Question

我想使用HttpWebRequest上传图像。在chrome的开发人员工具中，我找到了上传图片的数据，如下所示：

    ------WebKitFormBoundaryhEPEtLPHtg3KILWo
Content-Disposition: form-data; name="upload"; filename="photo-1556228720-9b1e04f13f63.jpg"
Content-Type: image/jpeg


------WebKitFormBoundaryhEPEtLPHtg3KILWo--

这是我的代码：

public static void HttpUploadFile(string url, string file, string paramName, string contentType)
    {
        string boundary = "------WebKitFormBoundary" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = CredentialCache.DefaultCredentials;
        wr.CookieContainer = cc;
        wr.Accept = "application/json, text/javascript, */*; q=0.01";
        wr.UserAgent = "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/76.0.3789.0 Safari/537.36 Edg/76.0.159.0";
        wr.Headers.Add("X-Requested-With", "XMLHttpRequest");

        Stream rs = wr.GetRequestStream();

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, file, contentType);
        byte[] headerbytes = System.Text.Encoding.ASCII.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

        FileStream fileStream = new FileStream("C:\\Users\\liang\\Desktop\\" + file, FileMode.Open, FileAccess.Read);
        byte[] buffer = new byte[4096];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            rs.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();

        byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n" + boundary + "--\r\n");
        rs.Write(trailer, 0, trailer.Length);
        rs.Close();

        WebResponse wresp = null;
        try
        {
            wresp = wr.GetResponse();
            Stream stream2 = wresp.GetResponseStream();
            StreamReader reader2 = new StreamReader(stream2);
            string sr = reader2.ReadToEnd();
        }
        catch (Exception ex)
        {
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }

它总是导致500服务器错误的异常。我尝试了很多次，找不到原因。我还检查了接收图像的api地址的源代码。这是php代码，只接收图片。

$file = $this->request->file("upload");
$name = $_FILES["upload"]["name"];
if ($file) {
                $newpath = ROOT_PATH . "/public/upload/images/";
                $info = $file->move($newpath, time().rand());
                if ($info) {
                    $imgname = $info->getFilename();

                    $imgpath = "/upload/images/" . $imgname;

                    $data = [
                        "code" => 0,
                        "msg" => "",
                        "data" => $imgpath
                    ];

                    return $data;
                }
            }

我认为导致错误的原因之一是名称不正确。

在边界中我仅设置文件名为图像名称，但在代码中我设置了文件的所有路径。我不知道这是否是导致错误的原因：

FileStream fileStream = new FileStream("C:\\Users\\liang\\Desktop\\" + file, FileMode.Open, FileAccess.Read);

我不认为这是不对的，但我不明白哪里错了。请告诉我谢谢你。响应是：

The remote server returned an error: (500) Internal Server Error.

谢谢。

我上载图片时，HttpWebrequest出现500服务器错误？

0 个答案: