Hi there stackoverflow!
In my laravel views I am using a default layout by calling
#layout('layouts.default')
to the same controller I am sending Ajax Requests yet I can't put 'if' to #layout if its a ajax call. Because if ajax request has made to controller it also produce header, footer and content(header and footer are in default layout). So to avoid this I made one copy without #layout of my view.
However its so boring to edit two files for making changes. Can't I add a code to my view something like that?:
#if(!$ajaxrequest)
#layout('layouts.master')
#endif
I want this because my codes in controllers are too messy
A slight variation is to put the logic for the layouts in your main layout template. e.g.
layouts/app.blade.php:
#if (Request::ajax())
#include('layouts.ajax-app')
#else
#include('layouts.default-app')
#endif
Your views just extend the main layout as usual e.g.
#extends('layouts.app')
#section('content')
Content goes here...
#endsection
And then create a default layout file (default-app.blade.php) and an ajax layout file (ajax-app.blade.php). The advantage of doing it this way is that any of your blade templates can be loaded via Ajax, without having to clutter up controller methods with lots of duplicated logic.
You can't have the #layout call after the if statement like that (see the notice in red under "Blade Templating" in the docs. You'll have to set the public $layout and call $this->layout->nest instead of View::make (see "The Basics" on the page linked to above).
You can use something like this in your view template:
#extends( 'layouts.' . (isset($layout) ? $layout : 'default'))
Also apply check in your controller(or Supercontroller) for AJAX request, if it is set $layout variable to needed layout. Otherwise "default" layout will be taken.
Related
I have a {{!! Form::model!!}} where I am passing a variable $values from the Controller and displaying it inside of the Form select those values. In mine main blade i have 3 additional blades I am including through
#if($jobs->open)
#include('jobs.review.open')
#endif
I am trying to load the data $jobs from controller only when someone clicks , I figured to do some by create an ajax call with onclick event, so i have the data $jobs back. An only now trying to include jobs.review.open blade.
Any help will be appreciated
One common strategy is to put the rendering in the AJAX call. The server returns a ready-to-go chunk of HTML in its reply, and the JavaScript code simply inserts it into the DOM at the appropriate place with innerHTML. The server uses a blade to prepare the HTML that it then returns.
The initial page HTML that is displayed doesn't include the content, but does include dummy placeholder <div>s which mark the places where the HTML will be inserted.
I have a view mypage.blade.php and a route.
The url is like : https://example.com/mypage/param1/param2. The route use param1 and param2 and generate the page.
Question 1
In that page, I try to get its HTML code. Is there a way to do it?. I tried render() but I don't get what I want.
Question 2
In the view, can I get the HTML code of an other view by specifying a path ?
You had the right idea. Not sure why it wouldn't work for you.
In the controller, set the view into a variable:
$view = view('myBaseView', compact('people', 'places', 'things'));
Now, if you dump the rendered view variable, you have the page's HTML:
dd($view->render());
To get the html of another view by specifying the path and using the internal controller, you would need to set up some kind of a wrapper or catch so that the view variable is not returned as a view, but rendered out to html as above. Your method would need to trap whatever the original controller was sending before it pushed out the view.
Of course, old school php can get the other page's rendered html too possibly if your server is set to allow this:
$html = file_get_contents('http://mypage.com/');
Something else you might find handy is the Laravel sections method. If you just want to render part of the page you can do so by calling whatever section you want from a partial view:
$sections = $view->renderSections(); // returns an associative array of 'content', 'pageHeading' etc
dd($sections['modalContent']); // this will only dump whats in the content section
I don't know what you want to do with this html, but if you wish to display it on a page, once you send it (you'd possibly want to return the view along with a compact of the variable $view... as a normal variable if so), remember to use this format:
{!! $view !!}
HTH
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 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 view called admin.users. In that view I include header and footer using #include directive. I write all my views-including stuff in routes.php, so for admin.users it is:
Route::get('users', function() {
// ...
return View::make('admin.users')->with('num', $usersNum);
}
And in users.blade.php:
#include('admin.partials.header')
// ....
#include('admin.partials.footer')
Is it possible to pass "users num" to header view in order to show that variable? And is it a good practice the way I'm combining views, because I read about controller layouts but actually I decided to have only rest controllers while I include view only in routes.php (like load static markup and after that communicate with server by ajax)
Have a look at 'ViewComposers' (http://laravel.com/docs/responses#view-composers). This is a great way to share data with your views and keep your routes file clean.
In global.php (or any other place really) add:
View::composer('admin.users', function($view)
{
// Do your $usersNum logic here
...
$view->with('num', $usersNum);
});
If at some point you want this data to be available in admin.dashboard as well, just rewrite to:
View::composer(array('admin.users', 'admin.dashboard'), function($view)
As I said in my comment (and what seems to have fixed your issue) you should create an admin.template view with your header and footer in it, and start your views with #extends('admin.template').