I try to send data form ajax to cakephp cotroller
function loadtooltip(obj, $user_id) {
//AJAX
var req = Inint_AJAX();
req.onreadystatechange = function () {
if (req.readyState==4) {
if (req.status==200) {
displaytooltip(obj, req.responseText);
}
}
};
req.open("POST", "http://127.0.0.1/cakeplate/tooltips/tooltip/", true);
req.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
req.send($user_id);
};
this controller
<?php
Class TooltipsController extends AppController{
var $name = 'Tooltips';
var $uses = array('Reply','User');
var $component = array('RequestHandler','Javascript','Ajax');
var $layout = 'tooltip';
function tooltip($user_id=NULL){
if(!empty($user_id)){
$tooltip = $this->Reply->User->findById($user_id);
$this->set('tooltip',$tooltip);
}
}
}
?>
I need somebody to help me to modified code
the way you're doing at the moment in the controller, you won't me able to get the user_id, because it is a var passed through GET method of http.
This variable would be accessible if you make a GET request for example for this url:
http://example.com/cakeplate/tooltips/tooltip/1 where 1 would be your $user_id.
If you send the request as POST, you can access the values in this var $this->data
This way you will be able to process the request based in the var that you pass to the controller.
Another problem that you will face that this controller will need to render a view, so i suggest that you take a look at http://book.cakephp.org/view/1238/REST, there you can see how you can create a route that will make the controller parse another view, it a different custom layout, like the json (the one i suggest in this case), and then you can show in this view only the json value.
Last, but important as well, i would suggest to that you use jQuery to do the javascript part, i think it will be easier, you can check it at http://api.jquery.com/jQuery.get
Related
I'm following a tutorial on Laravel, adding to a DB via a form. At the end of a function that saves to a DB, it returns back to the page where the form is, but I want to be taken to another page where the information is displayed. In the tutorial I created a controller with a function that returns a view containing all the database info - that element works fine however I can't seem to find a way of calling this function directly after saving to the database. I can also return any other view which just displays static view ( just html with no data handling ). Is what I'm trying to achieve possible?
public function store(){
$li = new \App\LTest1();
$li->creator = request('creator');
$li->title = request('title');
$li->views = request('views');
$li->save();
return back(); // this works
// return view('info'); // this works
//return ('Listings#showList'); this doesnt work = how do i call a function in a controller???
}
// routing
Route::get('info', function () {
return view('info'); // i can get to this static page from my store() function
});
Route::get('thedataviewpage', 'Listings#showList'); // you can route to this but not from the store() function
Redirect is the thing you need here
public function store() {
$li = new \App\LTest1();
$li->creator = request('creator');
$li->title = request('title');
$li->views = request('views');
$li->save();
return redirect('info'); // Redirect to the info route
}
Take this example. Be sure to add the proper route name and a proper message.
return redirect()->route('put here the route name')->with('success', 'Created.');'
to return to a controller action just use
return redirect()->action('Listings#showList');
or you can use route to call that controller action
return redirect('/thedataviewpage');
here is my piece of sample code
function AuthenticateLogin() {
$.getJSON('http://localhost:52293/api/APILogin/', function (data) {
if (data != null) {
}
}
);
}
The hosting url is below, which will be used across the mvc application, might be the controller/action will be varied across the application.
http://localhost:52293
for example, here i have hard coded the above url in all places, If I'm moving the application to other machine,then it is not good to change the url again in each and every places. so, is there any way to handle this ?
Give your API action a static name:
[RoutePrefix("api/APILogin")]
public class APILoginApiController {
[Route("", Name = "Login")]
public ActionResult Login(string userName) {
// ...
}
}
Then in your Razor JavaScript, you can utilize the UrlHelper by calling Url.HttpRouteUrl to dynamically build your URL for you.
$.getJSON('#Url.HttpRouteUrl("Login", new {})', function (data) {
// ...
});
The advantage of this approach is that if you change anything about how the route is formulated, it's in the [Route] attribute on the action. Matching the name like that will use the routing engine to always create the correct path. Otherwise, you're still stuck with (partial) hard-coded paths throughout your JavaScript.
If your route requires any variables, then that is provided within the empty anonymous object as the second parameter for HttpRouteUrl().
You should not hardcode the full absolute url like that. You may consider using the relative url. To generate relative url, you may consider using the Url helper methods
If your code is inside an external js file, you should consider using the helper method to generate the relative url in your razor view(s) and store it in a js variable which you can use in your external js files.
