Question

已经过了几天，我仍然坚持使用此图片上传功能，但我无法使用此功能。我不会上传我的任何源代码，因为它看起来很可怜而且不起作用所以我不得不求助一个可以让我为我工作的人。我已经谷歌搜索了一段时间，但似乎无法使它们中的任何一个工作。

基本上我要做的是将Bitmap上传到服务器，它必须作为fileStream发送，如下所示。这就是全部。

无论如何，这是Web服务的源代码。当从Android设备发出HttpPost请求时，这将被激活。请注意，这不是我的代码。其他人负责这个。

public Stream FileUpload(string fileName, Stream fileStream){
       var serverPath = System.Web.Hosting.HostingEnvironment.MapPath("~/FileUpload/");
       if (File.Exists(serverPath + fileName)) File.Delete(serverPath + fileName); // delete file if already used


       //FileStream fileToupload = new FileStream("D:\\FileUpload\\" + fileName, FileMode.Create);
       FileStream fileToupload = new FileStream(serverPath + fileName, FileMode.Create);

       byte[] bytearray = new byte[10000];//
       int bytesRead, totalBytesRead = 0;
       do
       {
           bytesRead = fileStream.Read(bytearray, 0, bytearray.Length);
           totalBytesRead += bytesRead;
       } while (bytesRead > 0);

       fileToupload.Write(bytearray, 0, bytearray.Length);
       fileToupload.Close();
       fileToupload.Dispose();

       FileStream fs = File.OpenRead(serverPath + fileName);
       WebOperationContext.Current.OutgoingResponse.ContentType = "image/jpeg";
       return fs;
   }

非常感谢任何帮助。

Answer 1

我就是这样做的，你当然是在后台任务中运行的

    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "***1234321**";
    String filePath = "path/to/file";
    String fileName = new File(filePath).getName();
    StringBuffer response =new StringBuffer();

    try {
        URL connectUrl = new URL((String) params[0]);
        FileInputStream fileInputStream = new FileInputStream(filePath);
        HttpURLConnection conn = (HttpURLConnection) connectUrl.openConnection();
        conn.setDoInput(true);
        conn.setDoOutput(true);
        conn.setUseCaches(false);
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Connection", "Keep-Alive");
        conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);

        DataOutputStream dos = new DataOutputStream(conn.getOutputStream());

        dos.writeBytes(twoHyphens + boundary + lineEnd);
        dos.writeBytes("Content-Disposition: form-data; name=\""+ params[2] +"\"; filename=\"" + fileName +"\"" + lineEnd);
        dos.writeBytes("Content-Type: application/octet-stream" + lineEnd);
        dos.writeBytes("Content-Transfer-Encoding: binary" + lineEnd);
        dos.writeBytes(lineEnd);

        // create a buffer of maximum size
        int bytesAvailable = fileInputStream.available();
        int maxBufferSize = 1024;
        int bufferSize = Math.min(bytesAvailable, maxBufferSize);
        byte[] buffer = new byte[bufferSize];

        // read file and write it into form...
        int bytesRead = fileInputStream.read(buffer, 0, bufferSize);

        while (bytesRead > 0) {
            dos.write(buffer, 0, bufferSize);
            bytesAvailable = fileInputStream.available();
            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        }

        // send multipart form data necessary after file data...
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

        // close streams
        fileInputStream.close();
        dos.flush();


        dos.close();

    } catch (Exception e) {

    }

图像上传问题从Android设备到.NET webservice

1 个答案: