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根据运行时权限文档，应用程序可以检查运行时权限并在尚未授予权限的情况下请求权限。所以我正在编写一个程序来一次处理运行时多个权限。当用户单击拒绝并选中“不再询问”复选框时，我做得正确但无法处理。在这种情况下，我想打开权限意图，如图 2 所示。当用户下次打开我的应用程序时。

当用户选中“不再询问”复选框并单击拒绝按钮时，我无法检测到这种情况。

在此处输入图像描述 在此处输入图像描述

这是我的代码：

final private int REQUEST_CODE_ASK_MULTIPLE_PERMISSIONS = 124;

private void insertDummyContactWrapper() {
    List<String> permissionsNeeded = new ArrayList<String>();

    final List<String> permissionsList = new ArrayList<String>();
    if (!addPermission(permissionsList, Manifest.permission.ACCESS_FINE_LOCATION))
        permissionsNeeded.add("GPS");
    if (!addPermission(permissionsList, Manifest.permission.READ_CONTACTS))
        permissionsNeeded.add("Read Contacts");
    if (!addPermission(permissionsList, Manifest.permission.WRITE_CONTACTS))
        permissionsNeeded.add("Write Contacts");

    if (permissionsList.size() > 0) {
        if (permissionsNeeded.size() > 0) {
            // Need Rationale
            String message = "You need to grant access to " + permissionsNeeded.get(0);
            for (int i = 1; i < permissionsNeeded.size(); i++)
                message = message + ", " + permissionsNeeded.get(i);
            showMessageOKCancel(message,
                    new DialogInterface.OnClickListener() {
                        @Override
                        public void onClick(DialogInterface dialog, int which) {
                            requestPermissions(permissionsList.toArray(new String[permissionsList.size()]),
                                    REQUEST_CODE_ASK_MULTIPLE_PERMISSIONS);
                        }
                    });
            return;
        }
        requestPermissions(permissionsList.toArray(new String[permissionsList.size()]),
                REQUEST_CODE_ASK_MULTIPLE_PERMISSIONS);
        return;
    }

    insertDummyContact();
}

private boolean addPermission(List<String> permissionsList, String permission) {
    if (checkSelfPermission(permission) != PackageManager.PERMISSION_GRANTED) {
        permissionsList.add(permission);
        // Check for Rationale Option
        if (!shouldShowRequestPermissionRationale(permission))
            return false;
    }
    return true;
}

and here is my callback method

@Override
public void onRequestPermissionsResult(int requestCode, String[] permissions, int[] grantResults) {
    switch (requestCode) {
        case REQUEST_CODE_ASK_MULTIPLE_PERMISSIONS:
            {
            Map<String, Integer> perms = new HashMap<String, Integer>();
            // Initial
            perms.put(Manifest.permission.ACCESS_FINE_LOCATION, PackageManager.PERMISSION_GRANTED);
            perms.put(Manifest.permission.READ_CONTACTS, PackageManager.PERMISSION_GRANTED);
            perms.put(Manifest.permission.WRITE_CONTACTS, PackageManager.PERMISSION_GRANTED);
            // Fill with results
            for (int i = 0; i < permissions.length; i++)
                perms.put(permissions[i], grantResults[i]);
            // Check for ACCESS_FINE_LOCATION
            if (perms.get(Manifest.permission.ACCESS_FINE_LOCATION) == PackageManager.PERMISSION_GRANTED
                    && perms.get(Manifest.permission.READ_CONTACTS) == PackageManager.PERMISSION_GRANTED
                    && perms.get(Manifest.permission.WRITE_CONTACTS) == PackageManager.PERMISSION_GRANTED) {
                // All Permissions Granted
                insertDummyContact();
            } else {
                // Permission Denied
                Toast.makeText(MainActivity.this, "Some Permission is Denied", Toast.LENGTH_SHORT)
                        .show();
            }
            }
            break;
        default:
            super.onRequestPermissionsResult(requestCode, permissions, grantResults);
    }
}
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1 回答 1

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不要同时请求多个权限。这使用户体验不愉快。如果用户单击'Do not ask again'，由于第一次未显示该复选框，他可能不想使用您的应用程序的该功能。但是，如果他确实尝试使用它，只需显示一个 toast 或小吃吧，说明该功能无法使用，因为用户尚未授予权限。

目前无法以编程方式打开权限屏幕，如果您尝试，您将获得安全异常。

于 2016-02-15T15:13:58.240 回答