multipart form - Web API

Question

I have a ASP.NET MVC project that needs an API controller which will accept a posted multipart form and extract the data out of the <formroot> xml tag (which is highlighted)

I am struggling on getting this working any help would be greatly appreciated

Currently I have a controller called UploadController and this is the code I currently have

public class UploadController : ApiController
{
    public async Task<HttpResponseMessage> PostFormData()
    {
        if (!Request.Content.IsMimeMultipartContent())
        {
            throw new HttpResponseException(HttpStatusCode.BadRequest);
        }

        string root = HttpContext.Current.Server.MapPath("~/App_Data");
        var provider = new MultipartFormDataStreamProvider(root);

        try
        {
            //Need to get the data from within the formroot tag


            return Request.CreateResponse(HttpStatusCode.OK);
        }
        catch (Exception e)
        {
            return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
        }
    }
}

I am unsure the best way to get the data from within the formroot, also forgive me if any of the code above is not correct.

Where exactly are you stuck? Writing the POST? The controller? Analyzing the message? What code do you have already? — Patrick Hofman, Oct 31 '16 at 07:58
Apologies my question was vague, I will edit and flesh out now — Paul Coan, Oct 31 '16 at 07:58
It looks like you want to fetch the uploaded file data? Is this correct? — Hypnobrew, Oct 31 '16 at 08:06
Yes, as you can see from the image the XML from the post contains a bunch of fields that I need to extract the data from. — Paul Coan, Oct 31 '16 at 08:08
Possible duplicate of http://stackoverflow.com/questions/5193842/file-upload-asp-net-mvc-3-0 — Peter B, Oct 31 '16 at 08:19

score 0 · Answer 1 · answered Oct 31 '16 at 10:43

I always use solution below:

<form action="/Home/Upload" enctype="multipart/form-data" id="upload" method="post">
      @Html.AntiForgeryToken()
      <input type="file" class="file" id="file" name="file" onchange="javascript:upload(this);" />
</form>

PS: "upload()" javascript function uses jquery to post the form.

function upload(obj) {
    var p = $(obj).parent();
    if (p.get(0).tagName != 'FORM') {
        p = p.parent();
    }
    p.submit();
}

In my controller i use as a modelbinder "HttpPostedFileBase".

[HttpPost]
[ValidateAntiForgeryToken]
public RedirectResult Upload(HttpPostedFileBase file)
{
    try
    {
        //physical path there you will save the file. 
        var path = @"c:\temp\filename.txt";
        file.SaveAs(path);
    }
    catch (UploadException ex)
    {

    }

    var url = "put here same url or another url";

    return RedirectResult(url);
}

score 0 · Accepted Answer · answered Nov 01 '16 at 06:10

Inside the web API controller you can access the XML file by using the code below :-

HttpPostedFile xmlFile = HttpContext.Current.Request.Files[0];

If you have more than one files posted, replace Files[0] with respective count 1 or 2 etc. Now you can load the file into a XmlDocument Object and extract the required node from it like :-

XmlDocument doc = new XmlDocument();
doc.Load(xmlFile.InputStream);
XmlNamespaceManager nsmgr = new XmlNamespaceManager(doc.NameTable);
nsmgr.AddNamespace("ab", "www.w3.org/2001/XMLSchema-instance");
XmlNode node = doc.SelectSingleNode("//ab:formroot", nsmgr);

Then you can perform whatever you functionality is provided with the node.

you sir are awesome! got me to where I needed thanks – Paul Coan Nov 01 '16 at 23:48 — Paul Coan, Nov 01 '16 at 23:48

multipart form - Web API

2 Answers2