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
Related
This is an extension from a post I made a few days ago: Change a page without refreshing - Laravel / Ajax
Basically, I'm trying to replicate the URL Structure of Soundcloud where you can click on a link, it'll load the content without refreshing the page however if you land directly on that page, it won't replicate design and effecitvely break.
I've been thinking of ways on how I can check in Laravel if the page is requested via Ajax or has been landed on without an Ajax call.
What's happening at the moment is that when I call the page, the view has a master template that's extended thus creating duplicate master templates on the one view.
I was thinking if I done something like
#if(!Request::ajax())
#extends('masterlayout')
#endif
It would work but tried and no luck.
Any help as always is greatly appreciated!
Thanks
Looks like #extends directive is always executed even in falsy if
You can add an empty layout and perform check like this:
#extends(Request::ajax()? 'layouts.empty' : 'layouts.master')
and you need add only this in layout empty.blade.php:
#yield('content')
or you can add ajax check in the master layout
#if(!Request::ajax())
// all layout code
#else
#yield('content')
#endif
You can check that in your controller action.
public function index(Request $request)
{
if($request->ajax()){
return "This is an AJAX call!";
}
return "Not an AJAX call...";
}
Thanks for all the answers, I eventually got this fixed by using #if(!Request::ajax()) on section that didn't need to be shown when the page was loaded via ajax.
Thank you again! :D
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
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.
I have a main.blade.php which contains my layout. As part of this layout, I need to show the categories from the database.
I would like to avoid doing the Eloquent call in every single controller.
How can I retrieve all categories to show on main.blade.php regardless of the controller?
You can do it in the base controller or the view layout you have.
In the view it would be:
#foreach(Category::all() as $category
and for a base controller, it would be something like:
$this->layout->categories = Category::all();
Hope this helps!
http://laravel.com/docs/templates
It would be great to call single action in template right? :)
You can check this package (check out layout facade section). Hope it helps.
I've been using Codeigniter in order to get accustomed to the Model-View-Controller architecture, and to try and speed up the process of making and implementing sites.
I keep seeing references to "Partial Views" but can't find a definition for the term.
Can anyone tell me what a partial view is, and where it is used?
A partial view is just a sub-view that you can include in a parent view. Let's take a look at a common example:
// Controller:
$data['myvar'] = array('element1', 'element2', 'element3');
$this->load->view('myview', $data);
// Myview:
<ul>
foreach ($myvar as $var) {
$this->load->view('partialview', array('var', $var));
}
</ul>
// Partialview:
<li><?= $var ?></li>
This is useful to repeat content according to a list.
Note that nothing differs between a view and a partialview, it's just the way you include it that defines the term.
The best way to describe a "partial view" is to think of it as a template, it displays a chunk of html with Model data passed to it.
Good examples of where to use one would be where you plan on displaying the same html over and over, like a menu or a page header or even better yet use them to display content requested using ajax.
Basically you call an action on the controller that returns the partial view from lets say jQuery and then put the returned markup into a select or div tag. Here is an example of doing that from my blog easy ajax with aspnet mvc and jquery, yes I know it asp.net mvc not php and codeigniter, but the principal is the same.