cakephp authenticate user with repeated entries in the Database table (manual authentication?) - cakephp

I'm creating an authentication system for a group of websites. The problem is that I have to use a pre-existing Database, which has a users table already full of entries, and that one user can have several accounts. Basically, a user has one account per website he has access to (it's not the best way to do this, but I can't change it). Each account is represented by an entry in the users table, with login, password, name... and the important field: website_id. This field tells the system what website that account has access to.
The big problem is that some users with more than one account have the exact same login/password information for all of them. For example, one user has 3 accounts:
account1: login = charly / pwd = 1234 / name = Charles ... website_id = 1
account2: login = charly / pwd = 1234 / name = Charles ... website_id = 2
account3: login = charly / pwd = 1234 / name = Charles ... website_id = 3
So if he goes to the website that has id = 2 and uses those credentials, he's granted access. If he goes to the website that has id = 4, he's denied access.
My problem is that since CakePHP does the login automatically, when a user tries to login, CakePHP checks only the first entry in the Database that matches the login/password submited in the form. So if a user is currently in the website with website_id = 3 and tries to login, Cake finds the first entry (account1), compares its website_id (1 in this case) to the current website's id (3), and since they're different, the access is not granted, but it should. _Please note that the comparison of the website_id vs the account's website_id is already being made manually in the login() function_.
This how the login() function looks like now:
function login() {
$userInfo = $this->Auth->user();
if ( isset($userInfo) ) {
if ($userInfo['User']['website_id'] == $this->website_id) {
//Users gets access to a website that he has an account for
}
else {
//User is denied access because his account is not registered for the current website
$this->Session->destroy();
$this->Session->setFlash(__('You don't have access to this website', true));
$this->redirect($this->Auth->logout());
}
}
}
What I would like is to be able to manually authorize the access to the current website by using the login/password submitted by the user to manually search in the users table, and if I find a match in one of the user accounts, grant the access, or otherwise deny access. To sum up, avoid all the automagic of Auth's component.

If the Auth component's login method fails, control is transferred back to the custom login action (e.g. UsersController::login()). I've used this to authenticate using either username or email address, but it could be easily adapted for this purpose. Same idea, different criteria. I offered what I think is a reasonably thorough response (with code) to a similar question. It may help you as well.

Related

Exception in Site.createExternalUser in Apex RESTclass: Site.ExternalUserCreateException: [That operation is only allowed from within an active site.]

I have a Non-Salesforce Auth System which holds usernames and passwords for a few thousand users. I am willing to migrate these users to Salesforce and give access to these users to my Experience Cloud site. I am developing an apex REST Resource which will take username and password as arguments and create a user with that username and password with a community profile. I am planning to call this API from my Non-Salesforce system and migrate all these users. I am using Site.createExternalUser method in this API. I am getting the exception
Site.ExternalUserCreateException: [That operation is only allowed from within an active site.]
The reason I am using Site.createExternalUser is because I don't want to send the welcome email/reset password email to my users since they already have signed up successfully long ago.
I am open to any alternatives for achiving this.
Below is my code:
#RestResource(urlMapping='/createUser/*')
global with sharing class createUserRestResource {
#HttpPost
global static String doPost(){
Contact con=new Contact();
con.Firstname="First";
con.LastName= "Last";
con.Email="first.last#example.com";
con.AccountId='/Add an account Id here./';
insert con;
usr.Username= "usernameFromRequest#example.com";
usr.Alias= "alias123";
usr.Email= "first.last#example.com";
usr.FirstName= "First";
usr.IsActive= true;
usr.LastName= "Last";
usr.ProfileId='/Community User Profile Id/';
usr.EmailEncodingKey= 'ISO-8859-1';
usr.TimeZoneSidKey= 'America/Los_Angeles';
usr.LocaleSidKey= 'en_US';
usr.LanguageLocaleKey= 'en_US';
usr.ContactId = con.Id;
String userId = Site.createExternalUser(usr, con.AccountId, 'Password#1234', false);
return userId;
}
}
You can suppress sending emails out in whole org (Setup -> Deliverability) or in the Community config there will be way to not send welcome emails (your community -> Workspaces -> Administration -> Emails).
Without running on actual Site I don't think you can pull it off in one go. In theory it's simple, insert contact, then insert user. In practice depends which fields you set on the user. If it's Partner community you might be setting UserRoleId too and that's forbidden. See MIXED DML error. In Customer community you might be safe... until you decide to assign them some permission sets too.
You might need 2 separate endpoints, 1 to create contact, 1 to make user out of it. Or save the contact and then offload user creation to #future/Queueable/something else like that.

