I want to have something like
Routes
GET /endpoint pathToTemplate.templateName.scala.html
In order to avoid the need to create a controller just to serve this template.
I need a template because I am serving up values using an imported scala library so this can't just be static html
#import tool.values._
<p>
#Tool.getValue()
<p>
It is impossible to achieve your desire solution but you can to make a workaround with one controller and the dynamic URL parts mapping.
Firstly create a controller which serves a view for any provided path. Route definitions must be placed in code for example with usage of a simple hash map instead of the route file.
object GlobalController extends Controller {
private val getRouterMap = Map(
"view1" -> views.html.view1(),
"view2" -> views.html.view2(),
"sub/view3" -> views.html.view3()
)
def route(path: String) = Action { implicit request =>
Ok(getRouterMap.getOrElse(path, views.html.notFound()))
}
}
Secondly at the end of the route file define a mapping for the created action as follow.
GET /*path controllers.GlobalController.route(path)
It is very important to put it as the last line. Otherwise it will shadow all other mappings defined below.
Hint
Anyway if I ware you I would reconsider your design. Singleton objects aren't easily testable. Sooner or later they will make your life really painful.
Related
I would like to show some data from the database in the menubar of my web page. To get the data, I have a data-access-object (DAO) which is usually created with Guice injection.
How can I use such an (injected) object in my Scala templates?
I could pass it as a parameter to the template, but I had to do this on every single page (because it should be displayed in the menubar). I'm looking for another solution where I don't have to pass it everywhere. Currently I'm creating a new object inside the template, whenever it is rendered (which gets me a cleaner code but a worse performance).
You can kinda-sorta fake this without too much effort.
First, create a Scala object that provides access to your DAO (this can contain as many things as you want, just repeat the pattern within the top-level object and the Implicits object).
package com.example.stuff
object ViewAccessPoint {
private[stuff] val myDaoCache = Application.instanceCache[MyDao]
object Implicits {
implicit def myDao(implicit application: Application): MyDao = myDaoCache(application)
}
}
In your view, you can then import the Implicits object into your template and get hold of the DAO created by Guice.
#import com.example.stuff.ViewAccessPoint.Implicits._
#import play.api.Play.current
myDao.whatever()
This works for both Java and Scala projects.
You can see this in practice here:
Access point
Template
On a side note, I would consider if you really want to be doing data access in your template layer.
I have splitted my application into two main areas.
Part(A)
PartStashContainer(B)
The content of A should be set based on what user wants.
So basically i can have 1..N classes which could be used in Class URI of Part in application model.
I don't know if i should replace the whole Part(A) with new dynamically created Part(C) which has content i want, or i should somehow to modify the existing Part (call setContributionURI, or setObject methods on Part object?).
It does make more sense to me to modify the existing Part, because it is defined in Application model and therefore already describing the location where the content should be.
Possible solutions:
Modify the Part object so it "reload" its content based on new setup (But how? Can setContributionURI or setObject methods help?)
Remove the old Part and add dynamically on same place in Application model the new Part (using EModelService and EPartService).
other solution??
If you want to reuse the Part then do something like:
MPart part = find or inject your part
MyClass myClass = (MyClass)part.getObject();
... call a method of MyClass to change the contents
MyClass is the class you specify for the object in the application model. You should add a method to that to let you change the contents.
Don't try to call setObject, this is really only for use by Eclipse. I don't think setContributionURI would do anything after the part is created (but I am not sure).
If you want to use different classes for the different data then you really should use different Parts.
I am currently active PlayFramework learner who came from world of PHP.
For example I have a Head block object in my app, which should hold title, charset encoding, meta information, etc. Something similar to Magento blocks, but without XML declaration
package blocks.Page
object Head {
var title: String = "";
}
In Application.index() method I have
blocks.Page.Head.title
Ok(views.html.application.index());
And finally in html template
#import blocks.Page.Head
<title>#Head.title</title>
However, blocks.Page.Head object is defined for entire application scope, not for single request. This object is the same for each request.
What is the right way to do, what I am trying to do? I can create container with all blocks and instantiate it with each request, then just pass to all templates. But I have a feeling that this is wrong way.
Just use usual class instead of object and pass instance to template as parameter.
Like this:
package blocks.Page
case class Head(title: String = "")
Controller:
val head = Head("Blah")
Ok(views.html.application.index(head))
And template will looks like:
#(head: blocks.Page.Head)
...
<title>#head.title</title>
I know the feeling when coming from a request-oriented language like PHP :). However, consider application-wide access as a gift of a VM (in PHP we need to go the extra mile of using some bytecode and data caching tool like APC or eAccellerator).
I would probably create a blockManager class which gives you static access to blocks by name/tag/id from the template: Block.get("MyBlock"). Then you can define and later modify your caching / storing strategy (holding in memory vs. loading from storage) without affecting your templates.
How can I eliminate to write $object = new Application_Model_Database() in every controller?
For example for an article controller, I have to type $articles = new Application_Model_Articles() for every controller. Should I put it under viewer controller, action helpers, or any other way?
