    string filepath = Attachment;
                    byte[] filedata = System.IO.File.ReadAllBytes(Attachment);
                    string contentType = MimeMapping.GetMimeMapping(Attachment);

                    System.Net.Mime.ContentDisposition cd = new System.Net.Mime.ContentDisposition
                    {
                        FileName = filename,
                        Inline = true,
                    };

                    Response.AppendHeader("Content-Disposition", cd.ToString());

                    return File(filedata, contentType);          
                }
                else { return Content("<h3> Patient Clinical Document Not Uploaded</h3>"); }

            }

从Core 3.2下的字节[]，你可以使用:

public ActionResult Img(int? id) {
    MemoryStream ms = new MemoryStream(GetBytes(id));
    return new FileStreamResult(ms, "image/png");
}

解决方案1:从图像URL在视图中呈现图像

你可以创建自己的扩展方法:

public static MvcHtmlString Image(this HtmlHelper helper,string imageUrl)
{
   string tag = "<img src='{0}'/>";
   tag = string.Format(tag,imageUrl);
   return MvcHtmlString.Create(tag);
}

然后这样使用它:

@Html.Image(@Model.ImagePath);

解决方案2:从数据库中渲染图像

创建一个返回图像数据的控制器方法，如下所示

public sealed class ImageController : Controller
{
  public ActionResult View(string id)
  {
    var image = _images.LoadImage(id); //Pull image from the database.
    if (image == null) 
      return HttpNotFound();
    return File(image.Data, image.Mime);
  }
}

并在视图中使用它:

@ { Html.RenderAction("View","Image",new {id=@Model.ImageId})}

要在任何HTML中使用此actionresult渲染的图像，请使用

<img src="http://something.com/image/view?id={imageid}>

为什么不简单一点，使用波浪号~操作符呢?

public FileResult TopBanner() {
  return File("~/Content/images/topbanner.png", "image/png");
}

使用MVC的发布版本，下面是我所做的:

[AcceptVerbs(HttpVerbs.Get)]
[OutputCache(CacheProfile = "CustomerImages")]
public FileResult Show(int customerId, string imageName)
{
    var path = string.Concat(ConfigData.ImagesDirectory, customerId, "\\", imageName);
    return new FileStreamResult(new FileStream(path, FileMode.Open), "image/jpeg");
}

显然，我在这里有一些关于路径构造的应用程序特定的东西，但返回FileStreamResult很好，很简单。

我做了一些性能测试，针对您每天对图像的调用(绕过控制器)，平均值之间的差异仅为3毫秒(控制器的平均值为68ms，非控制器的平均值为65ms)。

我尝试了答案中提到的其他一些方法，性能的影响要大得多……一些解决方案的响应高达非控制器的6倍(其他控制器平均340ms，非控制器65ms)。

稍微解释一下迪兰的回应:

有三个类实现了FileResult类:

System.Web.Mvc.FileResult
      System.Web.Mvc.FileContentResult
      System.Web.Mvc.FilePathResult
      System.Web.Mvc.FileStreamResult

它们都是不言自明的:

For file path downloads where the file exists on disk, use FilePathResult - this is the easiest way and avoids you having to use Streams. For byte[] arrays (akin to Response.BinaryWrite), use FileContentResult. For byte[] arrays where you want the file to download (content-disposition: attachment), use FileStreamResult in a similar way to below, but with a MemoryStream and using GetBuffer(). For Streams use FileStreamResult. It's called a FileStreamResult but it takes a Stream so I'd guess it works with a MemoryStream.

下面是一个使用内容处理技术的例子(未测试):

    [AcceptVerbs(HttpVerbs.Post)]
    public ActionResult GetFile()
    {
        // No need to dispose the stream, MVC does it for you
        string path = Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "App_Data", "myimage.png");
        FileStream stream = new FileStream(path, FileMode.Open);
        FileStreamResult result = new FileStreamResult(stream, "image/png");
        result.FileDownloadName = "image.png";
        return result;
    }