【发布时间】:2016-09-28 06:10:26
【问题描述】:
我正在学习开发 Android 移动应用程序。我做了一个简单的应用程序来发送短信。我正在使用运行棉花糖的 ADB 模拟器。我已经阅读了关于它的新权限 here 并发现了很多与 answers 相关的问题,我的问题没有解决。
当我执行 CLEAN 和 BUILD 时,会收到以下消息:
Error:Gradle: java.io.FileNotFoundException: C:Examples\MySMSApplication\app\build\intermediates\res\resources-debug.ap_ (Access is denied)
注意:当我删除 uses-permission android:name="android.permission.SEND_SMS 时,我能够成功构建和运行应用程序。但是,当我尝试发送短信说另一个模拟器号码 5554 时,我总是会收到我的消息 Permission denied。
我有设置了两个用户权限的 Android 清单。
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android" package="com.example.fisher.app">
<uses-permission android:name="android.permission.SEND_SMS"></uses-permission>
<uses-permission android:name="android.permission.RECEIVE_SMS"></uses-permission>
<application
android:allowBackup="true"
android:icon="@mipmap/ic_launcher"
android:label="@string/mysmsapp"
android:supportsRtl="true"
android:theme="@style/AppTheme">
<activity android:name=".MainActivity">
<intent-filter>
<action android:name="android.intent.action.MAIN"/>
<category android:name="android.intent.category.LAUNCHER"/>
</intent-filter>
</activity>
</application>
</manifest>
然后在mainActivity中
public class MainActivity extends AppCompatActivity {
Button buttonSend;
EditText textPhoneNo;
EditText textSMS;
private static final int PERMISSION_REQUEST = 100;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
buttonSend = (Button) findViewById(R.id.buttonSend);
textPhoneNo = (EditText) findViewById(R.id.editTextPhoneNo);
textSMS = (EditText) findViewById(R.id.editTextSMS);
buttonSend.setOnClickListener(new View.OnClickListener() {
//this is the onclick listener of send button
@Override
public void onClick(View v) {
String phoneNo = textPhoneNo.getText().toString();
String sms = textSMS.getText().toString();
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
if (checkSelfPermission(Manifest.permission.SEND_SMS) != PackageManager.PERMISSION_GRANTED) {
if (shouldShowRequestPermissionRationale(Manifest.permission.SEND_SMS)) {
Snackbar.make(findViewById(R.id.myView), "You need to grant SEND SMS permission to send sms",
Snackbar.LENGTH_LONG).setAction("OK", new View.OnClickListener() {
@Override
public void onClick(View v) {
requestPermissions(new String[]{Manifest.permission.SEND_SMS}, PERMISSION_REQUEST);
}
}).show();
} else {
requestPermissions(new String[]{Manifest.permission.SEND_SMS}, PERMISSION_REQUEST);
}
} else {
sendSMS(phoneNo, sms);
}
} else {
sendSMS(phoneNo, sms);
}
}
});
}
private void sendSMS(String phoneNo, String message) {
SmsManager sms = SmsManager.getDefault();
sms.sendTextMessage(phoneNo, null, message, null, null);
}
@Override
public void onRequestPermissionsResult(int requestCode, String[] permissions, int[] grantResults) {
super.onRequestPermissionsResult(requestCode, permissions, grantResults);
if (grantResults[0] == PackageManager.PERMISSION_GRANTED) {
Snackbar.make(findViewById(R.id.myView), "Permission Granted",
Snackbar.LENGTH_LONG).show();
sendSMS(textPhoneNo.getText().toString(), textSMS.getText().toString());
} else {
Snackbar.make(findViewById(R.id.MyView), "Permission denied",
Snackbar.LENGTH_LONG).show();
}
}
}
这是 ref 的 Gradle 构建:
apply plugin: 'com.android.application'
android {
compileSdkVersion 23
buildToolsVersion "23.0.3"
defaultConfig {
applicationId "com.example.fisher.myapp"
minSdkVersion 23
targetSdkVersion 23
versionCode 1
versionName "1.0"
}
buildTypes {
release {
minifyEnabled false
proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
}
}
}
dependencies {
compile fileTree(include: ['*.jar'], dir: 'libs')
testCompile 'junit:junit:4.12'
compile 'com.android.support:appcompat-v7:23.4.0'
compile 'com.android.support:design:23.1.1'
}
【问题讨论】:
-
从 API 24 开始,您需要在运行时请求许可:developer.android.com/training/permissions/requesting.html
-
@AmanGrover 我注意到了相同的链接,所以当它说运行时不是我的 onClick 事件在做什么时?其次,我无法构建它。我相信在将应用程序加载到模拟器之前就从 Gradle 得到了错误。
标签: java android android-6.0-marshmallow access-denied