[英]Open a webpage and click a button in Android
我想做的是在我的Android應用程序中,當我按一個按鈕時,我想打開此網頁(以代碼形式),然后按“打開”按鈕。 我將如何在Java / Android中完成此任務?
<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>
我在Android中解決此問題的代碼-感謝@his的很大提示
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
}
如果那是沒有任何Javascript的簡單form
,那么您就不需要模擬瀏覽器或單擊。 您只需使用HTTPClient
來構建POST
請求並適當設置參數OPEN
。
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