Laravel 8 Fortify User UUID Login Problem

I am currently setting up a new project using Laravel 8. Out of the box, Laravel is configured to use auto-incrementing ID's for the user's ID. In the past I have overrode this by doing the following.
Updating the ID column in the user table creation migration to
$table->uuid('id');
$table->primary('id');
Adding the following trait
trait UsesUUID
{
protected static function bootUsesUUID()
{
static::creating(function ($model) {
$model->{$model->getKeyName()} = (string) Str::orderedUuid();
});
}
}
Adding the following to the user model file
use UsesUUID;
public $incrementing = false;
protected $primaryKey = 'id';
protected $keyType = 'uuid';
On this new project, I did the same as above. This seems to break the login functionality. When the email and password are entered and submitted, the form clears as though the page has been refreshed. Thing to note is there are no typical validation error messages returned as would be expected if the email and/or password is wrong.
To check that the right account is actually being found and the password is being checked properly, I added the following code to the FortifyServiceProvider boot method. The log file confirms that the user is found and the user object dump is correct too.
Fortify::authenticateUsing(function(Request $request) {
\Log::debug('running login flow...');
$user = User::where('email', $request->email)->first();
if ($user && Hash::check($request->password, $user->password)) {
\Log::debug('user found');
\Log::debug($user);
return $user;
}
\Log::debug('user not found');
return false;
});
Undoing the above changes to the user model fixes the login problem. However, it introduces a new problem that is the login will be successful but it wont be the right account that is logged in. For example, there are 3 accounts, I enter the credentials for the second or third account, but no matter what, the system will always login using the first account.
Anyone have any suggestions or ideas as to what I may be doing wrong, or if anyone has come across the same/similar issue and how you went about resolving it?
Thanks.
After digging around some more, I have found the solution.
Laravel 8 now stores sessions inside the sessions table in the database. The sessions table has got a user_id column that is a foreign key to the id column in the users table.
Looking at the migration file for the sessions table, I found that I had forgot to change the following the problem.
From
$table->foreignId('user_id')->nullable()->index();
To
$table->foreignUuid('user_id')->nullable()->index();
This is because Laravel 8 by default uses auto incrementing ID for user ID. Since I had modified the ID column to the users table to UUID, I had forgotten to update the reference in the sessions table too.

How to deal with External authentication for already existing local user or new user

I using ASP.Net Core 3 Identity with Identity Server 4 for authentication ...
On the AspNetIdentity template the External Authentication Controller Callback method calls the AutoProvisionUserAsync method which has the following code:
var email = claims.FirstOrDefault(x => x.Type == JwtClaimTypes.Email)?.Value ??
claims.FirstOrDefault(x => x.Type == ClaimTypes.Email)?.Value;
if (email != null) {
filtered.Add(new Claim(JwtClaimTypes.Email, email));
}
var user = new User {
UserName = Guid.NewGuid().ToString(),
};
var identityResult = await _userManager.CreateAsync(user);
Basically it creates a user with a Guid as Username ...
In my database I am using Email as Username ... Is there any reason to use a Guid?
I suppose most External authentication services (Google, Facebook, etc) provides an Email.
So my idea would be:
Check if there is an User in the database already with that email.
If no User exists create one with the email obtained from the External authentication service.
Also add the external authentication to the User in the database;
If there is a User with the email in the database check if it has that External Login.
If the user does not have the external login register and add it.
Does this make sense?
Check if there is an User in the database already with that email.
On callback, first call is to FindUserFromExternalProviderAsync, it search users using nameIdentifier, then if not found there is call to AutoProvisionUserAsync
Basically it creates a user with a Guid as Username ...
In my database I am using Email as Username ... Is there any reason to use a Guid?
The ApplicationUser's base class is IdentityUser, IdentityUser has a prop for ID and one for email by design. thats why most of libraries take advantage of having GUID as ID in addition of email for extensibility. You can use the email for ID if you like to.

How to: CakePHP logging in without password?

I'm trying to find a way to log in user without password.
The reason is that I have phpBB3 forums in my site and the users already log in there. So I'm now building an expansion to the site to have more than just the forum (Using CakePHP). I thought that I could attach automatic account creation to CakePHP when user creates an account to forums (And ofcourse other link for the existing users). So the users would get CakePHP account that has the same username that they have registered in forums. That means that the only way to register to CakePHP part of the site would be to register to the forums first.
Now I'd like to handle the whole logging thing by phpBB3 login so users would still login to forums, and then I'd attach a piece of code that would also login them to CakePHP part of the site with the username they used to login to forums.
This way I could do also put users to their own ACL groups by their status in forums.
Thats what I'm after and I need to know the way to login users this way. I'm not looking for complete code I'm just looking for an answer that explains how I log in users in CakePHP without them having passwords at all.
I have also looked http://bakery.cakephp.org/articles/wilsonsheldon/2009/01/13/phpbb3-api-bridge but it just doesn't quite look what I'm looking for...
As far as I recall, Auth requires two pieces of info for a login.
You can change which fields in the users table are checked by auth with.
$Auth->fields = array(
'username' => 'username',
'password' => 'password'
);
So if you you want to be able to log in users according to their nickname and shoesize:
$Auth->fields = array(
'username' => 'nickname',
'password' => 'shoesize'
);
IMPORTANT:
The AuthComponent expects the password value stored in the database to be hashed instead of being stored in plaintext.
(I think it is a sha1 of the password and Security.salt)
In the above example, if any entries already existed in the database you'd have to overwrite the shoesize field for each of them with hashed versions of the shoesizes.
To generate a hashed password yourself you can use $Auth->password('A Password');
Quick and Dirty
If you fill the password fields in your users table with the return value of:
$Auth->password(null);
Then you can use the following:
$Auth->login(
array(
'User'=>array(
'username'=> USERNAME_FROM_PHPBB3,
'password'=>null
)
)
);
Less Quick and Dirty
When creating a new user.
Set the password field to the md5 hash of some random input.
$this->authUser[$this->User->alias][$Auth->fields['password']] = $Auth->password(md5(rand().rand()));
Use the Username from phpBB3 to retrieve the relevant record
from the users table in the database.
$this->authUser = $this->User->findByUsername( USERNAME_FROM_PHPBB3 );
If the query was successful Log in the user
if($this->authUser){
if($Auth->login($this->authUser)){
// Login Successful
}
}
From your cakephp app you can check if a user exist in the phpbb forums table and you can use the phpbb session to check if a user is logged in.
This function will solve your problem:
public function forceLogin($userName = NULL) {
$this->_setDefaults();
$this->User = ClassRegistry::init('User');
$this->User->recursive = 0;
$user = $this->User->findByUsername($userName);
if (!empty($user['User'])) {
$this->Session->renew();
$user['User']['id'] = null;
$user['User']['password'] = null;
$this->Session->write(self::$sessionKey, $user['User']);
}
return $this->loggedIn();
}

app on GAE | Restricted Google account authentications | listed google accounts not all |

I am quite new to this Google app engine. I am learning things every day.
I am have a forum on google app engine.
But I want is to be having private or restricted parts.
Some features should be locked for certain google account users, which are in some sort of access control list.
I plain words I can say, only those user who are in list of access can see the forum rest will be redirect to the "contact to admin" page.
As I am new I wanna know that is it something possible.
if yes, how can I achieve it ?
Thanks,
Alok
If you are using the built-in Users API, you can check users.is_current_user_admin() as an access control mechanism. Administrators can be managed via the dashboard.
If you need more granular, application-specific authorization logic, generally you would create a User model in the datastore that references the built-in UserProperty and also holds a list of roles or whatever else you need to check authorization.
To follow up Drew's reply, I use a similar system in my app, so my server code has something like the following class definition (simplified here for clarity)
class myUser(db.Model):
user = db.UserProperty(required=True)
rights = db.StringProperty(required=True, choices=set(["public", "private"]))
created = db.DateTimeProperty(auto_now_add=True)
lastaccess = db.DateTimeProperty(auto_now=True)
and then I have code like this where I handle queries
def checkUserRights(user):
q = db.GqlQuery("SELECT * from myUser WHERE user = :1", user)
u = q.get()
if not u:
# create a new 'public access' user if we haven't seen this person before
u = myUser(user=user, rights="public")
# always update the user record after the source is fetched (updates the lastaccess field)
db.put( u )
return u.rights
rights = checkUser(users.get_current_user())
if isPrivateArea and rights == "private":
....
This way I create a user for EVERY visitor, and then I have an admin interface to change the rights of selected users - you may decide, for example, to not create a record for every visitor
def checkUserRights(user):
q = db.GqlQuery("SELECT * from myUser WHERE user = :1", user)
u = q.get()
if not u:
# grant default public rights to anyone...
return "public"
# always update the user record after the source is fetched (updates the lastaccess field)
db.put( u )
return u.rights
This is, of course, on a page where the app.yaml specifies "login: required"

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