In your razor view
<script>
var myApp = myApp || {};
myApp.siteBaseUrl = "#Url.Content("~")"; // Get the app root
</script>
Now in your external js files
$.getJSON(myApp.siteBaseUrl+'api/APILogin/', function (data) {
// do something
});
You can also use Url.RouteUrl helper method to generate the urls to the api endpoints. For example
var myApp = myApp || {};
myApp.productsApiUrl = "#Url.RouteUrl("DefaultApi",
new { httproute = true, controller = "Products"})";
Now somewhere else in the js codde, you can use it like
$.getJSON(myApp.productsApiUrl , function (data) {
// do something with products data
});
This approach allows you to pass route values when you make the call and the helper method will build the url for you (based on the route definition)
myApp.productsDetailsUrl = "#Url.RouteUrl("DefaultApi",
new { httproute = true, controller = "Products", id= 210 })";
I trying to pass some json to a controller in cakePHP 2.5 and returning it again just to make sure it is all going through fine.
However I getting no response content back. Just a 200 success. From reading the docs I am under the impression that if I pass some json then the responseHandler will the return json as the response.
Not sure what I am missing.
Data being passed
var neworderSer = $(this).sortable("serialize");
which gives
item[]=4&item[]=3&item[]=6&item[]=5&item[]=7
appController.php
public $components = array(
'DebugKit.Toolbar',
'Search.Prg',
'Session',
'Auth',
'Session',
'RequestHandler'
);
index.ctp
$.ajax({
url: "/btstadmin/pages/reorder",
type: "post",
dataType:"json",
data: neworderSer,
success: function(feedback) {
notify('Reordered pages');
},
error: function(e) {
notify('Reordered pages failed', {
status: 'error'
});
}
});
PagesController.php
public function reorder() {
$this->request->onlyAllow('ajax');
$data = $this->request->data;
$this->autoRender = false;
$this->set('_serialize', 'data');
}
UPDATE:
I have now added the following to the routes.php
Router::parseExtensions('json', 'xml');
and I have updated my controller to
$data = $this->request->data;
$this->set("status", "OK");
$this->set("message", "You are good");
$this->set("content", $data);
$this->set("_serialize", array("status", "message", "content"));
All now works perfectly.
A proper Accept header or an extension should to be supplied
In order for the request handler to be able to pick the correct view, you need to either send the appropriate Accept header (application/json), or supply an extension, in your case .json. And in order for extensions to be recognized at all, extension parsing needs to be enabled.
See http://book.cakephp.org/...views.html#enabling-data-views-in-your-application
The view only serializes view vars
The JSON view only auto-serializes view variables, and from the code you are showing it doesn't look like you'd ever set a view variable named data.
See http://book.cakephp.org/...views.html#using-data-views-with-the-serialize-key
The view needs to be rendered
You shouldn't disable auto rendering unless you have a good reason, and in your case also finally invoke Controller:render() manually. Currently your action will not even try to render anything at all.
CakeRequest::onlyAllow() is for HTTP methods
CakeRequest::onlyAllow() (which btw is deprecated as of CakePHP 2.5) is for specifying the allowed HTTP methods, ie GET, POST, PUT, etc. While using any of the available detectors like for example ajax will work, you probably shouldn't rely on it.
Long story short
Your reorder() method should look more like this:
public function reorder() {
if(!$this->request->is('ajax')) {
throw new BadRequestException();
}
$this->set('data', $this->request->data);
$this->set('_serialize', 'data');
}
And finally, in case you don't want/can't use the Accept header, you need to append the .json extension to the URL of the AJAX request:
url: "/btstadmin/pages/reorder.json"
and consequently enable extension parsing in your routes.php like:
Router::parseExtensions('json');
ps
See Cakephp REST API remove the necessity of .format for ways to use the JSON view without using extensions.
Output your json data
public function reorder() {
$this->request->onlyAllow('ajax');
$data = $this->request->data;
$this->autoRender = false;
$this->set('_serialize', 'data');
echo json_encode($data);
}
I'm trying to generate ajax specific responses from my controllers by using the Request::ajax() method, which is working just fine. The only problem is that the way I have it set up right now isn't really a nice looking solution.
My controller:
class HomeController extends BaseController {
protected $layout = 'layouts/main';
public function __construct()
{
$this->beforeFilter('auth');
}
public function getIndex()
{
$view = View::make('content.home.index');
if(Request::ajax()) return $view; //For ajax calls we only want to return the content to be placed inside our container, without the layout
$this->layout->menu = 'content.menu';
$this->layout->content = $view;
}
}
So right now, for every method I define within my controllers I need to add the code snippet that checks for an AJAX request and returns a single view if the statement returns true.
