How to make a laravel 5 view composer - model-view-controller

I'm still learning Laravel and I'm working on a small project to help me understand better. In the project, I am in need of a global array, so that I may display it or its attributes on every view rendered. sort of on a notification bar, so that each page the user visits, he/she can see the number of notifications (which have been fetched in the background and are stored in the array).
I have done some research, and realized that I have to fetch and compile the array in a view composer I think. But everywhere I go, I cant seem to understand how to make a view composer.
I need to fetch the relevant rows from the database table, and make the resulting array available to each view rendered (I'm thinking attaching it somehow to my layouts/default.blade.php file.). Please help, any and all advice is greatly appreciated:)

You can now inject services on your view
More info here: https://laracasts.com/series/whats-new-in-laravel-5-1/episodes/2

You have to use Sub-Views of laravel blade. I guess your functionality is like a sidebar or like a top bar which will be rendered at every page.
//Your Controller pass data
class YOUR_CONTROLLER extends Controller {
public function index()
{
$data = YOUR_DATA;
return view('YOUR_VIEW_FILE', get_defined_vars());
}
}
//In Your View File
#extends('LAYOUTS_FILE')
#section('YOUR_SECTION')
#include('YOUR_SUB_VIEW_FOR_NOTIFICATION')//You need not pass any data passed all data will be available to this sub view.
#endsection
In your sub view
//Do what ever you want looping logic rendering HTML etc.
//In your layout file just yield or render the section that's it
#yield('YOUR_SECTION')
More explanation can be found Including Sub-Views

Related

Laravel view into another view

Just started learning Laravel not long ago and I've got a question about views.
I have a 2 view files (blade).
The one which shows the categories from database.
The other one is showing the results of as search in products.
I'd like to use them in the same view, in the first section would be the searcher area and the area below is where I would like to list the categories.
I tried a couple of things to make it work, but I'm not sure which is the right approach:
I made a function in a controller where I listed the categories and handled the search and sent that to the view, but i didn't like this approach because I'm handling two different logic in one function.
I tried to yield or #include the view file into the main view. I got an error, because the database query didn't happen that case. I think yield and include only available with static data.
So, I'm don't know how to handle that issue properly.
Can someone suggest me a solution for this problem?
Thanks the help in advance!
There can be several way.
You can try to fetch category data inside blade file.
So, for example, in category.blade, you can fetch category list from database.
Then it would be independent to other views file and controllers.
And you can include category.blade.php for example.
The second way, is to fetch category data via ajax.
In this case, you can not insert script in all pages of using category.
So, in this case, you can use template structure.
For example.
In template.blade.php
<!DOCTYPE html>
<html class="no-js css-menubar" lang="en">
#stack('style')
#include('admin.layouts.head')
#include('admin.layouts.header')
#include('admin.layouts.page')
#include('admin.layouts.footer')
#stack('script')
</html>
In page.blade.php
<div class="page">
#yield('page-content')
</div>
In individual page,
#extends('admin.layouts.template' ,['menu'=>'coach'])
#section('insert-css')
your custom css content here
#endsection
#section('page-content')
Your page content here
#endsection
You must get $categories just once and pass $categories to all views (or views that need to category or just pass it to categories.blade.php).
for this goal, I suggest that create a partial blade called categories.blade.php and show all cateogories in it and include it wherever necessary, but before before that go to app/Provider/AppServiceProvider.php and in boot() method get categories and share it in categories blade that you created it, like below :
public function boot() {
// pass to first param, address categories blade
View()->composer('categories', function ($view){
// set a name for variable that get it in categories view
$view->with(['categories'=>$categories]);
});
}
If you would like to know more about subscriptions, visit this link or Laravel docs

