可以通过编程方式关闭WIFI

Question

如何以编程方式关闭/打开wifi,以及为此需要root或系统应用程序.

Answer 1

权限是必需的.

我刚刚写了这个应用程序切换Wifi.

表现

<?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>

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());
      }
    });
  }