HttpURLConnection POST，conn.getOutputStream() 抛出异常

Question

我想使用HttpURLConnection进行POST请求。我正在尝试两种方法，但无论哪种方式，当执行conn.getOutputStream();时，我总是会得到一个异常。

我在两种情况下都收到相同的异常消息：

java.net.SocketException: 操作超时: 连接: 可能由于无效地址

function1：

public void makePost(String title, String comment, File file) {
    try {
        URL servlet = new URL("http://" + "www.server.com/daten/web/test/testupload.nsf/upload?CreateDocument");            
        HttpURLConnection conn=(HttpURLConnection)servlet.openConnection();
        conn.setDoOutput(true);
        conn.setDoInput(true);
        conn.setUseCaches(false);
        String boundary = "---------------------------7d226f700d0";
        conn.setRequestProperty("Content-type","multipart/form-data; boundary=" + boundary);
        //conn.setRequestProperty("Referer", "http://127.0.0.1/index.jsp");
        conn.setRequestProperty("Cache-Control", "no-cache");

        OutputStream os = conn.getOutputStream(); //exception throws here!
        DataOutputStream out = new DataOutputStream(os);
        out.writeBytes("--" + boundary + "\r\n");
        writeParam(INPUT_TITLE, title, out, boundary);
        writeParam(INPUT_COMMENT, comment, out, boundary);
        writeFile(INPUT_FILE, file.getName(), out, boundary);
        out.flush();
        out.close();

        InputStream stream = conn.getInputStream();
        BufferedInputStream in = new BufferedInputStream(stream);
        int i = 0;            
        while ((i = in.read()) != -1) {
            System.out.write(i);            
        }            
        in.close();
    } catch (Exception e) {  
        e.printStackTrace();
    }
}

或者函数 2：

public void makePost2(String title, String comment, File file) {

    File binaryFile = file;
    String boundary = Long.toHexString(System.currentTimeMillis()); // Just generate some unique random value.

    URLConnection connection = null;
    try {
        connection = new URL("http://" + "www.server.com/daten/web/test/testupload.nsf/upload?CreateDocument").openConnection();
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    connection.setDoOutput(true);
    connection.setDoInput(true);
    connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
    PrintWriter writer = null;
    try {
        OutputStream output = connection.getOutputStream(); //exception throws here
        writer = new PrintWriter(new OutputStreamWriter(output, CHARSET), true); // true = autoFlush, important!

        // Send normal param.
        writer.println("--" + boundary);
        writer.println("Content-Disposition: form-data; name=\""+ INPUT_TITLE +"\"");
        writer.println("Content-Type: text/plain; charset=" + CHARSET);
        writer.println();
        writer.println(title);

//        Send binary file.
        writer.println("--" + boundary);
        writer.println("Content-Disposition: form-data; name=\""+ INPUT_FILE +"\"; filename=\"" + binaryFile.getName() + "\"");
        writer.println("Content-Type: " + URLConnection.guessContentTypeFromName(binaryFile.getName()));
        writer.println("Content-Transfer-Encoding: binary");
        writer.println();
        InputStream input = null;
        try {
            input = new FileInputStream(binaryFile);
            byte[] buffer = new byte[1024];
            for (int length = 0; (length = input.read(buffer)) > 0;) {
                output.write(buffer, 0, length);
            }
            output.flush(); // Important! Output cannot be closed. Close of writer will close output as well.
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            if (input != null) try { input.close(); } catch (IOException logOrIgnore) {}
        }
        writer.println(); // Important! Indicates end of binary boundary.

        // End of multipart/form-data.
        writer.println("--" + boundary + "--");
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        if (writer != null) writer.close();
    }


}

- AdrianS

---

嗨@Blitzkr1eg，你能添加你的方法“writeParam”吗？ - Omar Barrera Valentin

不，这是从2011年的。 - AdrianS

Answer 1

该网址无法访问。可能是因为网址错误或DNS服务器无法解析主机名。尝试连接一个众所周知的网址，排除其他原因，例如：

InputStream response = new URL("http://stackoverflow.com").openStream();
// Consume response.

---

更新：根据评论，您需要为HTTP连接使用代理服务器。您还需要在Java端进行配置。在尝试连接到URL之前，请添加以下行。

System.setProperty("http.proxyHost", "proxy.example.com");
System.setProperty("http.proxyPort", "8080");

在运行时只需要执行一次即可。

另请参阅:

• Java指南 - 网络和代理

- BalusC

---

java.net.SocketException: Operation timed out: connect:could be due to invalid address 这句话已经说明了一切。 - Christian Kuetbach

我在一个HTML文件中有一个表单，它向完全相同的URL进行POST请求，并且它可以正常工作。这是怎么回事？ - AdrianS

你的意思是在普通的网络浏览器中吗？如果是这样，它是否使用HTTP代理服务器？如果是这样，你也应该在Java端进行配置。 - BalusC

是的！就是这样！现在尝试。 - AdrianS

好的，我更新了答案，添加了更多关于如何在Java端实现这一点的细节。 - BalusC

Answer 2

在建立连接之前（在这种情况下需要执行1个更多的步骤即连接），传输是不可能的。在配置连接后（即在对连接上的set***()操作完成后），应调用connect()。

缺少的是：

conn.connect();

- Zeena