Question

我想打开一个URL并向其提交以下参数，但只有将BufferedReader添加到我的代码中才能起作用。那是为什么？

Send.php是一个脚本，它会为我的数据库添加带有时间的用户名。

以下代码不起作用（它不会向我的数据库提交任何数据）：

        final String base = "http://awebsite.com//send.php?";
        final String params = String.format("username=%s&time=%s", username, time);
        final URL url = new URL(base + params);

        final HttpURLConnection connection = (HttpURLConnection) url.openConnection();
        connection.setRequestProperty("User-Agent", "Agent");
        connection.setRequestMethod("POST");
        connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

        connection.connect();

但是这段代码确实有效：

        final String base = "http://awebsite.com//send.php?";
        final String params = String.format("username=%s&time=%s", username, time);
        final URL url = new URL(base + params);

        final HttpURLConnection connection = (HttpURLConnection) url.openConnection();
        connection.setRequestProperty("User-Agent", "Agent");
        connection.setRequestMethod("POST");
        connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

        connection.connect();

        final BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
        String line;
        while ((line = in.readLine()) != null) {
            System.out.println(line);
        }

        connection.disconnect();

Answer 1

据我所知。当您调用connect()函数时，它只会创建连接。

您需要至少调用getInputStream()或getResponseCode()来提交连接，以便url指向的服务器能够处理请求。

为什么HttpURLConnection不发送HTTP请求

1 个答案: