Question

我正在尝试制作一个程序，将图像上传到接受多部分文件上传的网络服务器。

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

我做了很多尝试，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。我得到的回应看起来像是一个GET。

（我想在没有任何第三方库的情况下这样做）

更新：非工作代码在这里（没有错误，但似乎没有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);
  }

Answer 1

两件事：

请确保call setRequestMethod to set the HTTP request to be a POST。应该警告您手动执行多部分POST请求很困难且容易出错。
如果你在* NIX上运行，工具netcat对于调试这些东西非常有用。运行
netcat -l -p 3000

并将您的程序指向端口3000;你会看到程序发送的确切内容（Control-C随后将其关闭）。

Answer 2

我已经使用过它，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);

来自java的multipart文件上传发布请求

2 个答案: