Cisco VPN Client Error "Reason 442 Failed to Enable Virtual Adapter"

Cisco VPN Client Error:
Secure VPN Connection terminated locally by the Client
Reason 442 Failed to Enable Virtual Adapter

In Windows 8/8.1 (32/64 bit ) you may get an error when trying to connect to VPN from Cisco VPN Client saying Reason 442 Failed to Enable Virtual Adapter, to fix this error please make the required changes in the registry as shown in this post.

Note: Make sure you do a backup before do any modification to key values in registry.(In the registry menu, go to File > Export)

1. Open Registry Editor (Windows Key + R Key on keyboard to open "Run" window and type "regedit" to launch registry editor.)
2. Navigate to this path in registry editor HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\CVirtA
3. Right click on DisplayName and select Modify from the options.
4. Based on the operating system you are working, change the values as follows:

Windows 8 32bit (x86)
Change value @oem5.inf,%CVirtA_Desc%;Cisco Systems VPN Adapter to Cisco Systems VPN Adapter

Windows 8 64bit (x64)
Change value @oem5.inf,%CVirtA_Desc%;Cisco Systems VPN Adapter for 64-bit Windows to Cisco Systems VPN Adapter for 64-bit Windows

Note: In this fix you are just removing the part "@oemX.inf,%CVirtA_Desc%" and the modified value should be starting from "Cisco"

Resize and save an image which uploaded using file upload control in c#

Here is the code for controller class.

public class FileUploadController : Controller
{
    //
    // GET: /FileUpload/

    public ActionResult Index()
    {
        return View();
    }
    public ActionResult FileUpload()
    {
        return View();
    }
    [HttpPost]
    public ActionResult Index(HttpPostedFileBase file)
    {
        WebImage img = new WebImage(file.InputStream);
        if (img.Width > 1000)
            img.Resize(1000, 1000);
        img.Save("path");
        return View();
    }

    [AcceptVerbs(HttpVerbs.Post)]
    public ActionResult FileUpload(HttpPostedFileBase uploadFile)
    {
        if (uploadFile.ContentLength > 0)
        {
            string relativePath = "~/img/" + Path.GetFileName(uploadFile.FileName);
            string physicalPath = Server.MapPath(relativePath);


            FileUploadModel.ResizeAndSave(relativePath, uploadFile.FileName, uploadFile.InputStream, uploadFile.ContentLength, true);

            return View((object)relativePath);
        }
        return View();
    }
}

Here is the code for model class

public class FileUploadModel
{
    [Required]
    public HttpPostedFileWrapper ImageUploaded { get; set; }

    public static void ResizeAndSave(string savePath, string fileName, Stream imageBuffer, int maxSideSize, bool makeItSquare)
    {
        int newWidth;
        int newHeight;
        Image image = Image.FromStream(imageBuffer);
        int oldWidth = image.Width;
        int oldHeight = image.Height;
        Bitmap newImage;
        if (makeItSquare)
        {
            int smallerSide = oldWidth >= oldHeight ? oldHeight : oldWidth;
            double coeficient = maxSideSize / (double)smallerSide;
            newWidth = Convert.ToInt32(coeficient * oldWidth);
            newHeight = Convert.ToInt32(coeficient * oldHeight);
            Bitmap tempImage = new Bitmap(image, newWidth, newHeight);
            int cropX = (newWidth - maxSideSize) / 2;
            int cropY = (newHeight - maxSideSize) / 2;
            newImage = new Bitmap(maxSideSize, maxSideSize);
            Graphics tempGraphic = Graphics.FromImage(newImage);
            tempGraphic.SmoothingMode = SmoothingMode.AntiAlias;
            tempGraphic.InterpolationMode = InterpolationMode.HighQualityBicubic;
            tempGraphic.PixelOffsetMode = PixelOffsetMode.HighQuality;
            tempGraphic.DrawImage(tempImage, new Rectangle(0, 0, maxSideSize, maxSideSize), cropX, cropY, maxSideSize, maxSideSize, GraphicsUnit.Pixel);
        }
        else
        {
            int maxSide = oldWidth >= oldHeight ? oldWidth : oldHeight;

            if (maxSide > maxSideSize)
            {
                double coeficient = maxSideSize / (double)maxSide;
                newWidth = Convert.ToInt32(coeficient * oldWidth);
                newHeight = Convert.ToInt32(coeficient * oldHeight);
            }
            else
            {
                newWidth = oldWidth;
                newHeight = oldHeight;
            }
            newImage = new Bitmap(image, newWidth, newHeight);
        }
        newImage.Save(savePath + fileName + ".jpg", ImageFormat.Jpeg);
        image.Dispose();
        newImage.Dispose();
    }
}

Also this code supports the crop, flip, watermark operation etc.

Upload a image and display on same page in asp.net mvc4

Once you save the uploaded file on the server from your controller action you could pass back the url to this file to the view so that it can be displayed in an <img> tag:

public class FileUploadController : Controller
{
    public ActionResult Index()
    {
        return View();
    }
    public ActionResult FileUpload()
    {
        return View();
    }
    [AcceptVerbs(HttpVerbs.Post)]
    public ActionResult FileUpload(HttpPostedFileBase uploadFile)
    {
        if (uploadFile.ContentLength > 0)
        {
            string relativePath = "~/img/" + Path.GetFileName(uploadFile.FileName);
            string physicalPath = Server.MapPath(relativePath);
            uploadFile.SaveAs(physicalPath);
            return View((object)relativePath);
        }
        return View();
    }

}

Then make your view strongly typed and add an <img> tag that will display the image if the model is not empty:

@model string
<h2>FileUpload</h2>

@using (Html.BeginForm("FileUpload", "FileUpload", FormMethod.Post, new { enctype = "multipart/form-data" })
{
    <input name="uploadFile" type="file" />
    <input type="submit" value="Upload File" />
}

@if (!string.IsNullOrEmpty(Model))
{
    <img src="@Url.Content(Model)" alt="" />
}