根据这个:http://developer.android.com/preview/features/runtime-permissions.html#coding一个应用程序可以检查运行时权限和请求权限，如果它还没有被授予。弹出如下对话框:
如果用户拒绝一个重要的权限，在我看来，应用程序应该显示一个解释为什么需要权限和什么影响拒绝。该对话框有两个选项:
重试(再次请求许可) 拒绝(应用程序将工作没有该许可)。
但是，如果用户选中“Never ask again”，则不应该显示带有解释的第二个对话框，特别是如果用户之前已经拒绝了一次。 现在的问题是:我的应用程序如何知道用户是否选中了Never ask again?IMO onRequestPermissionsResult(int requestCode, String[] permissions, int[] grantResults)没有给我这个信息。
第二个问题是:谷歌是否计划在权限对话框中包含一个自定义消息，以解释为什么应用程序需要权限?这样就不会出现第二个对话框，这肯定会带来更好的用户体验。
当前回答
您可以使用if (ActivityCompat。shouldShowRequestPermissionRationale(this, Manifest.permission.CAMERA)方法来检测never ask是否被检查。
更多参考:检查这个
检查是否使用了多个权限:
if (ActivityCompat.shouldShowRequestPermissionRationale(this, Manifest.permission.CAMERA)
|| ActivityCompat.shouldShowRequestPermissionRationale(this, Manifest.permission.WRITE_EXTERNAL_STORAGE)
|| ActivityCompat.shouldShowRequestPermissionRationale(this, Manifest.permission.ACCESS_FINE_LOCATION)
|| ActivityCompat.shouldShowRequestPermissionRationale(this, Manifest.permission.RECORD_AUDIO)) {
showDialogOK("Service Permissions are required for this app",
new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
switch (which) {
case DialogInterface.BUTTON_POSITIVE:
checkAndRequestPermissions();
break;
case DialogInterface.BUTTON_NEGATIVE:
// proceed with logic by disabling the related features or quit the app.
finish();
break;
}
}
});
}
//permission is denied (and never ask again is checked)
//shouldShowRequestPermissionRationale will return false
else {
explain("You need to give some mandatory permissions to continue. Do you want to go to app settings?");
// //proceed with logic by disabling the related features or quit the app.
}
解释()方法
private void explain(String msg){
final android.support.v7.app.AlertDialog.Builder dialog = new android.support.v7.app.AlertDialog.Builder(this);
dialog.setMessage(msg)
.setPositiveButton("Yes", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface paramDialogInterface, int paramInt) {
// permissionsclass.requestPermission(type,code);
startActivity(new Intent(android.provider.Settings.ACTION_APPLICATION_DETAILS_SETTINGS, Uri.parse("package:com.exampledemo.parsaniahardik.marshmallowpermission")));
}
})
.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface paramDialogInterface, int paramInt) {
finish();
}
});
dialog.show();
}
上面的代码还将显示对话框，这将重定向到用户的应用程序设置屏幕，在那里他可以给予权限，如果选中了“永不再问”按钮。
其他回答
shouldShowRequestPermissionRationale根据之前权限请求中的用户首选项返回true或false。
如果用户只是拒绝权限(不是永远)shouldShowRequestPermissionRationale将返回true。如果永远拒绝许可，则返回false。窍门是即使用户允许了权限shouldShowRequestPermissionRationale也会返回false。
因此，我们可以结合这两个条件来得到永不再问的选择与否。
因此，如果用户不允许权限，并且shouldShowRequestPermissionRationale返回false，那么这意味着用户选择永远不会再次请求权限。
https://stackoverflow.com/a/58114769/5151336
可能对某人有用:——
我注意到的是，如果我们在onRequestPermissionsResult()回调方法中检查shouldshowrequestpermissionration理()标志，它只显示两种状态。
状态1:-返回true:——任何时候用户单击Deny permissions(包括第一次)。
状态2:-返回false:-如果用户选择“永不再问”。
详细工作实例链接
我在Android m中写了一个权限请求的简写，这段代码还处理了对旧Android版本的向后兼容性。
所有丑陋的代码都被提取到一个片段中，该片段将自己附加到请求权限的活动上。PermissionRequestManager的使用方法如下:
new PermissionRequestManager()
// We need a AppCompatActivity here, if you are not using support libraries you will have to slightly change
// the PermissionReuqestManager class
.withActivity(this)
// List all permissions you need
.withPermissions(android.Manifest.permission.CALL_PHONE, android.Manifest.permission.READ_CALENDAR)
// This Runnable is called whenever the request was successfull
.withSuccessHandler(new Runnable() {
@Override
public void run() {
// Do something with your permissions!