This leads to my question that is probably more PHP related than it is to the framework;
Is there a way of executing my AJAX check on every method call, without actually placing it inside the method? Or is there some other solution to keep my code DRY?
Thanks in advance!
PS: This is my first post on stackoverflow, so feel free to correct me if I made any mistakes
Create a new barebone layout named 'layouts/ajax' (or any name you like).
<?php echo $content ?>
In your Base controller, override this setupLayout() function.
protected function setupLayout()
{
if ( ! is_null($this->layout))
{
$layout = Request::ajax() ? 'layouts/ajax' : $this->layout;
$this->layout = View::make($layout);
}
}
Change your getIndex() function to this.
public function getIndex()
{
$view = View::make('content.home.index');
$this->layout->menu = 'content.menu';
$this->layout->content = $view;
}
Now non-ajax requests will be rendered using layout set in the controller, where as ajax requests will receive whatever set to $this->layout->content.
Note : Controller will neglect the layout setup in setupLayout(), if the called method returns truthy value. So this method will not work for functions like below.
public function getIndex()
{
return View::make('content.home.index');
}
You could just change the layout property, in the constructor, if it's an ajax request:
public function __construct()
{
$this->beforeFilter('auth');
if(Request::ajax()) {
$this->layout = '';
}
}
If it doesn't work try setting it to NULL instead.
Why would you return a VIEW via ajax? Are you using it to create a SPA? If so there are better ways. I'm generally against returning HTML via AJAX.
The route I'd go in your position is probably opposite of how you're doing it. Render the view no matter what, if the request is ajax, pass the extra data back and have JS render the data on the page. That's essentially how most Javascript MVC frameworks function.
Sorry if I am totally missing the point here, just going on an assumption of your end goal with the info you provided.
In asp.net mvc3,
I am using a javascript API to dynamically render a user interface. Part of the input section is going to be dependent on how many items the user wants to enter data for. As a result, something like this wont work
#(Html.EditorFor(m => m.P[5].C.Description))
because that cannot be done during runtime. What type of process would I use to call that helper during runtime with AJAX? Would I have a controller action which only returned that information which was called using $.ajax()? Would it be in a different place than a controller action?
At run time you could perform an ajax get to a controller action that will render a view as a string, which in turn could be inserted / appended into the DOM.
Create a new action result that returns JSON as per below:
return new JsonResult
{
JsonRequestBehavior = JsonRequestBehavior.AllowGet,
Data = new { html = this.RenderPartialViewToString("YourPartialView", model) }
};
Note, the above makes use of the following controller extension:
public static string RenderPartialViewToString(this Controller controller, string viewName = null, object model = null)
{
if (string.IsNullOrEmpty(viewName))
{
viewName = controller.ControllerContext.RouteData.GetRequiredString("action");
}
controller.ViewData.Model = model;
using (var sw = new StringWriter())
{
ViewEngineResult viewResult = ViewEngines.Engines.FindPartialView(controller.ControllerContext, viewName);
var viewContext = new ViewContext(controller.ControllerContext, viewResult.View, controller.ViewData, controller.TempData, sw);
viewResult.View.Render(viewContext, sw);
return sw.GetStringBuilder().ToString();
}
}
For further reading regarding this extension method: http://craftycodeblog.com/2010/05/15/asp-net-mvc-render-partial-view-to-string/
All that would remain would be to perform a get passing a parameter signifying the number of items to render and append the returned content into your view. Perhaps something like:
$.getJSON('url', numberofitems, function (data) {
$('#somecontainer').html(data.html);
});
If you are pulling HTML and inserting it into the DOM, you don't have to go via JSON.
Just have your action return a PartialView. It's already in the form of Html, and ready to be inserted into your DOM
JS
$.getJSON('/someurl/GetMyView',{count:10}, function (data) {
$('#target').html(data);
});
Controller:
[HttpGet]
public ActionResult GetMyView(int count)
{
MyModel model = //Get the model from somewhere
return PartialView(model);
}
View:
#model MyModel
<div>
#Model.SomeProperty
<div>
If I understand you correctly, you want to dynamically insert fields on the client to allow users to add N-number of fields without having a bazillion pre-rendered fields on the form and you are trying to use ajax to do this?
I'm sure you could do it, by rendering the html on the server and pushing it up to the client... have you considered dynamically adding the to the page via javascript? Unlike Webforms, MVC does not care what elements were on the page when it was rendered, it only cares about the data it receives in the HttpPost.