如何从Java客户端处理服务器端HTTP文件上载

我想把一个文件从客户端上传到服务器

客户端：使用HTTP post的Java 服务器：Javaservlet

我在这里添加了客户端编码。但是，我对服务器端处理没有任何概念。请帮我写一段代码

 private String Tag = "UPLOADER";
    private String urlString = "http://192.168.42.140:8080/sampweb";
    HttpURLConnection conn;
    String exsistingFileName = "/sdcard/log.txt";
    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";
    try {
        // ------------------ CLIENT REQUEST

        Log.e(Tag, "Inside second Method");

        FileInputStream fileInputStream = new FileInputStream(new File(
                exsistingFileName));

        // open a URL connection to the Servlet

        URL url = new URL(urlString);

        // Open a HTTP connection to the URL

        conn = (HttpURLConnection) url.openConnection();

        // Allow Inputs
        conn.setDoInput(true);

        // Allow Outputs
        conn.setDoOutput(true);

        // Don't use a cached copy.
        conn.setUseCaches(false);

        // Use a post method.
        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: post-data; name=uploadedfile;filename="
                        + exsistingFileName + "" + lineEnd);
        dos.writeBytes(lineEnd);

        Log.e(Tag, "Headers are written");

        // create a buffer of maximum size

        int bytesAvailable = fileInputStream.available();
        int maxBufferSize = 1000;
        // int bufferSize = Math.min(bytesAvailable, maxBufferSize);
        byte[] buffer = new byte[bytesAvailable];

        // read file and write it into form...

        int bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);

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

        // send multipart form data necessary after file data...

        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

        // close streams
        Log.e(Tag, "File is written");
        fileInputStream.close();
        dos.flush();
        dos.close();

    } catch (MalformedURLException ex) {
        Log.e(Tag, "error: " + ex.getMessage(), ex);
    }

    catch (IOException ioe) {
        Log.e(Tag, "error: " + ioe.getMessage(), ioe);
    }

    try {
        BufferedReader rd = new BufferedReader(new InputStreamReader(conn
                .getInputStream()));
        String line;
        while ((line = rd.readLine()) != null) {
            Log.e("Dialoge Box", "Message: " + line);
        }
        rd.close();

    } catch (IOException ioex) {
        Log.e("MediaPlayer", "error: " + ioex.getMessage(), ioex);
    }
}

---

我是服务器编程新手。如果我犯了任何错误，请澄清我。谢谢

实现这一点的最佳库仍然是Apache Commons。它有很好的文档记录并且易于使用。如果遇到任何问题，请先检查。

这不起作用。Commons文件上传是针对服务器而不是客户端的。问题是：“我在这里添加了客户端编码。但是，我对服务器端处理没有任何概念。”@GergelyBasco，如果你阅读标题，问题就不清楚了。真正的问题已简化为一句话。谢谢你的指点。标题有误导性，但问题本身很清楚，要用Java实现服务器端，而不是客户端。当前链接的副本描述了客户端实现。您误用了

available（）

，在这里根本不需要使用它。