用于呼出电话号码检测的Android应用程序正在崩溃
这让我发疯,当我拨出电话时,应用程序崩溃了 主要活动类别:用于呼出电话号码检测的Android应用程序正在崩溃,android,detection,phone-number,phone-call,Android,Detection,Phone Number,Phone Call,这让我发疯,当我拨出电话时,应用程序崩溃了 主要活动类别: 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 v
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();
}
}
外接接收器广播接收器:
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();
}
}
舱单:
<?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>
也许这有点愚蠢,但我自己却看不出来。
有什么想法吗?从日志中将receiver类公开: 无法实例化receiver 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,使它对包外的所有内容都不可见
Android要实例化并与你的接收器交互,它需要公开。你的logcat怎么说?等等,我来看看。好的,logcat输出添加了。你是对的,即使Tanis.7x回答正确,我也会接受你的答案,因为你是第一个
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)