Android按钮调用www地址

Question

我需要制作一个带有按钮的简单应用程序，它将调用一些 URL。我在我的 Rpi 上有一个 apache 服务器，我用它来通过发送这样的 GET URL 来控制 GPIO

http://192.168.0.105/index.php?pin=2&status=0。

在 android studio 中我只做了一个 onclick 按钮。

问题是我不熟悉 java 和 android，所以这是我的问题 - 让这个按钮发送 URL 的最简单方法是什么？我找到了一些关于从服务器发送或接收数据的教程，但这不是我想做的。

Answer 1

1. 为清单添加权限：<uses-permission android:name="android.permission.INTERNET" />

2. 创建一个类：

   class RequestTask extends AsyncTask<String, String, String> {
       String response;
       ProgressDialog dialog;
   
       @Override
       protected String doInBackground(String... params) {
           try {
               DefaultHttpClient httpClient = new DefaultHttpClient();
               HttpGet httpGet = new HttpGet(params[0]); // or HttpPost if you need
   
               ResponseHandler<String> resHandler = new BasicResponseHandler();
               response = httpClient.execute(httpGet, resHandler);
          } catch (Exception e) {
              System.out.println("E: " + e);
          }
   
          return null;
       }
   
       @Override
       protected void onPreExecute() {
           dialog = new ProgressDialog(MainActivity.this);
           dialog.setMessage("Loading...");
           dialog.setIndeterminate(true);
           dialog.setCancelable(true);
           dialog.show();
           super.onPreExecute();
       }
   
       @Override
       protected void onPostExecute(String result) {
           dialog.dismiss();
           super.onPostExecute(result);
       }
   }
   

3. 来电使用：new RequestTask().execute("http://192.168.0.105/index.php?pin=2&status=0")

Answer 2

您可以在单击事件的按钮内使用此代码。点击后打开网址。

Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("<your URL>"));
startActivity(browserIntent);

此外，如果您想作为 Web 客户端连接，但不想在浏览器中打开 - 您可以使用：

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpGet httpGet = new HttpGet("http://www.someplace.com");

ResponseHandler<String> resHandler = new BasicResponseHandler();
String page = httpClient.execute(httpGet, resHandler);

Answer 3

最简单的方法是使用 volley 库。

在清单中声明应用程序标签之外的互联网权限：<uses-permission android:name="android.permission.INTERNET" />

在 build.gradle 文件的依赖项中添加这一行：implementation 'com.android.volley:volley:1.0.0'

创建一个名为 VolleySingleton 的 java 文件并从 file in github 复制文件内容，只是不要复制包名。

在按钮内点击添加这些行：

url = "http://192.168.0.105/index.php";
StringRequest postRequest = new StringRequest(Request.Method.POST, url, 
    new Response.Listener<String>() 
    {
        @Override
        public void onResponse(String response) {
            // response
            Log.d("Response", response);
            //Do what ever you need to do with the response here.
            //If you don't need to return any response just echo OK from the php so that you become sure that the thing is working fine.
        }
    }, 
    new Response.ErrorListener() 
    {
         @Override
         public void onErrorResponse(VolleyError error) {
             // error
             Log.d("Error.Response", response);
       }
    }
) {     
    @Override
    protected Map<String, String> getParams() 
    {  
            Map<String, String>  params = new HashMap<String, String>();  
            params.put("pin", "2");  
            params.put("status", "0");

            return params;  
    }
};

VolleySingleton.getmInstance(getApplicationContext()).addToRequestQue(postRequest);

Answer 4

我已经创建了类似的功能

看看对你有没有帮助

您需要获得使用 Internet 的权限。在清单文件中，在应用程序标签之外声明互联网权限： <uses-permission android:name="android.permission.INTERNET" />

创建一个活动（WebViewActivity.java）

public class WebViewActivity extends AppCompatActivity {

private static final String TAG = "WebViewActivity";
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_web_view);

    Log.d(TAG, "onCreate: Started");

    Bundle extras = getIntent().getExtras();
    assert extras != null;
    String pin = extras.getString("pin"); //you can use getInt("pin") also, but afterwards in the url, it will be converted to String back, so it is not needed
    String status = extras.getString("status"); //you can use getInt("status") also, but afterwards in the url, it will be converted to String back, so it is not needed

    WebView webView = new WebView(this);
    setContentView(webView);
    WebSettings webSettings = webView.getSettings();
    webSettings.setJavaScriptEnabled(true); //This is required to enable javascript that is required by some pages
    String url = "http://192.168.0.105/index.php?pin" + pin + "&status=" + status;

    webView.loadUrl(url);

    Log.d(TAG, "url accessed: " + url);

    finish();//If you need to see the output after accessing the http page, then remove this line. Otherwise just copy as it is 
}

}

在MainActivity中，启动上面创建的activity即可。

public class MainActivity extends AppCompatActivity{
    int pin = 0, status = 2;// You can modify the values according to your requirements
    //....
    //....
    button.setOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View view) {
                    Intent intent = new Intent(getApplicationContext(), WebViewActivity.class);
                    intent.putExtra("pin", pin);
                    intent.putExtra("status", status);
                    startActivity(intent);
                    Toast.makeText(MainActivity.this, "URL Opened and pin, status values are successfully updated", Toast.LENGTH_SHORT).show();

                }
            });
    //....
    //....
}

如果您想知道在 xml 文件中放入什么 (activity_web_view.xml) 保持原样，我使用下面的代码发送值，就像你试图发送一样。

activity_web_view.xml

<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
tools:context=".WebViewActivity" />