I'm new in cakephp and I'm just wondering, how to test models and controllers without using views?
I have to simulate saving data using models and controllers without using froms from views. I was thinking about to make an array with the needed values, but maybe there is a better way to do that?
you can mock your model functions using code like:
$model = $this->getMockForModel('MyModel', array('save'));
$model->expects($this->once())
->method('save')
->will($this->returnValue(true));
You can output variables at any time from controllers (or models) without getting to the views. Yes, it's not how you should do things with an MVC framework, but for testing, it's pretty easy to whack this below your database call in the model/controller:
<? echo '<pre>'; print_r($my_array); exit; ?>
The other thing you can do is at the top of your action function in the controller put:
$this->layout = '';
$this->render(false);
... which will bypass the layout and skip the view rendering, so you can output whatever you like within that function without using the view.
At the beginning of your action, you may use:
$this->autoRender = false;
This will allow you to access your action directly by going to it's path (e.g. CONTROLLER/ACTION). Before passing your data array to save() or saveAll(), I recommend double-checking it with Debugger::dump(), and follow that with die(). This will make the array containing the save data print on your screen so you can verify it looks proper and follows Cake's conventions. The die() will prevent it from actually saving the data.
If everything looks correct, remove the dump() and die() and test it out again.
The first response, from Ayo Akinyemi, should also work well if you are Unit Testing your application.
Related
I'm evaluating frameworks for use with an API, and I'm taking a serious look at PHP Phalcon. It's looking promising - "lean" (load what you need), but with a lot of options.
I'm wondering... is it possible to not use views (templates, rather) with it? Do I have to set up a view, or can I just output .json?
Thanks!
There is a way in Phalcon to disable view in the action and avoid unnecessary processing:
public function indexAction() {
$this->view->disable();
$this->response->setContentType('application/json');
echo json_encode($your_data);
}
Depending on what you want to do you can disable the view as others have suggested and echo json encoded data, or you could use the built in response object as below:
$this->view->setRenderLevel(View::LEVEL_NO_RENDER);
$this->response->setContentType('application/json', 'UTF-8');
$this->response->setJsonContent($data); //where data is an array containing what you want
return $this->response;
There is also a tutorial in the docs that goes over building a simple REST api
http://docs.phalconphp.com/en/latest/reference/tutorial-rest.html
If you won't be using any views at all you can disable views at the very start.
$app = new \Phalcon\Mvc\Application();
$app->useImplicitView(false);
Even if you do this, I think you still have to set the view DI for the framework to work.
Also, if you want to output json there's a method for that:
$this->response->setJsonContent($dataArray);
$this->response->send();
Yeah, you can do it, I'm using PHP Phalcon.
To ignore view, in your controller your action should be like
public function indexAction() {
$var = array or other data
die(json_encode($var));
}
die(); in controller will not render any parent layout! :)
I have been trying to pass an array that is generated in view to controller.
Say, $abc=array(); How do I sent this array to controller, then process it and sent this array to another view?
Is it even feasible? Not getting any answer after googling! Please help.
I see no point to generate array from inside the view.
It depends how your framework passes the data between the layers.
For example the controller do pass the data to the View object.
I.e.
public function someMethodInController() {
$this->view->varName['arr_key'] = 'something';
}
But talking about view, it might not have a certain controller already instantiated (usually it's against the MVC paradigm)
So I see two major ways:
Instantiate a controller (I would not prefer this)
$controller = MyController();
$controller->abc = $abc;
Or use sessions. This way to transmit data is universal inbetween your domain, and does not care if you are using a certain pattern, or how do you transmit between layers.
$_SESSION['abc'] = $abc;
Once assigned, you can use it in the same session from each of your files.
I'm working with a PHP MVC Framework. Works really well. I like the separation of the business layer (model) with the business logic (controller). But i just stumbled upon a problem. Here's the thing:
Suppose i navigate to the following url:
http://localhost/user/showall/
In this case the userController.php is called and within that file there is a method showallAction() which gets executed.
In the showallAction() method i simply do a request to a model which gets all the users for me. Something like this:
public function showallAction()
{
// create userModel object
$users = new userModel();
// get all users and assign the data to a variable which can be accessed in the view
$this->view->users = $users->getAllUsers();
// render views
$this->view->render();
}
So this method gets all the users, assigns the data returned from the userModel to a variable and i can easily work with the returned data in my view. Just a typical MVC thing.
Now here comes the problem.
I also need to create a native iphone variant. Ofcourse the looks will be totally different. So all i actually want to do is to request this url:
http://localhost/user/showall/
And that it just gives me the array (in json format) back. So i can use that for the mobile development.
But this obviously can't be done right now because the showallAction() method assumes that it is for web browser display. It doesn't echo JSON formatted, instead it simply assings the array of users to a variable.
So that means i have to create another method "showallMobileAction()" in order to get the data, but specifically for the mobile device. But this is not an elegant solution. I'm sure that are better ways...
Anyone any idea how can i solve this problem??
In your situation i would modify the routing mechanism.
