I have a controller that has an ajax function and a seperate action that redirects the page. I want to know is it possible to make the ajax call when the link is clicked, but wait for the function to finish before the redirect action is called?
E.g.
<g:link controller="myController" action="myRedirectAction" before="saveData">link</g:link>
EDIT Added code
controller
def ajx_saveServiceGroup = {
//Code to save data to object
return
}
def saveConfigToRoLo = {
//code to save object to DB
redirect(action:"displayPDFSummary", id:orderId, params: [origSessionId: params.origSessionId, theSession: tempSession])
}
gsp
<g:link class="buttonSend" action="saveConfigToRoLo" id="${orderDataInstance.id}" params="[origSessionId: origSessionId, orderId: orderDataInstance.id, submitToBT: true]" before="ajx_saveServiceGroup">Submit</g:link>
you can have a remoteLink and then onSuccess callback you can change the windows.location to whatever you want !
A details description of what you are trying to achieve will help us provide better answers.
Related
I have a sign in page,the page send request to the controller and show errors after getting back the result.
This is the ajax call in sign in page
$.ajax({
url:"signin", //changed into temp.html page
data:{
username:$username,
password:$password,
email:$email },
success:function(errCode){
if(errCode==0){
alert('Error,wrong username/password');
}else{...}
},
type:'POST'
});
The problem is in controller def signin() I don't know how to return the data's value back to ajax,it either didn't work or wiped the page clean and printed just the data's value.How can I do it,is it the same if I want to return a javascript type file?
Edit:I found a way to work around it,I changed the ajax request destination to a temp page and return the value directly.Here is my view function in controller
def signin():
return dict()
def temp():
errCode=0
return errCode
I would like to be able to construct a query using laravel, and serialize it into a url string.
This would allow me to create routes which would unserialize a query builder, run the query, and make a view which displays the database results.
For example, to implement a button which refreshes a list of posts made by kryo:
http://example.com/ajax/posts.php?name=kryo&order_by=created_at&order_type=desc
Posts.php would simply be a route which unserializes, validates, and runs the query in the url params, and provides the results to a view.
Perhaps this is not useful in general, but I would personally find it handy specifically for ajax requests. If anyone knows how to implement this as a laravel plugin of some nature, that would be fantastic.
I'll try to give you a basic idea:
In Laravel you have to create a route to make a request to a function/method, so at first you need to create a route which will be listening for the ajax request, for example:
Route::get('/ajax/posts', array('uses' => 'PostController#index', 'as' => 'showPosts'));
Now, create a link in the view which points to this route, to create a link you may try this:
$url = to_route('showPosts');
If you use something like this:
<a class='ajaxPost' href="{{ $url }}?name=kryo&order_by=created_at&order_type=desc">Get Posts</a>
It'll create a ink to that route. So, make sure you are able to pass that $url to your JavaScript or manually you can write the url using /ajax/posts?name=.... Once you done creating the link then you need to create your JavaScript handler for this link (maybe using click event) then handle the click event from your handler, make ajax request, if it's jQuery then it could be something like this:
$('.ajaxPost').on('clcik', function(e){
e.preventDefault();
var url = $(this).attar('href');
$.getJSON(url, function(response){
$.each(response, function(key, value){
// loop... you may use $(this) or value
});
});
});
In your PostController controller class, create the index method:
class PostController extends BaseController {
public function index()
{
$name = Input::get('name');
$order_by = Input::get('order_by');
$created_at = Input::get('created_at');
$order_type = Input::get('order_type');
$posts = Post::whereName($name)->orderBy($order_by, $order_type)->get();
if(Request::ajax()) {
return Response::json($posts);
}
else {
// return a view for non ajax
}
}
}
If you want to send a rendered view from the server side to your JavaScript handler as HTML then change the getJson to get and instead of return Response::json($posts); use
return View::make('viewname')->with('posts', $posts);
In this case make sure that, your view doesn't extends the master layout. This may not be what you need but it gives you the idea how you can implement it.
sorry for asking this question ..i am working on a Cakephp 2.x ... i have a view page in my controller name folder e.g Controller/index.ctp ... and i have ajaxfiles are stored in app/webroot/ajax/ajaxfile.html
now on my index.php file i am acessing the ajax page like this
<a href="ajax-demo/ajaxfile.html" class="file-link">
<span class="icon file-png"></span>
Simple gallery</a>
Controller
public function index(){
}
now the problem is i want to send the variables to both of my pages ... index.ctp and ajaxfile ... how can i do this ??what is the best approach to tackle these things ....
do i have to move the ajaxfiles from webroot and paste under the controller name folder?
if is it so then how can i send variables to ajax files which has no model and controller
please if any one know the solution then please advice me. and give an example too
There are different way to achieve this, here I'm writing the simplest one
First you need to move your "index.ctp" file to your "View/YOUR CONTROLLER NAME/" folder.
