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 a

Answer 1

Sharing the code that worked for me, implementing @Shyju's answer.

View ( Razor Page ):

<form name="UploadForm" 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>

AJAX code added in Site.js (to make it a reusable):

// The function takes Form and the event object as parameter
function SubmitForm(frm, caller) {
caller.preventDefault();

var fdata = new FormData();

var file = $(frm).find('input:file[name="uploadfile"]')[0].files[0];
fdata.append("file", file);

$.ajax(
    {
        type: frm.method,
        url: frm.action,
        data: fdata, 
        processData: false,
        contentType: false,
        success: function (data) {
            alert(data);
        },
        error: function (data) {
            alert(data);
        }
    })

};

Answer 2

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);
    }
}