Dynamically created datasource not being passed to associated models in CakePHP - php

I have a Model - Car - the Car has several associated models, lets consider one of them which is linked with the hasMany relationship - Wheel
In my CarsController, I dynamically generate a datasource using the following code -
$schemaName = $this->Session->read('User.schema');
$config = ConnectionManager::getDataSource('default')->config;
$config['database'] = $schemaName;
ConnectionManager::create($schemaName, $config);
Then I set this datasource in my Car Model using the following line of code
$this->Car->setDataSource($schemaName);
After this I am able to query and operate on Car, however, if I try to operate on Wheel using the following statements - I get an error
$this->Car->Wheel->create();
$this->Car->Wheel->save($wheelData);
The error I get is -
Error: [MissingTableException] Table wheels for model Wheel was not found in datasource default.
For some reason the datasource is not being passed from Parent model to associated child models. If I explicitly set the datasource in Wheel using the following line then everything works fine.
$this->Car->Wheel->setDataSource($schemaName);
Can anyone help shed some light on this behavior and how to fix this? The reason I find this inconvenient is that my parent model has several associated models (which further have associated models) and setting datasource individually on each of them doesnt sound right.
Side Question - is there a way to check if a datasource already exists before trying to create one dynamically? I have a for-loop that wraps this entire code and each loop iteration will end up creating a new datasource
I am using CakePHP 2.5.4

Here is some code which will pass the setDatasource() call to associated models, if you want only the Car model to pass through its datasource put this code in the Car model, if you want all models to pass through, put it in AppModel.php
public function setDatasource($datasource = null) {
foreach (array_keys($this->getAssociated()) as $modelName) {
$this->{$modelName}->setDatasource($datasource);
}
parent::setDatasource($datasource);
}
To answer your comment, I would add public $dynamicDb = false in AppModel.php, meaning false would be the default value for this variable, then in your models override it by adding public $dynamicDb = true, then change the above function to:
public function setDatasource($datasource = null) {
foreach (array_keys($this->getAssociated()) as $modelName) {
if ($this->{$modelName}->dynamicDb === true) {
$this->{$modelName}->setDatasource($datasource);
}
}
parent::setDatasource($datasource);
}
(I haven't tested this amended function as I'm not on my dev PC right now, but its a fairly simple change and it should work)
To check if a datasource already exists before you create one, I can see two possible methods, one is by calling ConnectionManager::getDatasource($schemaName) and catching the exception if the datasource does not exist, or call ConnectionManager::sourceList() and check if your $schemaName is in the returned array, here's an implementation of the first option:
$schemaName = $this->Session->read('User.schema');
try {
ConnectionManager::getDatasource($schemaName)
} catch (MissingDatasourceException $e) {
$config = ConnectionManager::getDataSource('default')->config;
$config['database'] = $schemaName;
ConnectionManager::create($schemaName, $config);
}
and the second option:
$datasources = ConnectionManager::sourceList();
$schemaName = $this->Session->read('User.schema');
if (!in_array($schemaName, $datasources)) {
$config = ConnectionManager::getDataSource('default')->config;
$config['database'] = $schemaName;
ConnectionManager::create($schemaName, $config);
}
Hope this helps

Related

Make model dynamically using user supplied model name in laravel

I am stuck with it and i couldn't find any appropriate solution for this. what i want to achieve is in my admin panel based upon check box value i want to change active status of specific control and update database value using ajax. I want to make a common ajax function in controller to avoid repeatedly writing same ajax function for other controls like menu manager content manager, documents manager etc.So i want to send Model name to the ajax controller so that same function can be used. Ajax call is working perfectly but couldn't make appropriate models. for example:
$m = new App\Model.$request->model OR $m = 'App\Model\'.$request->model (adding last \ gives an error) or answer provided in Dynamically use model in laravel is not helping either. Is there any better ways if yes please suggest. I can do this below but it is hardcoded so i want to make model dynamic models
if($request->model ==='Menu')
$model = new \App\Http\Models\Menu;
else if($request->model === 'News')
$this->model = new \App\Http\Models\News;
else if($request->model === 'Document')
$this->model = new \App\Http\Models\Document;
Thankyou !!!
You can just use:
$modelName = '\\App\\Http\\Models\\'.$request->model;
$this->model = new $modelName;
However you should add validation to make sure only some allowed models would be used or at least do something like this:
if (! in_array($request->model, ['Menu', 'News', 'Document']))
{
throw new \Exception('Invalid model');
}
$modelName = '\\App\\Http\\Models\\'.$request->model;
$this->model = new $modelName;
This is because you don't want to expose probably all the models data for security reasons.
Try the below code
if($request->model){
$m = '\App'. '\' .$request->model;
//other code
}

Cakephp 3 newEntity() not working

I have to maintenance the cakephp 3 project and in this project, they have create newEntity using array variable and then data are not save into database. I set print_r to get result of userModel but it doesn't show anything.
I was using nginx server (not apache), mysql 5.6, cakephp 3.
Here is the code what they used.
$userModel = $this->newEntity($tableData);
print_r($userModel);
if (!$userModel->errors()) {
$result = $this->save($userModel);
if ($result)
{
return $result->id;
}
} else {
return false;
}
There are two things you need to check first.
Try outputting the $tableData variable and ensuring the elements in it match up against the data structure you have for the model object you're trying to create.
To manipulate the entity. You should be calling it using
$this->ModelName->newEntity($tableData);
$this->ModelName->save($userModel)
You need call the model name to generate a new entity and to save it.
If your model name are $tableData:
Replace your $this->newEntity($tableData) to $this->$tableData->newEntity();
and your $this->save($userModel) to $this->$tableData->save($userModel);

How do I connect to different databases at run time?

I am making a multi-tenant multi-database app that has one central database and many sub-databases.
The app creates an instance of an organisation in the central database, and to each organisation it creates a sub-database with different tables.
To achieve this, I have made a class class Setup that
Creates the Organisation
Creates its Database
Configures the connection to that database and connects to it
Runs the proper migrations to it.
All wrapped up in a constructor, so upon caling Setup::create all of this runs properly.
Most of the database configuration and connection are inspiried from this tutorial.
To test whether my logic goes as wanted, I interacted with my application via :
Tinker
Web Browser.
To my suprise, the outcome is different in both cases, and never as wanted as far as connecting to another database is concerned.
Interaction with tinker :
After creating calling my Setup::create and having output telling me everything went okay, I try to check for myself what database am I on right now Using:
DB::connection()->getDatabaseName()
It outputs the sub-database name we have just created for the organisation and connected to, which is logical and going accordingly.
However, I attempt to connect to another database by creating a new configuration for it and then connecting to it with the DB methods I have provided, it does not work, I am still on the sub-database I was on.
Interacting with the browser :
This time, having my Setup::create wrapped up properly in my controller's code, I attempt to test everything again, I also made a line in my layout to output me the current database :
<?php echo DB::connection()->getDatabaseName() ?>
At first, while I am still on the central database, its name appears, however after calling Setup::create, it switches to the sub-database -Which is expected- but then, after one refresh, I am on the central database again -Which is totally Unexpected-
So, what happens here? and how do I get to connect to all of my different databases how I wish when I wish?
Extra:
Testing in tinker, I have went to the point where I have commented out the migration code, and left the creation of the database and also the connection to it.
To my suprise, it does not connect to the database.
so I started thinking that the migration code has something to do with connecting to the database, or maybe tinker has different behaviors I completely ingore.
Important:
I have came across threads where solutions using QueryBuilders were mentioned
Please, do not provide such answers because my aim is to switch databases entirely to the point where I can use eloquent model's events with no problem.
By that I mean, I want to be able to use Model::create after having connected to the database instead of DB::connection()->....
Technical details:
I am using Laravel 5 with mysql-server, on Ubuntu Machine.
I stumbled upon this question and it had my answer.
I made a class called DatabaseConnection:
class DatabaseConnection extends Model
{
static $instances=array();
protected $database;
protected $connection;
public function __construct($options = null)
{
// Set the database
$database = $options['database'];
$this->database = $database;
// Figure out the driver and get the default configuration for the driver
$driver = isset($options['driver']) ? $options['driver'] : Config::get("database.default");
$default = Config::get("database.connections.$driver");
// Loop through our default array and update options if we have non-defaults
foreach($default as $item => $value)
{
$default[$item] = isset($options[$item]) ? $options[$item] : $default[$item];
}
$capsule = new Capsule;
$capsule->addConnection($default);
$capsule->setEventDispatcher(new Dispatcher(new Container));
$capsule->setAsGlobal();
$capsule->bootEloquent();
// Create the connection
$this->connection = $capsule->getConnection();
DatabaseConnection::$instances[] = $capsule;
return $this->connection;
}
}
So, whenever I am in a controller that manipulates tables of a sub-database, I simply go this way:
public function RandomActionInMyController()
{
$db_connection = new DatabaseConnection(['database' => 'name_of_db']);
$someModel = new Model/Model::find()..// Basically anything
return myreturnstuff;
}
Extra Bonus:
The use of the static attribute $instances in my DatabaseConnection
boils down to retrieving my latest database connection for ease uses.
For example, if I ever wanted to retrieve it, it would be wrapped in a function such as
function CurrentOrLatestDbConnection()
{
if( !empty(DatabaseConnection::$instances) )
{
return end(DatabaseConnection::$instances)->getConnection()->getDatabaseName();
}
}
Notes :
If you encounter errors such as Unknown class 'Container' or Capsule or anything of that kind, make sure you check the question link I have provided, and use use statements properly.
Concerning upcoming answers :
It seems to me that this database connection lives within the the brackets of the controller's action, so when I proceed to another action that specifies no connection, it returns to the central database automatically.
Which has got me thinking that there must be a way to set the database connection to the sub-database in a 'global' way to the whole function, such as a middleware or something.
I would love to see an answer, implementing such thing.
Update :
I came up with a neater way to do it.
I assume you are on the same ground as me, wanting to change databases conditionally in accordance with each controller... say, each of your controllers requires a different database, just for the sake of the argument.
What we will be using to solve this is `Middlewares.
First, to explain what we are about to do..
We are going to check for the name of the controller (and even action) and then set the proper database we wish to set.
Go to your command-line , type in:
php artisan make:middleware SetDatabaseConnectionMiddleware
To create a middleware with ready boilerplate.
Or, if you like it the hard way, go to your app_name/app/Http/Middleware and create one manually.
Go to your helper methods file( if you already have one, if not, dude make one!)
function getControllerAndActionName()
{
$action = app('request')->route()->getAction();
$controller = class_basename($action['controller']);
list($controller, $action) = explode('#', $controller);
return ['action' => $action, 'controller' => $controller];
}
This will return to you an array with both the action name and controller name, if you want to return restrictidly just the controller's name, feel free to remove 'action' => $action from the code.
Inside of your middleware, it will look this way :
namespace App\Http\Middleware;
use Closure;
use DatabaseConnection;
class SetProperDatabase
{
/**
* Handle an incoming request.
*
* #param \Illuminate\Http\Request $request
* #param \Closure $next
* #return mixed
*/
public function handle($request, Closure $next)
{
$database_name = '';
$controllerAndActionName = getControllerAndActionName();
$controller_name = $controllerAndActionName['controller'];
$action_name = $controllerAndActionName['action'];
if($controller_name == 'my_controller_nameController')
{
$database_name = 'your_proper_database_name';
}
else
{
$database_name = 'other_db';
}
$database_connection = new DatabaseConnection(['database' => $database_name']);
return $next($request);
}
}
4.Now, that you have created properly your middleware, let us tell your app where to find it and under what name.
Go to your app_name/app/Http/Kernel.php
In your $routeMiddleware variable, add this line
'set_proper_database' => \App\Http\Middleware\SetProperDatabase::class,
This way we know how to call it.
Finally, setting it up.
Go to your Controller.php (the Abstract class from which all of your controller's inherit)
public function __construct()
{
$this->middleware('set_proper_database');
}
And this should do it for you.
If you have any further questions, please feel free to comment.
// Resources :
1.Controller And Action Name
2.Middleware Documentation
3.Further Middleware Documentation
Notes :
I'd appreciate some edition concerning my styling and code indenting, since it seems I struggled to style my code properly in here but in vain, the indentions I used had no effeft.