1) To access the variable in view you need to set it from your controller's method like this
public index(){
$this->set('yourVariable', 'Your Value');
}
2) To access the value in your view file (index.ctp), you need to call this variable like this
$yourVariable;//If you want to print this then you can write like this
echo $yourVariable;
3) To call a ajax file from your index.ctp the simplest method is to call a onclick event on this anchor, the onclick event will call a JAVASCRIPT method which will further make a ajax call and will place the output in an element in your index.ctp, The ajax call will further call your controller method (implement your html related logic here)
For example,
<span class="icon file-png"></span>Simple gallery
<div id="yourAjaxFileOutputReplaceMentDiv"></div>
4) create a javascript method in your JS file, this JS file must be loaded in your layout file.
function yourAjaxCallMethod(BaseURL,yourVarible)
{
//Initialize Ajax Method
var req = getXMLHTTP();//Let's this method Initialize your Ajax
if (req)
{
req.onreadystatechange = function() {
if (req.readyState == 4)
{
if (req.status == 200)
{
document.getElementById('yourAjaxFileOutputReplaceMentDiv').innerHTML=req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
var URL = BaseURL+yourVarible+'/'+Math.random();
req.open("GET", URL, true);
req.send(null);
}
}
5) Your AJAX file related method in your controller "yourController". Set autoRender to False
public function ajaxMethod(){
$this->autoRender = false;
//Check $this->request['pass'] for arguments send from ajax call
$retreivedVariable = $this->request['pass'][0];
echo 'I retrieved variable'.$retreivedVariable;
}
However instead of writing core javascript and ajax method you can call the inbuild Ajax Helper for same.
I am making an ajax call to controller to post data from view to controller.And in the receiving controller I am updating my model with new values.Now I want to bind this new model to view again in success call of $.ajax post.Please Suggest.
one way to do this is to return a partial view from the controller. You can replace the contents of your previous view with the new html content. Lets expand on this...
so, here is your controller action
[HttpPost]
public ActionResult SomeMethod(params...){
....
var model = some model;
...
return PartialView("ViewName",model);
}
and in the ajax, use
$.ajax({
url : #Url.Create("Action","Controller"),
type : 'POST',
data: { ... your data params ..},
success : function(result){
$("#ContainerId").html(result);
}
})
in the html you would need a div with the id = "ContainerId". The content would get swapped out by the html passed back in the success function.
The Model is only used in RAZOR when rendering the page. Once you get to the point where you are using AJAX, the model is no longer available to you.
What, exactly, are you trying to accomplish? Maybe there is another way to do it?
since I'm new in jquery, can you tell me how to redirect a page to another action method ?
I develop a MVC web application. I use jquery post method to do some validation, and when it return true, it will be redirect page to another one.
My problem is..when when I redirect page using window.location, it's works well in IE (IE 9). but didn't work on firefox & chrome.
So, I try to using jquery post method to redirect page from action method in my controller. I call redirect post method in a jquery post call back.
it is my code :
$.post(posturl, formData, function (result) {
if (result == 'True') {
$.post("/Controller/RedirectMethod", {_Action: 'Index', _Controller: 'Home'}, null);
}
else
alert('failed');
}
);
and this is my RedirectMethod :
[HttpPost]
public ActionResult RedirectMethod(string _Action, string _Controller)
{
return RedirectToAction(_Action, _Controller);
}
so how I should create a nested post in another post callback ?
or there is another way to redirect page ?
thanks,
You won't be able to do a RedirectToAction inside an Ajax request.
If you need the web page to change to a different location inside the Ajax response, use window.location as Ben suggested.
One thing to bear in mind is that you'll need to remove the 'HttpPost' action filter.