Submit a Form using AJAX in ASP.Net Core MVC

Question

I am working with ASP.Net Core 2.1, and trying to upload a file while returning it's url, without refreshing the page.

I am trying to write the JavaScript in site.js as the _RenderPartial("scripts") renders all scripts at the end of the page and hence directly using script tag in the razor view is not working. Secondly, adding it to site.js gives me an opportunity to call the script across the site views.

My Controller action looks like :

    [HttpPost]
    [DisableRequestSizeLimit]
    public async Task<IActionResult> Upload()
    {
      // Read & copy to stream the content of MultiPart-Form
      // Return the URL of the uploaded file
      return Content(FileName);
    }

My view looks like :

<form id="FileUploadForm" action="~/Resources/Upload" method="post" enctype="multipart/form-data">
<input name="uploadfile" type="file" />
<button name="uploadbtn" type="submit" onclick="SubmitForm(this.parentElement, event)">Upload</button>

The site.js currently looks like :

function SubmitForm(form, caller) {
caller.preventDefault();
$.ajax(
    {
        type: form.method,
        url: form.action,
        data: form.serialize(),
        success: function (data) { alert(data); },
        error: function (data) { alert(data); }
    })}

Presently, the code bypasses the entire script and the file is uploaded and new view displaying the file name is returned. I need help to create the javascript.

Answer 1

Unfortunately the jQuery serialize() method will not include input file elements. So the file user selected is not going to be included in the serialized value (which is basically a string).

What you may do is, create a FormData object, append the file(s) to that. When making the ajax call, you need to specify processData and contentType property values to false

<form id="FileUploadForm" asp-action="Upload" asp-controller="Home" 
                                              method="post" enctype="multipart/form-data">
    <input id="uploadfile" type="file" />
    <button name="uploadbtn" type="submit">Upload</button>
</form>

and here in the unobutrusive way to handle the form submit event where we will stop the regular behavior and do an ajax submit instead.

$(function () {
    $("#FileUploadForm").submit(function (e) {
        e.preventDefault();

        console.log('Doing ajax submit');

        var formAction = $(this).attr("action");
        var fdata = new FormData();

        var fileInput = $('#uploadfile')[0];
        var file = fileInput.files[0];
        fdata.append("file", file);

        $.ajax({
            type: 'post',
            url: formAction,
            data: fdata,
            processData: false,
            contentType: false
        }).done(function (result) {
            // do something with the result now
            console.log(result);
            if (result.status === "success") {
                alert(result.url);
            } else {
                alert(result.message);
            }
        });
    });
})

Assuming your server side method has a parameter of with name same as the one we used when we created the FormData object entry(file). Here is a sample where it will upload the image to the uploads directory inside wwwwroot.

The action method returns a JSON object with a status and url/message property and you can use that in the success/done handler of the ajax call to whatever you want to do.

public class HomeController : Controller
{
    private readonly IHostingEnvironment hostingEnvironment;
    public HomeController(IHostingEnvironment environment)
    {
        _context = context;
        hostingEnvironment = environment;
    }
    [HttpPost]
    public async Task<IActionResult> Upload(IFormFile file)
    {
        try
        {
            var uniqueFileName = GetUniqueFileName(file.FileName);
            var uploads = Path.Combine(hostingEnvironment.WebRootPath, "uploads");
            var filePath = Path.Combine(uploads, uniqueFileName);
            file.CopyTo(new FileStream(filePath, FileMode.Create));
            var url = Url.Content("~/uploads/" + uniqueFileName);
            return Json(new { status = "success", url = url });
        }
        catch(Exception ex)
        {
            // to do : log error
            return Json(new { status = "error", message = ex.Message });
        }
    }
    private string GetUniqueFileName(string fileName)
    {
        fileName = Path.GetFileName(fileName);
        return Path.GetFileNameWithoutExtension(fileName)
                  + "_"
                  + Guid.NewGuid().ToString().Substring(0, 4)
                  + Path.GetExtension(fileName);
    }
}