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Home/ Questions/Q 3930908
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T23:18:07+00:00 2026-05-19T23:18:07+00:00

I have the following web service that allows me uploading files: [WebService(Namespace = http://tempuri.org/)]

  • 0

I have the following web service that allows me uploading files:

[WebService(Namespace = "http://tempuri.org/")]
    [WebServiceBinding(ConformsTo = WsiProfiles.BasicProfile1_1)]
    [System.ComponentModel.ToolboxItem(false)]
    // Pour autoriser l'appel de ce service Web depuis un script à l'aide d'ASP.NET AJAX, supprimez les marques de commentaire de la ligne suivante. 
    [System.Web.Script.Services.ScriptService]
    public class Upload : System.Web.Services.WebService
    {
        [WebMethod]
        public bool UploadFile(string PictureName, byte[] PictureStream)
        {
            FileStream fileStream = null;
            BinaryWriter writer = null;
            string filePath;

            try
            {
                filePath = HttpContext.Current.Server.MapPath(".") + ConfigurationManager.AppSettings["PictureUploadDirectory"] + PictureName;

                if (PictureName != string.Empty)
                {
                    fileStream = File.Open(filePath, FileMode.Create);
                    writer = new BinaryWriter(fileStream);
                    writer.Write(PictureStream);
                }

                return true;
            }
            catch (Exception)
            {
                return false;
            }
            finally
            {
                if (fileStream != null)
                    fileStream.Close();
                if (writer != null)
                    writer.Close();
            }
        }
    }

It works great when it gets invoked from my Windows form.

Now, I’m trying to make it work with HTML/JavaScript. The index.html file is put on the server side (to avoid the cross-domain error) and contains the following code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <title></title>
        <link rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.8/themes/base/jquery-ui.css" type="text/css" media="all" /> 
        <link rel="stylesheet" href="http://static.jquery.com/ui/css/demo-docs-theme/ui.theme.css" type="text/css" media="all" /> 
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js" type="text/javascript"></script> 
        <script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.8/jquery-ui.min.js" type="text/javascript"></script>
        <script type="text/javascript">
            $(document).ready(function () {
                $("#btnSend").click(function () {
                    var fileStream = $("#fileToSend").val();
                    $.ajax({
                        type: "POST",
                        data: { PictureName: 'foobar', PictureStream: fileStream },
                        dataType: "image/gif",
                        url: "http://localhost/WebServices/Upload.asmx/UploadFile",
                        contentType: "application/x-www-form-urlencoded",
                        success: function () {
                            alert('yes');
                        },
                        error: function () {
                            alert('bouh');
                        }
                    });
                });
            });
        </script>
    </head>
    <body>
        <input type="file" id="fileToSend"/>
        <br />
        <input type="button" value="Envoyer" id="btnSend"/>
    </body>
</html>

When I try to upload a file, I get the following response from the server:

System.ArgumentException: Unable to convert C:\fakepath\logo.gif into System.Byte.

How can I avoid this error?

Thanks, regards.

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  1. Editorial Team
    Editorial Team
    2026-05-19T23:18:07+00:00Added an answer on May 19, 2026 at 11:18 pm

    You can’t upload a file with Ajax. You can use an iframe which simulates a similiar effect, but you still have to sumbit the form. (by $("form").submit(); or clicking the submit button)

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