Question

我正在使用此代码将参数发送到网页并从中获得正确的响应。

System.Net.WebClient oWeb = new System.Net.WebClient();
oWeb.Proxy = System.Net.WebRequest.DefaultWebProxy;
oWeb.Proxy.Credentials = System.Net.CredentialCache.DefaultCredentials;
oWeb.Headers.Add("Content-Type", "application/x-www-form-urlencoded");
byte[] bytArguments = System.Text.Encoding.ASCII.GetBytes("value1=123&value2=xyz");
byte[] bytRetData = oWeb.UploadData("http://website.com/file.php", "POST", bytArguments);
response = System.Text.Encoding.ASCII.GetString(bytRetData);

但是现在我想发送一个像（.doc）这样的文件+上面的参数（value1，value2），但我不知道该怎么做。< / p>

Answer 1

使用WebClient.QueryString传递与请求关联的名称/值对。

NameValueCollection parameters = new NameValueCollection();
parameters.Add("value1", "123");
parameters.Add("value2", "xyz");
oWeb.QueryString = parameters;
var responseBytes = oWeb.UploadFile("http://website.com/file.php", "path to file");
string response = Encoding.ASCII.GetString(responseBytes);

Answer 2

    public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        string boundary = "---------------------------" + 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 = System.Net.CredentialCache.DefaultCredentials;
        Stream rs = wr.GetRequestStream();
        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            rs.Write(boundarybytes, 0, boundarybytes.Length);
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);
        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.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);
        FileStream fileStream = new FileStream(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);
            result = reader2.ReadToEnd();
        }
        catch (Exception ex)
        {
            System.Windows.MessageBox.Show("Error occurred while converting file", "Error!");
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }

从SO复制但不记得它的链接。这就是它的使用方式

        NameValueCollection nvc = new NameValueCollection();
        nvc.Add("parm1", "value1");
        nvc.Add("parm2", "value2");
        nvc.Add("parm3", "value3");
        HttpUploadFile("http://www.example.com/upload.php",@filepath, "file", "text/html", nvc);

此处@filepath是您要上传的文件的路径：c：\ file_to_upload.doc 而file是php中使用的字段的名称为$ _Files ['file']

在post请求中发送文件+参数

2 个答案: