I'm working on a small CakePHP application that is subject to the following constraint (awkward but out of my control): I need it to work on either of two identical databases, with the choice being based on URL. For example:
http://www.example.com/myapp/foo/action/param
http://www.example.com/myapp/bar/action/param
The first obvious solution is to have two identical CakePHP applications at myapp/foo and myapp/bar with different database configurations. This has a kludgy feel to it, though, so I'm trying to find an elegant way of creating a single application.
The approach I'm considering is this: Define routes such that myapp/foo and myapp/bar will be associated with the same controller. Then give my DATABASE_CONFIG class a constructor:
function __construct() {
$pathParts = explode('/', $_SERVER['REQUEST_URI']);
if (array_search('foo', $pathParts)) {
$this->default = $this->fooConfig;
} else if (array_search('bar', $pathParts)) {
$this->default = $this->barConfig;
}
}
(Where of course I've defined fooConfig and barConfig for the two databases.) I do have control over the URL, so I can be confident that there won't be extraneous occurrences of foo or bar in the URL.
My question is this: Is there a simpler, more elegant way of handling this odd situation? Maybe something in AppModel and/or AppController? Although I'm getting rid of duplicated code, I can't shake the feeling that I'm replacing one code smell with another.
There are a few ways to do this, here is one.
Write a sweet custom route in which you always match:
Router::connect('/:ds/*', array(), array('routeClass' => 'SweetDbRoute'));
Then have SweetDbRoutes set a class variable you can use everywhere, including in your model constructors. Then it should fail so you don't actually adjust the request.
App::import('SweetDbClass', array('file' => '/path/to/your/sweet_db_class.php'));
class SweetDbRoute extends CakeRoute {
// put your failing route code here, but use your SweetDbClass to track datasource ...
// see http://book.cakephp.org/view/1634/Custom-Route-classes
}
Then in your AppModel:
App::import('SweetDbClass', array('file' => '/path/to/your/sweet_db_class.php'));
class AppModel extends Model {
public function __construct($id = false, $table = null, $ds = null) {
$ds = SweetDbClass::$ds;
parent::__construct($id, $table, $ds);
}
}
So for example, after you perform an insert in one database, the two won't be "identical", right? Are these 2 DB somehow synced with each other? I don't know what do you need to do on those DB, but it's probably easier just to do 2 separate apps.
Yes, you can specify the DB configuration in the model: http://book.cakephp.org/view/922/Database-Configuration but you can't change it on-the-fly though (the models are not expected to change association to another table, I suppose). What you do is probably the only way.
I do have control over the URL, so I can be confident that there won't be extraneous occurrences of foo or bar in the URL
Yes, there can be "extraneous occurrences of foo or bar in the URL" :)) But it won't break your app.
Related
we are using subdirectories in our projects no separete views and controllers but in models we didn’t learn yet. Recently I’ve found this https://github.com/cakephp/cakephp/issues/60451 and actually routes and plugins we are already using, we just want to separete our models like this:
Model
-Entity
–Financial
—Money.php
-Table
–Financial
—MoneyTable.php
I’ve tryed put like this then controller is not able to find his model. How can I do to organize it, and make it work?
Things that we've tried:
Use $this->setAlias('TableModel');
Call in controller:
$this->TableModel = $this->loadModel('Subfolder/TableModel');
didn't work for SQL build, and other classes.
CakePHP uses the TableRegister to load models. That class can be configured to use a class that implements the LocatorInterface, and CakePHP uses the TableLocator as the default.
The only thing you can do is configure your own LocatorInterface instance in your bootstrap.php. You would have to create your MyTableLocator and have it change the className for tables to point to subdirectories. What rules for this class name rewritting are used is purely up to you.
bootstrap.php:
TableRegister::setTableLocator(new MyTableLocator());
MyTableLocator.php:
class MyTableLocator extends TableLocator {
protected function _getClassName($alias, array $options = [])
{
if($alias === 'Subfolder/TableModel') {
return TableModel::class;
}
return parent::_getClassName($alias, $options);
}
}
The above isn't working code.
I'm just demonstrating what the function is you need to override, and that you need logic in place to return a different class name.
You can check if the $alias contains the / character, and if so. Return a class name by extracting the subfolder name from the $alias. Take a look at the TableLocator to see how it's using the App::className function.
I have an application in which we give a very friendly interface for managing data. This is done through many controllers' add/edit/view functions. But now the requirement has come that we should have "super admins" able to edit anything, and scaffolding will give them a quick and dirty manner of changing data. Since scaffolding uses add/edit/view by default, I've unintentionally overwritten the ability to scaffold.
I can't just go and change all my calls to edit/add for our "user friendly" data managing. So I want to essentially ignore the add/edit/view when, for example, a user has a flag of "yes, please let me scaffold". I imagined it would be something like:
public function edit($id) {
if (admin_user) {
$scaffold;
} else {
[user-friendly version code]
}
}
But no dice. How can I achieve what I want?
suppose you already have admin users and you want to scaffold only super-user:
Also suppose you store the information about beeing a super-user or not in a column named super in the users table
in your core.php
Configure::write('Routing.prefixes', array('admin', 'super));
in your appController
public $scaffold = 'super';
beforFilter() {
if($this->Auth->user('super') && !isset($this->params['super'])
$this->redirect(array('super' => true));
}
Now I can't try this code but the idea should work.
edit: we need to check if we are already in a super_action to avoid infinite redirect
I am working with CakePhp 2.x. I have three Columns:
User | Course | UserCourseRole
Each user can edit multiple courses and one course can be edited by multiple users. So far so good.
If a user wants to see an index of all the courses i want to show a 'edit'-link only next to the courses which he can in fact edit.
How can i realize this? I figured i would have to set some sort of extra field inside the CourseController and check for this field inside the view. Is this the right way to go?
My current Code is
CourseController.php
...
public function index() {
$courses = $this->Course->find('all', array('recursive' => 2));
$this->set('courses', $courses);
}
...
Courses/index.ctp
<!-- File: /app/View/Courses/index.ctp -->
...
<?php foreach ($courses as $course):?>
...
<?php
echo $this->Html->link('edit', array('action' => 'edit', $course['Course']['id']));
?>
...
In beforeRender() or beforeFilter() set $this->Auth->user() as a variable to the view, for example as userData.
$this->set('userData', $this->Auth->user());
Implement a (auth)helper that uses that variable (you can make it configurable as a helper setting) and do your checks like:
if ($this->Auth->hasRole($course['Course']['role']) { /* ... */ }
if ($this->Auth->isLoggedIn() { /* ... */ }
if ($this->Auth->isMe($course['Course']['user_id']) { /* ... */ }
Implement the hasRole() method according to whatever your specific requirements are.
Doing this as helper as a bunch of advantages, it is easy to reuse, overload and adapt to whatever your checks are and you don't use a component in a view plus that you should avoid calling statics and singletons a lot in your app. Also it is pretty easy to read and understand what the code does.
I think the good idea is set some variable or constans after logged (if user has privileges) and uses if statement for check.
if($allow === true) {
echo $html->link('Edit',...
}
or use AuthComponent::user() in Views.
This idea it's not good if we can many kind of admins (admin, moderator, reviewier, etc.)
Maybe someone will have a better solution
In my CakePHP application, I have a model like this:
class Duck extends AppModel {
var $name = 'Duck';
function get_table_name() {
$tbl_name = //compute default table name for this model
}
}
I would like to write the function get_table_name() that outputs the default table name for the model. For the example above, it should output ducks.
EDIT:
Several people have pointed out the use of $this->table.
I did small testing and found out the following:
In the question as I have put above, $this->table indeed contains the table name.
However, actually, my code looked more like this:
class Duck extends Bird {
var $name = 'Duck';
function get_table_name(){
$tbl_name = //comput default table name for this model
}
}
class Bird extends AppModel {
}
In this case $this->table is empty string.
I went with this approach because I wanted to share some code between two of my models. Looks like this is not a good way to share code between models which need some common functionality.
You're looking for the Inflector class.
Inflector::tableize($this->name)
(tableize calls two Inflector methods to generate the table name: underscore() and pluralize())
Edit:
According to the source code, $this->table should contain the name of the table that CakePHP will use for the model, but in my experience this isn't always set. I'm not sure why.
To get the name of the table that the model is currently using, you can use: $this->table. If you don't manually change the model's table conventions, this may be the most useful in the case of CakePHP ever changing its conventions to use table names using something other than Inflector.
CakePHP's Inflector
function get_table_name() {
$tbl_name = Inflector::pluralize($this->name);
}
OR the tableize method
function get_table_name() {
$tbl_name = Inflector::tableize($this->name);
}
Edit
This also addresses the apparent "ghost" issue with $this->table in the Model.
Digging around in the __construct for Model I discovered two things:
Cake uses Inflector::tableize() to get the table name. This alone is enough to warrant using tableize over pluralize. You'll get consistent results.
$this->table is not set by the Model::__construct() unless $this->useTable === false AND $this->table === false.
It appears that if you know you haven't set $this->useTable to false you should be able to use this over $this->table. Admittedly though I only briefly scanned the source and I haven't really dug deep enough to say why $this->table isn't working sometimes.
To get the full table name for a model you have to take the table prefix into account.
$table = empty($this->table) ? Inflector::tableize($this->name) : $this->table;
$fullTableName = $this->tablePrefix . $table;
I used to use inflector to get the table name from model's name
$tableName = Inflector::pluralize(Inflector::underscore($model));
but this is not really universal, using useTable looks better, by default it will contain table's name by convention, and if you have a table that does not match the conventions, then you should manually specify it by useTable. So, in both cases the result will be correct
$this->User->useTable
Little history; I hate the fact that I can't use enums in CakePHP, but I get it. However, another pet peev I have is that my Booleans return 0 or 1 and there is no way to universally turn them to yes' and no's.
So I though I would create a little function in the afterFind method of the AppModel to do this for me. The first step I wanted to take was to identify which columns where boolean (since some columns will return zeros and ones that do not need to be converted). I devised this little peace of code:
function __construct($id = false, $table = null, $ds = null) {
parent::__construct($id, $table, $ds);
foreach($this->_schema as $col => $colDetails){
if($colDetails['type'] == 'boolean')
$this->_booleans[] = $col;
}
}
However a quick debug($this) in the model show that only the current model's boolean columns are captured. When I hit those columns directly the $this->_booleans show up but again, not those of associated models.
I've looked though the manual and the API..I see no mention as to how to approach a solution.
What am I doing wrong?
Enums are not supported by CakePHP in order to make an application database type independent. Enums are not supported by many database engines. The simplest solution for your task is:
echo $model['boolField'] ? 'Yes' : 'No';
The problem is that $this->_booleans in the AppModel only contains the schema details of the current model. In fact, the code is probably working. You should check $this->_booleans and $this->Related->_booleans, and I bet you'll find what you're looking for.