CODEWITHSUNDEEP
c#jqueryasp.netweb-servicesdownload
Webman
Webman

Reputation: 1554

Download a file via web service and jquery?

I'm trying to create in asp.net 2.0 a web service to download a file (pop up window with Open or Save the file), in this way:

$.ajax({
        type: "POST",
        url: "webservice.asmx/download",
        data: "{}",
        contentType: "application/json; charset=utf-8",
        success: function (res) {
            console.log("donwloaded");
        }
    });

And in "webservice.asmx"

[WebMethod()]
public byte[] DownloadFile(string FName)
{
    System.IO.FileStream fs1 = null;
    fs1 = System.IO.File.Open(FName, FileMode.Open, FileAccess.Read);
    byte[] b1 = new byte[fs1.Length];
    fs1.Read(b1, 0, (int)fs1.Length);
    fs1.Close();
    return b1;
}

[WebMethod]
public void download()
{
    string filename = "test.txt";
    string path = "C:\\test.txt";

    byte[] ls1 = DownloadFile(path);
    HttpResponse response = Context.Response;

    response.Clear();
    response.BufferOutput = true;
    response.ContentType = "application/octet-stream";
    response.ContentEncoding = Encoding.UTF8;
    response.AppendHeader("content-disposition", "attachment; filename=" + filename);

    response.BinaryWrite(ls1);

    response.Flush();
    response.Close();
}

In this way I see the contents of the file (I can't to download the file whith a popup window).

Where I wrong? Is it possible to do this?

thanks in advance

Upvotes: 2

Views: 8691

Answers (2)

Artemiy
Artemiy

Reputation: 1979

a good way to do it is simply redirect the browser to the "webservice.asmx/download" instead of calling that via ajax. i.e.

window.location.href = 'webservice.asmx/download';

when browser will detect you returning octet-stream, it will ask to download the file but will not clear/eraze the page you were on.

Upvotes: 0

sino
sino

Reputation: 766

that is not a good way , to make file download thru popup window , just open a new pop and make the url of that window to point to your file or webservice url . 

window.open("fileurl"  )

Upvotes: 1

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