Your question almost sounds like an OOP best practices question as opposed to a Zend Framework specific question. Regardless of whether or not I'm using a framework, and regardless of what framework I choose, I base when and where I create new objects on testability how many times I have to write $object = new My_Random_Object();.
Speaking specifically to the Zend Framework: Objects I'm going to use everywhere, or almost everywhere, get created in Bootstrap.php. These objects generally include a database adapter, logger, view object, and any plugins I might use. To access these across the application, I'll create private properties in the appropriate controllers and assign the objects to those properties in the controller's init() method.
class ExampleController extends Zend_Controller_Action
{
public function init()
{
$bootstrap = $this->getInvokeArg('bootstrap');
$this->_db = $bootstrap->getResource('db');
$this->_log = $bootstrap->getResource('log');
// and so on, and so forth
}
}
Ideally, models, services, daos, etc, will all be relatively tightly grouped by controller and by action. In my experience, and this is speaking generally, if I have the same model or service class showing up across all of the controllers in my application, I have an organization problem. That being said, any model that shows up in only one action gets created in that action. If it's across actions in a controller, it gets created in the init() method and assigned to a property. If it shows up across multiple controllers, it gets created in my Bootstrap.php.
(Ideally, everything gets created in the Bootstrap.php, so you can swap out that bootstrap for testing purposes. Sadly, I don't always do that, and I most often use the principles I outlined above.)
Well do you really need it in every controllers? Because that's pretty much by design. You implement models when you need them. Its not that much code really.
Now if its to be used across actions from a controller you could always:
class MyController extends Zend_Controllers{
$protected $_articleModel;
...
and in your constructor or __init() function initialize it so you can use it in every action thru $this->_articleModel
If you REALLY want it everywhere in your application just initialize it in your bootstrap and store it in the registry.
public function __initModels(){
$articles = new Application_Model_Articles()
Zend_Registry::set('articles', $articles );
}
And access it in your controllers like so:
Zend_Registry::get('articles')->fetchAll();
But then your still writing a couple of characters.
Hope this help!
IF you want to use models in the controllers you must call it..anyway some shortcuts are here
1.You can initialize it in the init section of your controller like
public function init(){
$this->object = new Application_Model_Database();
}
So that the this->object is available in all the actions of that particular controller
2.Use Zend_registry as suggested in the above answer
Another possibility is to use a Dependency Injection container, such as the Symfony DI component. It takes care of instantiating your objects, and you get some additional benefits:
Separation of concerns. You have a component devoted to create your object tree.
Easier testability of the objects.
Last, but not least, the performance benefits given by lazy instantiation (objects are created only when you ask for them). Thus, if some object is not used by the particular controller serving your request, it's not instantiated).
It's a bit more laborious than the above solutions, but much more flexible if you need to maintain and extend your application in the future.
Hope that helps,
If you are using this object to just display data in your view and are using your controller to grab the data and assign it to your view, like so:
//someControllerAction
$object = new Application_Model_Articles();
$object->fetchAll();
//assign to view
$this->view->articles = $object;
You might be better off making a view helper similar to:
//Articles.php put in /application/views/helpers
class Zend_View_Helper_Articles extends Zend_View_Helper_Abstract {
public function Articles() {
$articles = new Application_Model_Articles();
$articles->fetchAll();
//return rowset object
return $articles;
Then in your view (phtml) you could do something like:
//someView.phmtl
<?php $articles = $this->Articles(); ?>
<h1><?php echo $this->escape($articles->title); ?></h1>
<p><?php echo $this->escape($articles->body); ?></p>
building a view helper allows you to bypass the controller completely if you just need to display data from the model. This is a very simple example and can be used with partials and partialLoops.
REF:ZF reference Custom View Helper
ZF partial view helper reference
I am using zend framework for my project and I have a requirement to route the path where I want it. For example:
I have a path www.example.com/module/controller/action1 and I want to internally route in to www.example.com/module/controller/action2.
But the main problem is I do not want to use the function in which I have to specify the module, controller and action [$this->_forward('action2', 'controller', 'module');], simply I something like this: $this->_forward('module/controller/action2');.
Please suggest me if anybody has its solution. its urgent need of my project.
Thanks,
Jitu
The controller method _forward() can accept only the action if it is in the same controller and module as the first action you are forwarding from.
So in your action1Action() method, you can simply call:
$this->_forward('action2');
If your second action is not in the same module/controller, you could subclass Zend_Controller_Action into your own "base" application controller that all your other action controllers inherit from (a good practice on ZF projects I find) and then create another one called _forwardFromUrl() or something like that, which breaks your URL apart and passes it to _forward() (or create an action controller helper if you just need this one extra thing).
This example is simplified and assumes your $url will always be in the format module/controller/action:
protected function _forwardFromUrl($url)
{
$parts = array_reverse(explode("/",$url));
$this->_forward($parts[0],$parts[1],$parts[2]);
}
Hope that helps!