Problem with AJAX

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Problem with AJAX

Asked 10 years ago

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I am developing a site in ASP.NET MVC, I wanted that when I changed screen only be updated my div .ESTRUTURA here is the code:

Javascript:

function loadPage(page) {
    $.ajax({
        type: 'GET',
        url: page,
        dataType: 'html',
        cache: true,
        async: true,
        success: function (data) {
            console.log($('.ESTRUTURA').html(data));
            $('.ESTRUTURA').html(data);
        }
    });
}

HTML (In this case it is the _Layout.cshtml)

<body>
    <div class="bg">
        @Html.Partial("_Header")
        <div class="ESTRUTURA"> 
            @RenderBody()
        </div>
        @Html.Partial("_Footer")
    </div>
    <script src="~/Scripts/jquery-1.8.2.min.js"></script>
    <script src="~/Scripts/vendors/underscore.js"></script>
    <script src="~/Scripts/vendors/backbone.js"></script>
    <script src="~/Scripts/app.js"></script>
</body>

Inside my DIV .ESTRUTURA has the @RenderBody(), I believe the problem is there. The variable data, receives the entire HTML of the page (but should not).

So your problem is in the construction of the variable data. You must review what you are getting in your request ajax

– CesarMiguel

2014/11/14 at 12:25
The variable receives the return of @Renderbody(), I was able to fix this problem by changing the return of the controllers to Partialview(), but the first page I load (Home), has to be View(), otherwise the other content of the page like Header Footer and navbar

– Danilo Oliveira

2014/11/14 at 12:29
That’s right, your home has to be a view. However I see your problem... The variable data receive the return of the Renderbody??

– CesarMiguel

2014/11/14 at 12:34
Yes, that in the case when I put Partialview(), the Renderbody() only and returns content, but when I put View() it returns me the whole page, I need him to only run the Home View() the first time, the others need to be like Partialview()

– Danilo Oliveira

2014/11/14 at 12:47

1 answer

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by Diego Vieira • **3,782** points · Answer 1 · 2014-11-14T13:00:57+00:00

Do it like this, it can work in two ways.

The first is by making the action just return to partial view, that is, the view without the master page (_Layout.cshtml).

Controller (First form)

public ActionResult MinhaAction() {
   return PartialView();
}

But we also have a problem, if the user decides to access the URL directly will only return the partial part, in which case I would work as below.

Controller (Second form)

public ActionResult MinhaAction() {
   return View();
}

[HttpPost, ActionName("MinhaAction")]
public ActionResult MinhaActionPartial() {
  return PartialView();
}

And you should change the request method to POST instead of GET, so will make every request POST returns the partial part and GET the full page.

Jquery

function loadPage(page) {
    $.ajax({
        type: 'POST',
        url: page,
        dataType: 'html',
        cache: true,
        async: true,
        success: function (data) {
            console.log($('.ESTRUTURA').html(data));
            $('.ESTRUTURA').html(data);
        }
    });
}