启动隐藏的应用程序时，秘密code拨打 [英] launch hidden app when secret code is dialed

问题描述

我的要求是要推出一个隐藏的应用程序时，一个秘密code为dailed。

my requirement is to launch an hidden app when a secret code is dailed.

MainActivity.java

MainActivity.java

public class MainActivity extends
        BroadcastReceiver {

    String dialed_number;

    @Override
    public void onReceive(Context context, Intent intent)
    {
        dialed_number = intent.getStringExtra(Intent.EXTRA_PHONE_NUMBER);

        if(dialed_number.equals("*0*1235#"))
        {
            Intent appIntent = new Intent(context, MainActivity.class);
            appIntent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
            context.startActivity(appIntent);
            setResultData(null);
        }
    }

}

AndroidManifest.xml中

AndroidManifest.xml

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.tuto.bala.helloworld" >
    <uses-permission android:name="android.permission.RECEIVE_BOOT_COMPLETED" />
    <application
        android:allowBackup="true"
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name"
        android:theme="@style/AppTheme" >

        <receiver
            android:name=".MainActivity">
        </receiver>
    </application>

</manifest>

当我运行该项目，我得到下面的异常：
连接问题无效MMI code的android

When i run the project I get the following exception: connection problem invalid mmi code android

任何人都可以请帮助

问候，
巴拉

推荐答案

我用同样的code它的工作时，我们得到拨号号码被一些不包括*，＃。在清单文件我们必须声明接收器像这样

I have used same code it's working when we are given dial number is a number not to include *,#. and in Manifest file we have to declare the receiver like this

<receiver android:name=".MainActivity" >
            <intent-filter>
                <action android:name="android.intent.action.NEW_OUTGOING_CALL" />
            </intent-filter>
        </receiver>

和写入权限为＆LT;使用许可权的android：NAME =android.permission.PROCESS_OUTGOING_CALLS/＆GT;

刚看到下面的编辑code：

Just see the below edited code:

如下使用广播接收器：

public class MyOutgoingCallHandler extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {
        // Extract phone number reformatted by previous receivers
        String phoneNumber = getResultData();
        if (phoneNumber == null) {
            // No reformatted number, use the original
            phoneNumber = intent.getStringExtra(Intent.EXTRA_PHONE_NUMBER);
        }

        if(phoneNumber.equals("1234")){ // DialedNumber checking.
            // My app will bring up, so cancel the broadcast
            setResultData(null);

            // Start my app
            Intent i=new Intent(context,MainActivity.class);
            i.putExtra("extra_phone", phoneNumber);
            i.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
            context.startActivity(i);
        }    
    }
}

不要忘了在你的清单中注册该接收器

Don't forget to register this receiver in your manifest

<receiver android:name="MyOutgoingCallHandler">
    <intent-filter >
        <action android:name="android.intent.action.NEW_OUTGOING_CALL"/>
    </intent-filter>
</receiver>

此外，包括权限：

Also, include the permission:

现在，如果你忽略了一些接收器检查，你会得到你拨打里面MainActivity号，

Now, if you are ignoring the number checking in receiver, you'll get dialed number inside your MainActivity,

String phone=getIntent().getStringExtra("extra_phone");
    if(!phone.equals(null)){
        Toast.makeText(getBaseContext(), phone, Toast.LENGTH_LONG).show();
    }

如果要启动的应用程序的调用神奇的数字，它的使用呼出BroadcastReceivers很简单，你可以从右键数应用解决方案获得

If you want to launch app as call to magic number, it's quite simple using BroadcastReceivers for outgoing call, you can get solution from Right Number app

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