可以通过编程方式关闭WIFI

时间:2011-04-23 19:52:40

标签: android wifi

如何以编程方式关闭/打开wifi,并且需要root或系统应用程序。

3 个答案:

答案 0 :(得分:6)

需要权限。

我刚刚写了这个应用程序来切换Wifi。

<强>清单

<?xml version="1.0" encoding="utf-8"?>
<manifest
    xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.stackoverflow.q5766518"
    android:versionCode="1"
    android:versionName="1.0">
    <uses-sdk
        android:minSdkVersion="3" />

    <uses-permission
        android:name="android.permission.ACCESS_WIFI_STATE" />
    <uses-permission
        android:name="android.permission.CHANGE_WIFI_STATE" />
    <uses-permission
        android:name="android.permission.WAKE_LOCK" />

    <application
        android:icon="@drawable/icon"
        android:label="@string/app_name">
        <activity
            android:name=".Main"
            android:label="@string/app_name">
            <intent-filter>
                <action
                    android:name="android.intent.action.MAIN" />
                <category
                    android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>

    </application>
</manifest>

<强>布局

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout
    xmlns:android="http://schemas.android.com/apk/res/android"
    android:orientation="vertical"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent">
    <Button
        android:id="@+id/my_button"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:text="Toggle Wifi" />
</LinearLayout>

主要活动

  @Override
  public void onCreate(Bundle savedInstanceState)
  {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    final Button myButton = (Button) findViewById(R.id.my_button);
    myButton.setOnClickListener(new View.OnClickListener()
    {
      @Override
      public void onClick(View v)
      {
        final WifiManager wifi = (WifiManager) getSystemService(Context.WIFI_SERVICE);
        wifi.setWifiEnabled(!wifi.isWifiEnabled());
      }
    });
  }

答案 1 :(得分:1)

WIFI_ON is a secure setting;只有系统固件签名的应用程序才能保留适当的权限并使用SDK进行调整。


<强>更新

正如评论中所指出的,

setWifiEnabled()可能支持这一点。我没有看到需要许可的证据,但是如果有的话,你会得到一个堆栈跟踪,指出需要什么。我为忘记这条道路而道歉。

答案 2 :(得分:0)

是的,它可能。 wifimangr.setWifiEnabled(假);

  
    
      
        
          
            
              
                
                  
                    
                      
                        
                          
                            
                              
                                
                                  
                                    
                                      
                                        
                                          
                                            
                                              
                                                
                                                  

创建Wifimanager的对象..并将方法setWifiEnabled调用为&#34; false&#34;。                                                    wifimangr.setWifiEnabled(false);

                                                
                                              
                                            
                                          
                                        
                                      
                                    
                                  
                                
                              
                            
                          
                        
                      
                    
                  
                
              
            
          
        
      
    
  

你需要CHANGE_WIFI_STATE权限。