mvc upload file with model - second parameter posted file is null

Question

I have a simple model with 1 string property which I render on a simple view.

the view looks like the following:

@using (Html.BeginForm(\"UploadFile\         


        

Answer 1

Why not add the uploaded files to your model like this:

public class UploadFileModel 
{
    public UploadFileModel()
    {
        Files = new List<HttpPostedFileBase>();
    }

    public List<HttpPostedFileBase> Files { get; set; }
    public string FirstName { get; set; }
    // Rest of model details
}

Then change your view to this:

@using (Html.BeginForm("UploadFile", "Home", FormMethod.Post, new { encType="multipart/form-data" }))
{
    @Html.TextBoxFor(m => m.FirstName)
    <br /><br />

    @Html.TextBoxFor(m => m.Files, new { type = "file", name = "Files" })<br /><br />
    <input type="submit" value="submit me" name="submitme" id="submitme" />
}

Then your files will be posted back as follows:

public ActionResult UploadFile(UploadFileModel model)
{
    var file = model.Files[0];
    return View(model);
}

Answer 2

Alternatively, (if acceptable) remove the [Required] validation annotation for your file from your model and check for the file in your Controller action, and add an error if not found:

[HttpPost]
[ValidateAntiForgeryToken]
public async Task<ActionResult> ActionWithFileUpload(ViewModel viewModel)
{
    if (ModelState.IsValid)
    {
        if (Request.Files.Count > 0)
        {
            var postedFile = Request.Files[0];
            if (postedFile != null && postedFile.ContentLength > 0)
            {
                string imagesPath = HttpContext.Server.MapPath("~/Content/Images"); // Or file save folder, etc.
                string extension = Path.GetExtension(postedFile.FileName);
                string newFileName = $"NewFile{extension}";
                string saveToPath = Path.Combine(imagesPath, newFileName);
                postedFile.SaveAs(saveToPath);
            }
        }
        else
        {
            ModelState.AddModelError(string.Empty, "File not selected.");
        }
    }

    return RedirectToAction("Index"); // Or return view, etc.
}

Answer 3

Change your name file to fileUpload and enctype it's work

@using (Html.BeginForm("UploadFile", "Home", FormMethod.Post, new { enctype="multipart/form-data" }))
{
    @Html.TextBoxFor(m => m.FirstName)
    <br /><br />

    <input type="file" name="fileUpload" /><br /><br />
    <input type="submit" value="submit me" name="submitme" id="submitme" />
}

[HttpPost]
public ActionResult UploadFile(UploadFileModel model, HttpPostedFileBase fileUpload)
{
   // DO Stuff
   return View(model);
}

Answer 4

For dealing with a single file input you can define a HttpPostedFileBase property within the ViewModel:

public class SomeModel() 
{ 
    public SomeModel() 
    {
    }

    public HttpPostedFileBase SomeFile { get; set; }
}

And then implement it in the following way:

View:

@model SomeModel

@using (Html.BeginForm(
    "Submit", 
    "Home", 
    FormMethod.Post, 
    new { enctype="multipart/form-data" }))
{
    @Html.TextBoxFor(m => m.SomeFile, new { type = "file" })
    <input type="submit" value="Upload" 
        name="UploadButton" id="UploadButton" />
}

Controller:

[HttpPost]
public ActionResult Submit(SomeModel model)
{
    // do something with model.SomeFile

    return View();
}

In case you need to deal with multiple files, you can either:

• create multiple properties and implement them separately just like the one above;
• change the public HttpPostedFileBase SomeFile property to something like public List<HttpPostedFileBase> SomeFiles and then span multiple @Html.TextBoxFor(m => m.SomeFile, new { type = "file" }) controls to have them all within that list.

In case you need additional info check out this blog post that I wrote on the topic.