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我想在有来电时启动应用程序。1）每当有来电时，用户都会弹出一个窗口，询问用户您要调用哪个应用程序 i）默认电话应用程序或 ii）MyApplication 我正在使用以下代码

public class Record extends BroadcastReceiver {
    @Override
    public void onReceive(Context arg0, Intent arg1) {
        Bundle bundle=arg1.getExtras();
        String state=bundle.getString(TelephonyManager.EXTRA_STATE);

        if(state.equalsIgnoreCase(TelephonyManager.EXTRA_STATE_RINGING))
        {
            Intent i=new Intent(arg0,TempDemoActivity.class);
            arg0.startActivity(i);
        }       
    }
}

显现

<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="temp.demo"
    android:versionCode="1"
    android:versionName="1.0" >

    <uses-sdk android:minSdkVersion="7" />

    <uses-permission android:name="android.permission.READ_PHONE_STATE" />

    <application
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name" >

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

        <receiver android:name=".Record" >
            <intent-filter>
                <action android:name="android.intent.action.PHONE_STATE" />
            </intent-filter>
        </receiver>
    </application>

</manifest>
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1 回答 1

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Intent intent = new Intent(context, DisplayScreen.class);
intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
context.startActivity(intent);

试试这个

于 2013-05-01T19:37:02.943 回答