CODEWITHSUNDEEP
javascriptphpajaxyii
Stev_k
Stev_k

Reputation: 2126

Pass variables from javascript to yii via ajax

I am writing a Yii application (or I should say building on top of someone else's Yii application!). I need to use quite a lot of javascript, because I'm using the Leaflet map library. So a lot of this is external to the PHP code and called with registerScriptFile, registerScript etc.

To get data from the javascript into the main Yii application, I have been using AJAX calls.

$.ajax({
    type: "POST",
    url: "xxx/index.php/site/updateajax",
    data: {recordID: feature.properties.id}
}).success(function(result){
    $("#recordTable").replaceWith(result);
});

and 

public function actionUpdateAjax()
{
    $data = $_POST['recordID'];
    $this->renderPartial('/map/_legend_layers', array('data' => $data),false,false);
}

in the php. I discovered I needed to call $("#element").replaceWith(result); because otherwise the rendering was being returned but not being rendered.

In this case this works OK because I can call replaceWith, but I have a nagging feeling this isn't the right way to do it, and also I have another AJAX call that renders a whole page so I can't really replace that.

My question is how can I either force the html returned by these AJAX calls to render the page, or more likely I suppose, how can I just use AJAX to update a variable, return back a success code or something, then let the yii code carry on and do it's thing - I'm only using AJAX as a variable passing mechanism.

Thanks

Upvotes: 0

Views: 914

Answers (2)

Stev_k
Stev_k

Reputation: 2126

This worked for me when I needed to replace the whole page with the html from an AJAX call:

$.ajax({
type: "POST",
url: "xxx",
success: function(result){
document.open();
document.write(result);
document.close();
});

The only issue I've found so far is that it doesn't update the url...

Upvotes: 0

ernie
ernie

Reputation: 6356

This looks normal to me . . . you're using renderPartial to generate HTML, and then you're using the callback in your AJAX to put that HTML where you want it via replaceWith. For the second use case where you mention you want a whole page, it sounds like you'll want to render the page, rather that do a renderPartial . . .

The one change I'd make is to use a success callback in the actual jQuery.ajax() call, rather than the jqXHR.success() method that's going to be deprectated in 1.8, i.e.:

$.ajax({
    type: "POST",
    url: "xxx/index.php/site/updateajax",
    data: {recordID: feature.properties.id},
    success: function(result){ $("#recordTable").replaceWith(result); }
});

Upvotes: 1

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