It would be useful, if you could add extension at the end of URL, which represents the format you expect, like :
http://foo.bar/news/latest >> HTML document
http://foo.bar/news/latest.html >> HTML document
http://foo.bar/news/latest.rss >> you RSS feed
http://foo.bar/news/latest.json >> data in JSON format
It's a simple pattern to recognize. And you can later expand this to add .. dunno .. pdf output, or Atom feeds.
Additionally , two comments :
Model is not a type of objects. Instead it is a layer, containing objects responsible for business logic, and objects responsible for data storage/retrieval.
View should be a full blown object, to which you bind the domain objects (objects responsible for business logic).
You could pass parameters to your url:
/user/showall/json
and get the third URL segment with a custom function or a built-in one. For instance, with CodeIgniter: $this->uri->segment(3).
Some frameworks will pass the additional parameters to your method. Just try this with the URL I wrote above:
public function showallAction()
{
print_r(func_get_args());
}
I'm not familiar with PHP MVC but in general terms I'd use the "accepts" HTML header field to request the response in either "text/html" or "text/json", the controller would check for the accepts type and return the response accordingly.
For a project I used some logic in my view, this is not the way to go so I want to get it out.
The problem is that it can't be done from a class method of my model because Zend will make 10000 of queries from 10000 of instances to the database and it becomes very slow.
So I have to do it on a way that it loads all data at once, then processes it and returns the data back to the view. In my view it works the way I do it, the only problem is that it is IN the viewfiles.
What is the way to go? Just make a class in the model that inputs the values and returns required data?
Thanks
Here is the way i would go to display data from a MVC perspective
Controller
function someAction(){
$someTable = new Model_DbTable_SomeTable();
$allData = $someTable->fetchAll();
$arrayFormattedData = DataProcessor::process($allData);
$this->view->data = $arrayFormattedData;
}
You have to do your logic processing in a model (in the example above its done in the static class DataProcessor throught the process method (Not neccessarly the way to go, but it could be a good start)
View
echo $this->dataParser($this->data); // using a view helper to parse data to be displayed
or
echo $this->partialLoop('partialLoop.phtml', $this->data); // using the partial loop view helper built in in ZF
Finally, you should try to make your models as flexible as possible to make them reusable which is the key in oop development.
I have a CakePHP application that in some moment will show a view with product media (pictures or videos) I want to know if, there is someway to include another view that threats the video or threats the pictures, depending on a flag. I want to use those "small views" to several other purposes, so It should be "like" a cake component, for reutilization.
What you guys suggest to use to be in Cake conventions (and not using a raw include('') command)
In the interest of having the information here in case someone stumbles upon this, it is important to note that the solution varies depending on the CakePHP version.
For CakePHP 1.1
$this->renderElement('display', array('flag' => 'value'));
in your view, and then in /app/views/elements/ you can make a file called display.thtml, where $flag will have the value of whatever you pass to it.
For CakePHP 1.2
$this->element('display', array('flag' => 'value'));
in your view, and then in /app/views/elements/ you can make a file called display.ctp, where $flag will have the value of whatever you pass to it.
In both versions the element will have access to all the data the view has access to + any values you pass to it. Furthermore, as someone pointed out, requestAction() is also an option, but it can take a heavy toll in performance if done without using cache, since it has to go through all the steps a normal action would.
In your controller (in this example the posts controller).
function something() {
return $this->Post->find('all');
}
In your elements directory (app/views/element) create a file called posts.ctp.
In posts.ctp:
$posts = $this->requestAction('posts/something');
foreach($posts as $post):
echo $post['Post']['title'];
endforeach;
Then in your view:
<?php echo $this->element('posts'); ?>
This is mostly taken from the CakePHP book here:
Creating Reusable Elements with requestAction
I do believe that using requestAction is quite expensive, so you will want to look into caching.
Simply use:
<?php include('/<other_view>.ctp'); ?>
in the .ctp your action ends up in.
For example, build an archived function
function archived() {
// do some stuff
// you can even hook the index() function
$myscope = array("archived = 1");
$this->index($myscope);
// coming back, so the archived view will be launched
$this->set("is_archived", true); // e.g. use this in your index.ctp for customization
}
Possibly adjust your index action:
function index($scope = array()) {
// ...
$this->set(items, $this->paginate($scope));
}
Your archive.ctp will be:
<?php include('/index.ctp'); ?>
Ideal reuse of code of controller actions and views.
For CakePHP 2.x
New for Cake 2.x is the abilty to extend a given view. So while elements are great for having little bits of reusable code, extending a view allows you to reuse whole views.
See the manual for more/better information
http://book.cakephp.org/2.0/en/views.html#extending-views
Elements work if you want them to have access to the same data that the calling view has access to.
If you want your embedded view to have access to its own set of data, you might want to use something like requestAction(). This allows you to embed a full-fledged view that would otherwise be stand-alone.
I want to use those "small views" to
several other purposes, so It should
be "like" a cake component, for
reutilization.
This is done with "Helpers", as described here. But I'm not sure that's really what you want. The "Elements" suggestion seems correct too. It heavily depends of what you're trying to accomplish. My two cents...
In CakePHP 3.x you can simple use:
$this->render('view')
This will render the view from the same directory as parent view.