这让我发疯,当我拨打电话时,应用程序崩溃了。
MainActivity类:
import android.os.Bundle;
import android.widget.Toast;
import android.app.Activity;
import android.content.Intent;
import android.content.IntentFilter;
public class MainActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Toast.makeText(getBaseContext(), "started", Toast.LENGTH_LONG).show();
}
}
OutgoingReceiver BroadcastReceiver:
import android.content.BroadcastReceiver;
import android.content.Context;
import android.content.Intent;
import android.content.IntentFilter;
import android.widget.Toast;
class OutgoingReceiver extends BroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
String number = intent.getStringExtra(Intent.EXTRA_PHONE_NUMBER);
Toast.makeText(context, "Outgoing: "+number, Toast.LENGTH_LONG).show();
}
}
The Manifest:
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.secret"
android:versionCode="1"
android:versionName="1.0" >
<uses-permission android:name="android.permission.PROCESS_OUTGOING_CALLS"/>
<uses-sdk
android:minSdkVersion="12"
android:targetSdkVersion="19" />
<application
android:allowBackup="true"
android:icon="@drawable/ic_launcher"
android:label="@string/app_name">
<activity
android:name=".MainActivity"
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=".OutgoingReceiver" >
<intent-filter>
<action android:name="android.intent.action.NEW_OUTGOING_CALL" />
</intent-filter>
</receiver>
</application>
</manifest>
更新: Logcat输出
07-21 12:12:52.160: E/AndroidRuntime(3654): FATAL EXCEPTION: main
07-21 12:12:52.160: E/AndroidRuntime(3654): Process: com.example.secret, PID: 3654
07-21 12:12:52.160: E/AndroidRuntime(3654): java.lang.RuntimeException: Unable to instantiate receiver com.example.secret.OutgoingReceiver: java.lang.IllegalAccessException: access to class not allowed
07-21 12:12:52.160: E/AndroidRuntime(3654): at android.app.ActivityThread.handleReceiver(ActivityThread.java:2400)
07-21 12:12:52.160: E/AndroidRuntime(3654): at android.app.ActivityThread.access$1700(ActivityThread.java:135)
07-21 12:12:52.160: E/AndroidRuntime(3654): at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1272)
07-21 12:12:52.160: E/AndroidRuntime(3654): at android.os.Handler.dispatchMessage(Handler.java:102)
07-21 12:12:52.160: E/AndroidRuntime(3654): at android.os.Looper.loop(Looper.java:136)
07-21 12:12:52.160: E/AndroidRuntime(3654): at android.app.ActivityThread.main(ActivityThread.java:5017)
也许它有些愚蠢,但我自己也看不到。 有什么想法吗?
答案 0 :(得分:2)
从您的日志中
无法实例化接收器com.example.secret.OutgoingReceiver:java.lang.IllegalAccessException:不允许访问类
问题是您的OutgoingReceiver
没有访问修饰符,使其对包外的所有内容都不可见。
要让Android实例化并与您的接收器进行交互,它需要公开。
答案 1 :(得分:1)
将接收者类公开