Laravel 5 - getting data to a view

I think this is slightly different to the usual controller passing data to the view. I have a Project which has one DocumentOne. Within my app, the user creates a Project. This then redirects them to the show page for this project.
So with the project created, and the user on the show page for that project, I display the project ID. I then provide a select menu where the user can select a Document to display. So say I am in Project with the ID of 1, I then decide to show DocumentOne for this project. This displays a form with inputs for DocumentOne.
When the user fills in the form and submits, the data is saved to the database. The Project ID is the foreign key for DocumentOne. The following route is set up for DocumentOne
Route::resource('projects.documentOne', 'DocumentOneController');
Now I have data for DocumentOne which is linked to the Project with an ID of 1. However, if I now go back to the projects show page and then select Document One from the dropdown again, all I see is an empty form. This is obviously because the controller for this is
public function show(Project $project)
{
return view('projects.show', compact('project'));
}
So I am never passing it data for DocumentOne because theoretically it is not created when the Project is first shown. What I want to do is when the Document is selected in the Projects show page, is to have the form populated with whatever is in the database for that Document. If nothing is in the database, then the form will be empty. I have a DocumentOne Controller, but I dont know if I can link this to the Projects show page. I was thinking about doing something like this in the DocumentOne controller
public function show(DocumentOne $documentOne)
{
return view('projects.show', compact('documentOne'));
}
But not sure this will work. Hope I have not been too confusing and you understand what I am attempting, hoping someone can offer advice on how best to handle this situation.
Thanks
In my previous project, I also deal with such requirement, I thought so. Here my solution to solve such requirement.
Actual code calling from ajax.
Routes
get('setFlashData',function(Request $request){
$final_response = array();
$data_information = $request->except('_token');
$request->session()->flash('cmg_quick_create_data', $data_information);
if($request->session()->has('cmg_quick_create_data')){
$final_response['result']['success'] = true;
}
return response()->json($final_response);
});
But according to you requirement:
$data_information = $request->except('_token');
$request->session()->flash('cmg_quick_create_data', $data_information);
My basic functionality was, to share form data from Quick Create Section which is pop-up form to Full create form section, and whenever user click to "Go To Full Form" button from pop up, ajax call mentioned function which will set the flash data and than on destination side I only check weather its contain the flash data or not. and deal according to data.
#if (Session::has('cmg_quick_create_data')) {
{!! Form::model(Session::get('cmg_quick_create_data'),["class"=>"form-horizontal","data-parsley-validate"=>"data-parsley-validate",'role'=>'form','files'=>true]) !!}
#else
{!! Form::open(["class"=>"form-horizontal","data-parsley-validate"=>"data-parsley-validate",'role'=>'form','files'=>true]) !!}
#endif
I can understand this solution might be different from you requirement but hope full to figure out your solution. Look forward to hearing from you if still unclear from my side.

