googlesamples之easypermissions使用

使用

1. app/build.gradle

dependencies {    compile 'pub.devrel:easypermissions:0.3.0'}

2. 在Activity / Fragment实现PermissionCallbacks，回调方法：

@Overridepublic void onPermissionsGranted(int requestCode, List<String> list) {    // Some permissions have been granted    // ...    LogUtil.d("Some permissions have been granted=" + requestCode);}@Overridepublic void onPermissionsDenied(int requestCode, List<String> list) {    // Some permissions have been denied    // ...    LogUtil.d("Some permissions have been denied=" + requestCode);}

3. 在需要使用到权限之前调用methodRequiresTwoPermission方法，如相机和发送短信：

@AfterPermissionGranted(1010)private void methodRequiresTwoPermission() {    String[] perms = {Manifest.permission.CAMERA, SEND_SMS};    if (EasyPermissions.hasPermissions(this, perms)) {        // Already have permission, do the thing        // ...        LogUtil.d("Already have permission, do the thing");    } else {        // Do not have permissions, request them now        LogUtil.d("Do not have permissions, request them now");        EasyPermissions.requestPermissions(this, "camera_and_send_sms", 1010, perms);    }}

4. 还是要授权回调：

@Overridepublic void onRequestPermissionsResult(int requestCode, String[] permissions, int[] grantResults) {    super.onRequestPermissionsResult(requestCode, permissions, grantResults);    // Forward results to EasyPermissions    EasyPermissions.onRequestPermissionsResult(requestCode, permissions, grantResults, this);}

这样就over了。

源码地址

https://github.com/googlesamples/easypermissions ，点击底部「阅读原文」直达。

一点思考

就在《Android 6.0 RuntimePermission》文章里，我对Android 6.0 运行权限做了个简单封装，但没有考虑到Fragment使用和勾上不再询问应该去setting，还一个问题，封装只是针对单个权限，推荐官方easypermissions库，来看看easypermissions是如何解决我未考虑的问题。

Fragment使用

直接提供了Fragment可以调用的方法：

@SuppressLint("NewApi")public static void requestPermissions(@NonNull Fragment fragment,                                      @NonNull String rationale,                                      @StringRes int positiveButton,                                      @StringRes int negativeButton,                                      int requestCode,                                      @NonNull String... perms) {    if (hasPermissions(fragment.getContext(), perms)) {        notifyAlreadyHasPermissions(fragment, requestCode, perms);        return;    }    if (shouldShowRationale(fragment, perms)) {        RationaleDialogFragmentCompat                .newInstance(positiveButton, negativeButton, rationale, requestCode, perms)                .show(fragment.getChildFragmentManager(), DIALOG_TAG);    } else {        fragment.requestPermissions(perms, requestCode);    }}

不再询问去setting

我在《Android 6.0 RuntimePermission》说当勾上“不再询问”，只能选择拒绝，再次进入，shouldShowRequestPermissionRationale方法始终false，当onPermissionsDenied方法回调时，肯定是都是拒绝，再根据shouldShowRequestPermissionRationale方法不就知道是否勾上了“不再询问”，我咋没有想到这点呢：

@Overridepublic void onPermissionsDenied(int requestCode, List<String> perms) {    Log.d(TAG, "onPermissionsDenied:" + requestCode + ":" + perms.size());    // (Optional) Check whether the user denied any permissions and checked "NEVER ASK AGAIN."    // This will display a dialog directing them to enable the permission in app settings.    if (EasyPermissions.somePermissionPermanentlyDenied(this, perms)) {        new AppSettingsDialog.Builder(this).build().show();    }}@Overridepublic void onActivityResult(int requestCode, int resultCode, Intent data) {    super.onActivityResult(requestCode, resultCode, data);    if (requestCode == AppSettingsDialog.DEFAULT_SETTINGS_REQ_CODE) {        // Do something after user returned from app settings screen, like showing a Toast.        Toast.makeText(this, R.string.returned_from_app_settings_to_activity, Toast.LENGTH_SHORT)                .show();    }}

EasyPermissions的somePermissionPermanentlyDenied方法，就是根据shouldShowRequestPermissionRationale方法判断是否勾上了“不再询问”。

public static boolean somePermissionPermanentlyDenied(@NonNull Activity activity,                                                      @NonNull List<String> deniedPermissions) {    for (String deniedPermission : deniedPermissions) {        if (permissionPermanentlyDenied(activity, deniedPermission)) {            return true;        }    }    return false;}public static boolean permissionPermanentlyDenied(@NonNull Activity activity,                                                  @NonNull String deniedPermission) {    return !shouldShowRequestPermissionRationale(activity, deniedPermission);}

跳转setting代码：

Intent intent = new Intent(Settings.ACTION_APPLICATION_DETAILS_SETTINGS);Uri uri = Uri.fromParts("package", mContext.getPackageName(), null);intent.setData(uri);// Start for result//noinspection NewApi The Builder constructor prevents thisstartForResult(intent);

封装只针对单个权限

我应该封装多个权限，多权限是个数组，如果只有一个元素，不就是单个权限嘛！

转自：http://mp.weixin.qq.com/s/RKFP6rAfWuNcaATzxDmFlQ