Question

我想要做的是在我的Android应用程序中，当我按下按钮时，我想打开此网页（在代码中）并按下“打开”按钮。我将如何在Java / Android中实现这一目标？

<html>
<head>
<meta charset="UTF-8" />
<title>Webpage</title>
</head>


<?php
if (isset($_POST['OPEN']))
{
exec("<link here>");
}

?>


<form method="post">
<button name="OPEN"> Open</button><br>

</form>
</html>

Android中我的代码解决了这个问题 - 感谢@his

的非常大的提示

            String RPI_IP = "http://192.172.26.1/index.php";
    String param = "OPEN=";

      HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(RPI_IP);

        try {
            // Add your data
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            nameValuePairs.add(new BasicNameValuePair("OPEN", ""));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            // Execute HTTP Post Request
            HttpResponse response = httpclient.execute(httppost);

        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
        } catch (IOException e) {
            // TODO Auto-generated catch block
        }

Answer 1

如果这是一个没有任何Javascript的简单form，那么您不需要模拟浏览器或点击。您只需使用HTTPClient构建POST请求并相应地设置参数OPEN。

打开网页，然后单击Android中的按钮

1 个答案: