如何通过 url.openStream() 发送 POST 数据？

Question

我正在寻找教程或快速示例，我如何发送 POST 数据抛出 openStream。

我的代码是：

URL url = new URL("http://localhost:8080/test");
            InputStream response = url.openStream();
            BufferedReader reader = new BufferedReader(new InputStreamReader(response, "UTF-8"));

你可以帮帮我吗 ？

Answer 1

    URL url = new URL(urlSpec);
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();
    connection.setRequestMethod(method);
    connection.setDoOutput(true);
    connection.setDoInput(true);

    // important: get output stream before input stream
    OutputStream out = connection.getOutputStream();
    out.write(content);
    out.close();        

            // now you can get input stream and read.
    BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
    String line = null;

    while ((line = reader.readLine()) != null) {
        writer.println(line);
    }

Answer 2

使用 Apache HTTP 组件http://hc.apache.org/httpcomponents-client-ga/

教程： http : //hc.apache.org/httpcomponents-client-ga/tutorial/html/fundamentals.html

寻找 HttpPost - 有一些发送动态数据、文本、文件和表单数据的例子。

Answer 3

特别是Apache HTTP 组件，客户端将是最好的方法。 它抽象了您通常必须手动完成的许多讨厌的编码