CakePHP: authorizing actions based on belongsTo relationships

Let's keep it simple.
A project has just two models:
User (hasMany Project)
Project (belongsTo User)
Users are only allowed to perform actions on the projects which they own. No one else's.
I know how to manually check who the logged in user is and whether or not he/she owns a specific project, but is there a better, more global way to do this? I'm looking for a more D.R.Y. way that doesn't require repeating the same validation inside multiple actions. For example, is there a config setting like maybe...
Configure::write('Enforce_belongs_to', true);
...or maybe a setting/option on the Auth component.
Maybe this is crazy, but I figured I'd ask.
Adding to Nunser's answer, here would be a general concept of how the behavior would be. You can then attach it to the applicable model.
class StrongBelongBehavior extends ModelBehavior
{
public function beforeFind( Model $Model, $query = array() ) {
$query['conditions'] = array_merge( (array)$query['conditions'], array( $Model->alias.'.user_id' => CakeSession::read("Auth.User.id" ) );
return $query;
}
public function beforeSave( Model $Model ) {
$projectId = Hash::get( $Model->data, 'Poject.id' );
if( $projectId ) {
$Model->loadModel('UserProject'); // UserProject is a custom model
$canEdit = $Model->UserProject->projectIDExists( $projectId ); // returns true if projectId belongs to the current user
if ( ! $canEdit ) {
return false;
}
}
return true;
}
}
I'm not sure if what I'm answering is the best-utermost-dry-it's-almost-dehydrating approach, but is the simplest thing I could think of.
In the Project model, create a function that return an array of project ids associated to an user.
class Project extends AppModel {
public function getByUserId($userId) {
$projectsArray = array();
if ($userId != "valid")
//do all the checks, if it's not null, numeric, if the id exists, etc
$projects = $this->Project->find('all', array('conditions'=>
array('user_id'=>$userId)));
if (!empty($projects)) {
foreach($projects as $i => $project)
$projectsArray[] = $project['Project']['id'];
}
return $projectsArray;
}
}
You mention a find('first') in your comment, but I'm assuming you want all the projects related to the user instead of just the first. If not, it's a simple modification of that function. Also, I'm just getting the ids, but you may want an $id=>$name_project array, up to you.
Now, I don't know what you mean by "only allowed to perform actions", is it just edits that are restricted? Or lists or views should be restricted and not even shown to the user if the project is not his/hers?
For the first case, restrict editing, modify beforeSave.
public function beforeSave($options = array()) {
if(!$this->id && !isset($this->data[$this->alias][$this->primaryKey])) {
//INSERT
//not doing anything
} else {
//UPDATE
//check if project inside allowed projects array
$allowed = $this->getByUserId(CakeSession::read("Auth.User.id"));
if (!in_array($this->id, $allowed))
return false; //or throw error and catch it in the controller
}
return true;
}
The code is untested, but in general terms, you prevent the edit of a project that is not "the user's" just before the update of the record. I assume the insert of new projects is free for everyone. According to this post, all saving functions except saveAll pass through this filter first, so you will need to overwrite the saveAll function and add a validation similar to the one in beforeSave (as explained in the answer there).
And for the second part, filtering results so the user isn't even aware that there are other projects instead of his/hers, change beforeFind. The docs talk about restricting results based on user's roles, so I guess we're on the right track.
public function beforeFind($queryData) {
//force the condition
$allowed = $this->getByUserId(CakeSession::read("Auth.User.id"));
$queryData['conditions'][$this->alias.'.user_id'] = $allowed;
return $queryData;
}
Since the $allowed array has just id values, it'll work like an IN clause, but if you change that array structure, be sure to modify these functions accordingly.
And that's it. I'm thinking about the more basic cases here, edit, view, delete... Ups, delete... change the beforeDelete function also, to avoid any evil users who want to delete others property. The logic remains the same (check if project id is in allowed array, if not, return false or throw error), so I won't add the example of that function here. But that's the basic stuff. If for some reason you want to have the allowed projects in the controller, call the getByUserId function in beforeFilter and handle that ids array there. You can even store it in session, but you'll have to have in mind maintaining that session when adding or deleting projects.
If you want a superadmin that can see and edit everything, just add a condition in getByUserId that checks the role of the user, and if it is an admin, return all projects.
Also, keep in mind: maybe Project has many... subprojects (not much imagination), and so, the user related to the project can add subprojects, but the same evil user as before modifies the hidden project_id that subproject has and edits it. In that case, I recommend you also add a validation in Subproject to avoid actions on models related to Project that are not his. If you have the Security component in place and the edit and delete actions can just be reached by forms, this is a minor thing because Security Component well used avoids form tampering. But give it a thought to see if you need to validate "Subproject" instances also.
As Ayo Akinyemi mentioned, you can use all this as a behavior. I haven't personally done so, but it meets the requirements, all the callbacks modified here are what you modify in a behaviour. You'll have to encapsulate the logic, column names (need to be variable an not set hardcoded, like user_id), etc, but it will be reusable in any other cake project you'll have. Something like StrongBelongBehavior or MoreDRYBehavior. And share it if you do it :)
I'm not sure if Auth component has some way of doing what you want, but that would be the best option I guess. Until some enlightens me (I haven't investigate much this issue), this is the solution I'd use.

