Is there a way to change the Url generated by the LinkPager in Yii2?
I have a page with results that are loaded with ajax when the search field is updated. But my pagination buttons links to a new page with the ajax-call url.
I looked in the files, and it seems like this function generates the url
public function createUrl($page, $pageSize = null, $absolute = false)
and that is not option to change. What is the way to solve this problem?
My results page is at the front page, but the pagination buttons links to ajax/results?page=2&page-size=10.
What can be more elegant then extending a class and replacing the renderPageButton function with your own :).
Another way, probably more the yii2 way :) if you take a look at LinkPager you will see that it is not responsible of creating the URLS. Pagination is https://github.com/yiisoft/yii2/blob/master/framework/data/Pagination.php#L257 The createUrl function in Pagination creates the URL and it takes into consideration the $route defined for the pagination object. If there is no $route defined then it takes the current controller's route. So by setting up $route for the pagination you can set the URL the buttons will go to.
Related
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:
{!! $view !!}
HTH
Currently my users must get the visit form given by Route::get then fill it in to get back a result view given by Route::post. I need to create a shareable link such as /account/search/vrm/{vrm} where {vrm} is the VRM that is usually filled in on the form page. This VRM then needs to redirected to Route::post as post data. This needs to be done by my controller. How can I do this in my controller?
Routes:
// Shows form view
Route::get('/account/search', 'User\AccountController#getSearch')->name('account.search');
// Shows result view
Route::post('/account/search', 'User\AccountController#runSearch');
// Redirect to /account/search as POST
Route::get('/account/search/vrm/{vrm}', function($vrm) { ???????? });
POSTs cannot be redirected.
Your best bet is to have them land on a page that contains a form with <input type="hidden"> fields and some JavaScript that immediately re-submits it to the desired destination.
You can redirect to a controller action or call the controller directly, see the answer here:
In summary, setting the request method in the controller, or calling a controller's action.
Ps: I don't want to repeat the same thing.
For those who comes later:
If you are using blade templating engine for the views, you can add '#csrf' blade directive after the form starting tag to prevent this. This is done by laravel to prevent cross site reference attacks. By adding this directive, you can get around this.
return redirect()->route('YOUR_ROUTE',['PARAM'=>'VARIABLE'])
I have different pages where users can add comments. I want to save the url of the page where the comment was made so that I can see it afterwards. Is this possible without hiding the the current url to the comment form? For example has the request() such a property? Or is hiding the url to the form the only way for that.
Yes there is a global helper for that
url()->previous();
url()->current();
url()->full();
If you prefer to use $request:
$request->url();
https://laravel.com/docs/5.4/helpers#urls
I have a working project on Codeigniter 3. Now I have to build a FAQ page and I had this doubt: do I need a Controller for every URL?
It is, the FAQ page is a static page, but CodeIgniter generally routes URLs to Controllers, like domain/controller/method. But it seems a waste to build a Controller to only load the View.
No, it's not right way to make controller for every page. Just make one function which shows page by fetching data from database.
First of all make a table named pages in your database then save page_content, page_name, permalink for your different pages.
Now suppose your default controller is home, make a function in it with name page as below.
function pages( $permalink )
{
// get page data based on page_name passed in URL
$this->db->where( array( 'permalink' => $permalink ) );
$data['page'] = $this->db->get( 'pages' )->result();
// load view and pass page object to view
$this->load->view( 'view_file', $data );
}
Now same function will show different page content based on permalink passed in URL.
For example if URL is www.example.com/index.php/home/pages/faq then content of faq page will be shown.
Ok, this idea might seem quite a bit crazy and it kindo' is (at least for me at my level).
I have a fairly standarad rails app (some content pages, a blog, a news block, some authentication). And I want to make it into a single page app.
What I want to accomplish is:
All the pages are fetched through AJAX like when using turbolinks, except that the AJAX returns only the view part (the yield part in the layout) withought the layout itself, which stays the same (less data in the responces, quicker render and load time).
The pages are mostly just static html with AngularJS markup so not much to process.
All the actual data is loaded separately through JSON and populated in the view.
Also the url and the page title get changed accordingly.
I've been thinking about this concept for quite a while and I just can't seem to come up with a solution. At this point I've got to some ideas on how this actualy might be done along with some problems I can't pass. Any ideas or solutions are greatly appreciated. Or might be I've just gone crazy and 3 small requests to load a page are worse then I big that needs all the rendering done on server side.
So, here's my idea and known problems.
When user first visits the app, the view template with angular markup is rendered regularly and the second request comes from the Angular Resource.
Then on ngClick on any link that adress is sent to ngInclude of the content wrapper.
How do I bind that onClick on any link and how can I exclude certain links from that bind (e.g. links to external authentication services)?
How do I tell the server not to render the layout if the request is comming from Angular? I though about adding a parameter to the request, but there might be a better idea.
When ngInclude gets the requested template, it fires the ngInit functions of the controllers (usually a single one) in that template and gets the data from the server as JSON (along with the proper page title).
Angular populates the template with the received data, sets the browser url to the url of the link and sets the page title to what it just got.
How do I change the page title and the page url? The title can be changed using jQuery, but is there a way through Angular itself?
Again, I keep thinking about some kind of animation to make this change more fancy.
Profit!
So. What do you guys think?
OK, in case enyone ever finds this idea worth thinking about.
The key can be solved as follows.
Server-side decision of whether to render the view or not.
Use a param in the ngInclude and set the layout: false in the controller if that param is present.
Have not found an easier way.
Client-side binding all links except those that have a particular class no-ajax
Here's a directive that does it.
App.directive('allClicks', function($parse) {
return {
restrict: 'A',
transclude: true,
replace: true,
link: function(scope, element, attrs) {
var $a = element.find('a').not($('a.no-ajax')),
fn = $parse(attrs['allLinks']);
$a.on('click', function(event) {
event.preventDefault();
scope.$apply(function() {
var $this = angular.element(event.target);
fn(scope, {
$event: event,
$href: $this.attr('href'),
$link: $this
});
});
});
}
};
})
And then use it on some wrapper div or body tag like <body ng-controller="WrapperCtrl" all-links="ajaxLink($href)"> and then in your content div do <div id="content" ng-include="current_page_template">
In your angular controller set the current_page template to the document.URL and implement that ajaxLink function.
$scope.ajaxLink = function(path) {
$scope.current_page_template = path+"?nolayout=true";
}
And then when you get your JSON with your data from the server don't forget to use history.pushState to set the url line and document.title = to setr the title.