Update template inside a view rendered from another controller

I am looking for the way to refresh a template inside a view rendered from another controller than the template's controller, I mean:
I got two controllers AdminController & UserController. And two gsps /admin/listUsers & /user/_searchResult.
Then a want to render view listUsers who have inside the template _searchResult and all right.
Now, i want to refresh the template _searchResult, but cant find how. I tryed calling render(view:"/admin/listUsers", template:"/user/_searchResult", model:[searchResult:result])
AdminController.groovy
#Secured(['ROLE_ADMIN'])
def listUsers(){
//...
}
UserController.groovy
#Secured(['ROLE_ADMIN'])
def search(){
//search users for the givven params and send result by chain if there's an action or update a template if it's needed
//in my case this method need to update the template _searchResult
}
#Secured(['ROLE_ADMIN'])
def searchResult(){
//...
[searchResult:result]
}
listUsers.gsp
//...
<formRemote name="searchForm" url="[action:"search", controller:"user"]">
//Some fields for the search
//I need to place here some hidden inputs to send which
//template i want to update or action to redirect
</formRemote>
<g:render template="/user/_searchResult"/>
//...
_searchResult.gsp
//Just itterate and print the search result in a table
I hope I have explained the problem correctly, thanks!
I don't think I entirely understand your question, but I think the source of your confusion is that the way you are naming things doesn't follow regular conventions and you're not using the right tools for the job. Let me explain...
The methods on Controllers are called Actions. They send some data (the Model) to a View to be rendered into HTML. Views can be composed from smaller, reusable fragments called Templates. (sorry if I sound like I'm being condescending here, but I'm just trying to make sure we're all on the same page).
Now, what you've called templateA is actually a View, not a Template. You're correct that templateA (your View) can call templateB to render some markup, but then having the templateB try to call a method on another Controller doesn't make sense. That's not how things flow.
If you have some logic that needs to be executed after you've sent your Model to the View, you want to use a Tag Library (http://grails.org/doc/latest/guide/theWebLayer.html#taglibs).
To summarise, here's a quick recap.
A request should only call one Action, which sends the model to only one view.
If you need to reuse logic between Controllers, move that code to a Service.
If you need to reuse markup between Views, move that markup to a Template.
If you have logic that you want to have executed after you've sent the Model to the View, use a Tag Library.
Hopefully this will point you in the right direction.
--- UPDATE ---
OK, with the real code I can see better what you're trying to achieve. Firstly, as you're using the <g:formRemote> tag, you should have a good read of the docs at http://grails.org/doc/latest/ref/Tags/formRemote.html to understand what it does.
What you will have here is 2 separate requests. The first will be a regular page load by your browser, which is handled by the listUsers() action. Once the page is then finished loading, the user will enter a search term and hit the submit button. This will fire off a second ajax request, which will be handled by the search() action. This action could use the _searchResult.gsp template to render a HTML table to display the search results. When the browser get this, it will insert it into the DOM where you've told it to put it using the "update" attribute of the <g:formRemote> tag.
The important thing here is that from the server's perspective, these are 2 separate requests that are completely independent. They both first call an action, then send a model (a Map containing some data) to a view, which renders/merges the data with HTML and sends it back to the browser.
The difference between the 2 is that the first is a complete page load by the browser, whereas for the second request, the browser only loads a small chunk of HTML (the search results table) and updates the page content without reloading it.
So your code would look more like this...
AdminController.groovy
#Secured(['ROLE_ADMIN'])
def listUsers() {
render(view:"/admin/listUsers")
}
listUsers.gsp
<g:formRemote name="searchForm" update="insertSearchResultsHere"
url="[controller: 'user', action:'search']">
<input name="searchTerm" type="text" />
</g:formRemote>
<div id="insertSearchResultsHere"></div>
UserController.groovy
#Secured(['ROLE_ADMIN'])
def search() {
// use the search term to get a List<User>
render(template: "/user/searchResult", model: [users: users])
}
_searchResult.gsp
<table>
<g:each var="user" in="${users}">
%{-- Iterate through your search results --}%
</g:each>
</table>
I solved it by placing the attribute update and rendering the template alone:
gsp:
<formRemote name="searchForm" url="[action:"search", controller:"user"]" update="divToUpdate">
//Some fields for the search
</formRemote>
<div id="divToUpdate">
<g:render template="/user/_searchResult"/>
</div>
Controller:
def search(){
render(template:"/user/_searchResult", model:[searchResult:result])
}
When i asked this question, i was new on Grails community and i was confused with the use of remoteFunction and tags that use it like remoteForm. But i had this confusion because of i had not read the documentation. So in my case, the solution was search for documentation about how to use remote tags and render. Thanks to #AndrewW for show me the way.