Working with related model in CakePHP

This is a fairly basic question about CakePHP, but since my knowledge of this framework is rather rusty, it is making me lose a lot of time.
I have a ManyToMany relation between Guest and Present. Whenever a new Guest is created and associated with a present, I would like to mark the Present as taken. If the present is already taken, some error should arise. The reason why I am not just declaring that a Guest hasMany Presents is because in the future things may change and more than one guest could associate to a present, so I prefer to avoid a Db migration.
My Guest::add() action looks like follows. It is called with a POST with the data of a new Guest and the id of an existing Present.
public function add() {
if ($this->request->is('post')) {
$id = $this->request->data['Present']['id'];
$this->Guest->create();
$present = $this->Guest->Present->findById($id);
if ($present['Present']['taken']) {
throw new ForbiddenException();
}
if ($this->Guest->save($this->request->data)) {
if ($this->Guest->Present->saveField('taken', true)) {
// Give the guest a uuid and proceed with a welcome message
$this->Guest->read();
$this->set('uuid', $this->Guest->data['Guest']['uuid']);
}
}
}
else {
throw new ForbiddenException();
}
}
What happens is that a new Guest is created (correct) and associated with the given present (correct) but when I save the taken field a new present is created instead of modifying the given one.
What is the correct way to proceed to update the current Present?
If it is of any help, I am using CakePHP 2.0
For obtaining the model data by the primary key it's better to use theIn addition read method:
$present = $this->Guest->Present->read(null, $id);
The read method sets the model's id attribute so that further calls to other methods affect the same data record, rather than creating a new one. This should solve the problem you are having.
Model callbacks tend to be better suited for these situations. You could add a beforeSave callback to the Guest class to checks if the present is already taken, and not allow the creation if it is. This way the model logic is left in the model layer and you don't need to do any extra work e.g. if the constraint has to be enforced also when existing Guests are saved, or created from different controllers or actions.
It sounds like the ID of the model you are trying to save is losing scope. You should be able to resolve your issue by updating your code:
...
if ($this->Guest->save($this->request->data)) {
$this->Guest->Present->id = $id;
if ($this->Guest->Present->saveField('taken', true)) {
...

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