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Home/ Questions/Q 3237046
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T17:41:32+00:00 2026-05-17T17:41:32+00:00

I am trying to use the ImageShack API to upload images. To use it,

  • 0

I am trying to use the ImageShack API to upload images. To use it, I am supposed to POST the image using multipart/form-data. I did it like …

var postData = "";
var req = HttpWebRequest.Create("http://www.imageshack.us/upload_api.php");
req.Method = "POST";
req.ContentType = "multipart/form-data";
postData += "key=my_key_here&";
postData += "type=base64&";

// get base64 data from image
byte[] bytes = File.ReadAllBytes(@"D:\tmp\WpfApplication1\WpfApplication1\Images\Icon128.gif");
string encoded = Convert.ToBase64String(bytes);
postData += "fileupload=" + encoded;

byte[] reqData = Encoding.UTF8.GetBytes(postData);
using (Stream dataStream = req.GetRequestStream())
{
    dataStream.Write(reqData, 0, reqData.Length);
}

var res = (HttpWebResponse)req.GetResponse();
var resStream = res.GetResponseStream();
var reader = new StreamReader(resStream);
string resString = reader.ReadToEnd();
txt1.Text = resString;

but ImageShack is complaining that

<links>
    <error id="parameter_missing">Sorry, but we've detected that unexpected data is received. Required parameter 'fileupload' is missing or your post is not multipart/form-data</error>
</links>

FileUpload is present and I am using multipart/form-data whats wrong?

UPDATE:

New Code http://pastebin.com/TN6e0CD8

Post data http://pastebin.com/fYE9fsxs

UPDATE 2

i looked at the other question Multipart forms from C# client. modified my code with boundary, removed the expect 100 header still i cant get it working …

ServicePointManager.Expect100Continue = false;
var boundary = "-----------------------------28520690214962";
var newLine = Environment.NewLine;
var propFormat = boundary + newLine +
                    "Content-Disposition: form-data; name=\"{0}\"" + newLine + newLine + 
                    "{1}" + newLine + newLine;
var fileHeaderFormat = boundary + newLine +
                        "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"" + newLine;

var req = (HttpWebRequest)HttpWebRequest.Create("http://jm/php/upload.php");
req.Method = WebRequestMethods.Http.Post;
req.ContentType = "multipart/form-data; boundary=" + boundary;

using (var reqStream = req.GetRequestStream()) {
    var reqWriter = new StreamWriter(reqStream);
    var tmp = string.Format(propFormat, "str1", "hello world");
    reqWriter.Write(tmp);
    tmp = string.Format(propFormat, "str2", "hello world 2");
    reqWriter.Write(tmp);
    reqWriter.Write(boundary + "--");
    reqWriter.Flush();
}
var res = req.GetResponse();
using (var resStream = res.GetResponseStream()) {
    var reader = new StreamReader(resStream);
    txt1.Text = reader.ReadToEnd();
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-17T17:41:32+00:00Added an answer on May 17, 2026 at 5:41 pm

    I believe that you are not building the request body correctly.
    First, you need to include part boundary (random text) in content type header. For example,

    Content-Type: multipart/form-data;
    boundary=—-WebKitFormBoundarySkAQdHysJKel8YBM

    Now format of request body will be something like

    ------WebKitFormBoundarySkAQdHysJKel8YBM 
    Content-Disposition: form-data;name="key"
    
    KeyValueGoesHere
    ------WebKitFormBoundarySkAQdHysJKel8YBM 
    Content-Disposition: form-data;name="param2"
    
    ValueHere
    ------WebKitFormBoundarySkAQdHysJKel8YBM 
    Content-Disposition: form-data;name="fileUpload"; filename="y1.jpg"
    Content-Type: image/jpeg 
    
    [image data goes here]
    

    I will suggest you to use tool such as Fiddler to understand how these requests are built.

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