Dynamic layouts in CakePHP

Sorry about the question title, but I couldn't find a more appropriate way to phrase this.
I am currently building a CakePHP powered website and I'm not quite sure how to approach the following issue. The website looks something like the follwing mockup:
.
The greyed out areas are part of the layout, because their content does not change between views. In the sidebar, I have a collection of ads who are linked to several models. I need controller logic to determine the picture associated with an ad. Also, the ad list needs to be dynamic. Where should I put the logic for building the sidebar?
I've thought about:
putting the logic into the AppController (beforeFilter / afterFilter) - the problem is I can't use the controller logic I need (the other controllers inherit from AppController, I'm not sure how to use them there).
making a component - is it okay to build components that rely on controllers?
replicating the sidebar code in all controllers that render views - this seems kind of stupid to me.
What is the Cake way for this?
Update
After some reading and experimenting, I've gotten to refactoring most of it.
I obtained the best performance by moving the logic for building my ads in the model (eliminating the component that retrieved the pictures) and not using requestAction. It's almost three times faster and the code looks much better.
I've done something similar for data-driven navigation. I put my logic in AppController::beforeRender and haven't had any problems. I'm not sure I understand your concern related to controller inheritance. I retrieve my menus via:
$menus = $this->NavMenuItem->groupByMenu();
$this->set( compact( 'menus' ) );
I then created an element that renders the menu. It's executed by the layout via:
<?php echo $this->element( 'navigation', array( 'id' => 'secondary', 'menu' => $menus['SECONDARY'] ) ) ?>
If that doesn't help, maybe you can further explain your issue with controller inheritance in a comment.
I guess the answer is requestAction in case the results are cachable:
http://book.cakephp.org/view/434/requestAction
It can be done in this way:
Create an element that will help in layout of the Ad Block
Create one or more controller that will generate the data required for rendering of the block
Use requestAction for getting the data out of the models and into the element.
Check the cake book, there is an example of an element where data from Post Model is used to display top/latest 5 posts. Your requirement, I feel, is very similar to it.
Alex,
you're getting a SQL error because the build() function has to be in the Sidebar model, not controller. Also, you don't necessarily need to use $user = array('Sidebar'); you could calling Sidebar in all of your models with this:
$Sidebar = ClassRegistry::init('Sidebar'); and then $Sidebar->find();, $Sidebar->build(); etc.
Or, if you only need to call the build() function from the Sidebar model, you could do this:
$sidebar = ClassRegistry::init('Sidebar')->build();
$this->set('sidebar', $sidebar);
Cheers.

What is the best practice for displaying multiple views in Codeigniter?

I'm trying to determine the best practice for calling multiple views from the same method in a controller.
Is it preferable in the controller to make one view call, then have that view call all the views it needs, or call all the views you need in sequence in the controller?
Example:
function index(){
//set all data variables
//then send them to the view
$this->load->view($index_view, $data);
}
or
function index(){
//set variables
//then call each view
$this->load->view($header_view, $header_data);
$this->load->view($body_view, $body_data);
$this->load->view($footer_view, $footer_data);
The Codeigniter guide shows both ways, but does not seem to advise to the best practice...is there one?
I didn't like the way of including the header/footer within the view, and I didn't like loading the footer and header each time in every single Controller function.
To fix this, I extended the Controller class with my own view display function.
<?php
// in application/libraries/MY_Controller.php
class MY_Controller extends Controller {
function _displayPage($page, $data = array()) {
$this->view->load('header', $data);
$this->view->load($page, $data);
$this->view->load('footer', $data);
}
}
?>
// in application/controllers/home.php
<?php
class Home extends MY_Controller {
function index() {
$this->_displayPage('home/index', array('title' => 'Home'));
}
}
?>
Not sure if this is CodeIgniter "best practice" but it makes sense to me.
I don't think there is a definitive answer for that. Choose one and stick with it, it's important to be consistent.
Anyway, I'd prefer the second one.
I would say that the controller should only display one view. Then it's up to the view if it wants to show a header, footer, sidebar or whatever. The controller shouldn't have to care, its job is to get data from a model and hand it to a view. Not decide if the view should have a header and a footer.
Agree with Christian Davén: its view / display logic not data or business / logic. essentially its the same as using php includes for snippets like navigation, footer etc. you're just embedding markup.
This is expected behavior. Once variables are set they become available within the controller class and its view files. Sending an array in $this->load->view() is the same as sending an array directly to $this->load->vars() before calling the view file. This simplifies things for most people using multiple views in a controller. If you are using multiple view files in a single controller and want them to each have their own set of variables exclusively, you’ll need to manually clear out the $this->load->_ci_cached_vars array between view calls.
A code comment in the Loader class describes another situation showing why this is the desired default behavior:
You can either set variables using the dedicated $this->load_vars()
function or via the second parameter of this function. We'll merge
the two types and cache them so that views that are embedded within
other views can have access to these variables.

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