// This is called after the user has granted all
// permissions, we are one a older platform where
// the user does not need to grant permissions
// manually, or all permissions are already granted
}
})
// Optional, called when the user did not grant all permissions
.withFailureHandler(new Runnable() {
@Override
public void run() {
// This is called if the user has rejected one or all of the requested permissions
L.e(this.getClass().getSimpleName(), "Unable to request permission");
}
})
// After calling this, the user is prompted to grant the rights
.request();
来看看:https://gist.github.com/crysxd/385b57d74045a8bd67c4110c34ab74aa
你可以听漂亮。
侦听器
interface PermissionListener {
fun onNeedPermission()
fun onPermissionPreviouslyDenied(numberDenyPermission: Int)
fun onPermissionDisabledPermanently(numberDenyPermission: Int)
fun onPermissionGranted()
}
MainClass的权限
class PermissionUtil {
private val PREFS_FILENAME = "permission"
private val TAG = "PermissionUtil"
private fun shouldAskPermission(context: Context, permission: String): Boolean {
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
val permissionResult = ActivityCompat.checkSelfPermission(context, permission)
if (permissionResult != PackageManager.PERMISSION_GRANTED) {
return true
}
}
return false
}
fun checkPermission(context: Context, permission: String, listener: PermissionListener) {
Log.i(TAG, "CheckPermission for $permission")
if (shouldAskPermission(context, permission)) {
// Load history permission
val sharedPreference = context.getSharedPreferences(PREFS_FILENAME, 0)
val numberShowPermissionDialog = sharedPreference.getInt(permission, 0)
if (numberShowPermissionDialog == 0) {
(context as? Activity)?.let {
if (ActivityCompat.shouldShowRequestPermissionRationale(it, permission)) {
Log.e(TAG, "User has denied permission but not permanently")
listener.onPermissionPreviouslyDenied(numberShowPermissionDialog)
} else {
Log.e(TAG, "Permission denied permanently.")
listener.onPermissionDisabledPermanently(numberShowPermissionDialog)
}
} ?: kotlin.run {
listener.onNeedPermission()
}
} else {
// Is FirstTime
listener.onNeedPermission()
}
// Save history permission
sharedPreference.edit().putInt(permission, numberShowPermissionDialog + 1).apply()
} else {
listener.onPermissionGranted()
}
}
}
以这种方式使用
PermissionUtil().checkPermission(this, Manifest.permission.ACCESS_FINE_LOCATION,
object : PermissionListener {
override fun onNeedPermission() {
log("---------------------->onNeedPermission")
// ActivityCompat.requestPermissions(this@SplashActivity,
// Array(1) { Manifest.permission.ACCESS_FINE_LOCATION },
// 118)
}
override fun onPermissionPreviouslyDenied(numberDenyPermission: Int) {
log("---------------------->onPermissionPreviouslyDenied")
}
override fun onPermissionDisabledPermanently(numberDenyPermission: Int) {
log("---------------------->onPermissionDisabled")
}
override fun onPermissionGranted() {
log("---------------------->onPermissionGranted")
}
})
在activity或fragmnet中覆盖onRequestPermissionsResult
override fun onRequestPermissionsResult(requestCode: Int, permissions: Array<out String>, grantResults: IntArray) {
if (requestCode == 118) {
if (permissions[0] == Manifest.permission.ACCESS_FINE_LOCATION && grantResults[0] == PackageManager.PERMISSION_GRANTED) {
getLastLocationInMap()
}
}
}
请不要为这个解决办法向我扔石头。
这是可行的，但有点“俗气”。
当调用requestPermissions时，注册当前时间。
mAskedPermissionTime = System.currentTimeMillis();
然后在onRequestPermissionsResult中
如果结果不允许，请再次检查时间。
if (System.currentTimeMillis() - mAskedPermissionTime < 100)
由于用户不可能那么快地点击拒绝按钮，我们知道他选择了“永不再问”，因为回调是即时的。
使用风险自负。
