1

我编写了以下代码，用于将两个文件（一个 XML 和一个 DOCX 文件）发布到 Web 服务中：

public void postMultipleFiles(string url, string[] files)
    {
        string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
        HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(url);
        httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
        httpWebRequest.Method = "POST";
        httpWebRequest.KeepAlive = true;
        httpWebRequest.Credentials = System.Net.CredentialCache.DefaultCredentials;
        Stream memStream = new System.IO.MemoryStream();
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
        string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";
        memStream.Write(boundarybytes, 0, boundarybytes.Length);
        for (int i = 0; i < files.Length; i++)
        {
            string header = string.Format(headerTemplate, "file" + i, files[i]);
            //string header = string.Format(headerTemplate, "uplTheFile", files[i]);
            byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
            memStream.Write(headerbytes, 0, headerbytes.Length);
            FileStream fileStream = new FileStream(files[i], FileMode.Open,
            FileAccess.Read);
            byte[] buffer = new byte[1024];
            int bytesRead = 0;
            while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
            {
                memStream.Write(buffer, 0, bytesRead);
            }
            memStream.Write(boundarybytes, 0, boundarybytes.Length);
            fileStream.Close();
        }
        /*AJ*/
        httpWebRequest.ContentLength = memStream.Length;
        Stream requestStream = httpWebRequest.GetRequestStream();
        memStream.Position = 0;
        byte[] tempBuffer = new byte[memStream.Length];
        memStream.Read(tempBuffer, 0, tempBuffer.Length);
        memStream.Close();
        requestStream.Write(tempBuffer, 0, tempBuffer.Length);
        requestStream.Close();
        try
        {
            WebResponse webResponse = httpWebRequest.GetResponse();
            Stream stream = webResponse.GetResponseStream();
            StreamReader reader = new StreamReader(stream);
            response.InnerHtml = reader.ReadToEnd();
        }
        catch (Exception ex)
        {
            response.InnerHtml = ex.Message;
        }
        httpWebRequest = null;
    }

在网络服务中，我检索到与以下内容相同的内容：

public string MultiFilePost()
{
    string xmls = "";
    string txt = "";
    XmlDocument xmldoc = new XmlDocument();

    if (HttpContext.Current.Request.InputStream != null)
    {

        StreamReader stream = new StreamReader(HttpContext.Current.Request.InputStream);
        /*First File*/
        Stream xmlStream = System.Web.HttpContext.Current.Request.Files[0].InputStream;
        StreamReader rd = new StreamReader(xmlStream);
        xmls = rd.ReadToEnd();
        /*Second File*/
        Stream txtStream = System.Web.HttpContext.Current.Request.Files[1].InputStream;
        rd = new StreamReader(txtStream);
        txt = rd.ReadToEnd();
    }

    return xmls;
}

因此，我将输入流转换为字符串。我可以将字符串“xmls”转换为 XmlDocument 对象，但是在字符串 txt 中我收到了某种字符串，并且我需要一种方法来处理它，因为 XmlDocument 处理 XML 文件。

提前致谢。

4

1 回答 1

0

使用XmlDocument.Load 方法（流）：

// First file
var xmlStream = System.Web.HttpContext.Current.Request.Files[0].InputStream;
var xmlDocument = new XmlDocument();
xmlDocument.Load(xmlStream);

// The same code for the second file processing.
// Second file
var wordDocStream = System.Web.HttpContext.Current.Request.Files[1].InputStream;
var wordXmlDocument = new XmlDocument();
wordXmlDocument.Load(wordDocStream);

更新

要使用 Word 文档：

于 2013-02-11T09:26:30.570 回答