3

我正在尝试制作一个将图像上传到接受多部分文件上传的网络服务器的程序。

更具体地说，我想向http://iqs.me发出一个 http POST 请求，该请求在变量“pic”中发送一个文件。

我已经做了很多尝试，但我不知道我是否已经接近了。最难的部分似乎是让 HttpURLConnection 发出 POST 类型的请求。我得到的响应看起来像是一个 GET。

（我想在没有任何第三方库的情况下做到这一点）

更新：这里有非工作代码（没有错误，但似乎没有做 POST）：

  HttpURLConnection conn = null;
  BufferedReader br = null;
  DataOutputStream dos = null;
  DataInputStream inStream = null;

  InputStream is = null;
  OutputStream os = null;
  boolean ret = false;
  String StrMessage = "";
  String exsistingFileName = "myScreenShot.png";

  String lineEnd = "\r\n";
  String twoHyphens = "--";
  String boundary =  "*****";

  int bytesRead, bytesAvailable, bufferSize;
  byte[] buffer;
  int maxBufferSize = 1*1024*1024;
  String responseFromServer = "";
  String urlString = "http://iqs.local.com/index.php";


  try{
    FileInputStream fileInputStream = new FileInputStream( new File(exsistingFileName) );
    URL url = new URL(urlString);
    conn = (HttpURLConnection) url.openConnection();
    conn.setDoInput(true);
    conn.setDoOutput(true);
    conn.setRequestMethod("POST");
    conn.setUseCaches(false);

    conn.setRequestProperty("Connection", "Keep-Alive");
    conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);

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

    dos.writeBytes(twoHyphens + boundary + lineEnd);
    dos.writeBytes("Content-Disposition: form-data; name=\"pic\";" + " filename=\"" + exsistingFileName +"\"" + lineEnd);
    dos.writeBytes(lineEnd);

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

    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);
    }

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

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


  }catch (MalformedURLException ex){
    System.out.println("Error:"+ex);
  }catch (IOException ioe){
    System.out.println("Error:"+ioe);
  }

  try{
    inStream = new DataInputStream ( conn.getInputStream() );
    String str;
    while (( str = inStream.readLine()) != null){
      System.out.println(str);
    }
    inStream.close();
  }catch (IOException ioex){
    System.out.println("Error: "+ioex);
  }
4

2 回答 2

2

两件事情：

  1. 确保调用 setRequestMethod 将 HTTP 请求设置为 POST。应该警告您，手动执行多部分 POST 请求很困难且容易出错。

  2. 如果你在 *NIX 上运行，工具netcat对调试这些东西非常有用。跑
    netcat -l -p 3000

    并将您的程序指向端口 3000；你会看到程序正在发送的确切内容（Control-C 之后关闭它）。

于 2010-04-15T14:41:31.387 回答
0

我用过这个，发现它在多部分文件上传中很有用

File f = new File(filePath);
PostMethod filePost = new PostMethod(url);
Part[] parts = { new FilePart("file", f) };
filePost.setRequestEntity(new MultipartRequestEntity(parts,
filePost.getParams()));
HttpClient client = new HttpClient();
status = client.executeMethod(filePost);
于 2012-04-02T11:00:13.360 回答