使用AJAX将Zend PHP返回的数组转换为JQuery，而无需重新加载页面

问题描述：

I'm having a bit of a problem using AJAX with Zend. Apologies if I'm rambling a little. I'd rather give too much information than not enough.

OK, so I have a table of events in my view tpl that should be updated when when a set of dates is selected. In order to do this, when dates are selected, I have an AJAX request sent to "Controller/Action?dates=somedates". There is then some code done to break this up and send a mysql request through the model. The model then returns an array of events to the controller.

All of this so far works fine. The issue I have from here is getting this array back to JQuery so I can insert it into the table on my tpl. Whenever the PHP hits the return statement in the action, it tries to reload the page.

function getEvents(){
    var string = $('#datesselect .entrybox').find('.selected').attr('data-cleandate');

    $.ajax({
        url : '/Listing/getevents?dates=' + string,
        type : 'get',
        success : function(data){
            console.log(data);
        }
    });
}

Basically I just want the following: AJAX calls PHP > PHP does model call and returns array of results > Array is passed back to JQuery without reloading the page.

Is this possible? Am I doing something fundamentally wrong? Any ideas at all?

Thanks for your time.

答

Add this to your controller:

public function init()
{
    $ajaxContext = $this->_helper->getHelper('AjaxContext');
    $ajaxContext->addActionContext('getevents', array('json', 'html'))->initContext();
}

This will tell zend that you either return json or html format.

Then in your geteventsAction do the following:

$this->_helper->layout()->disableLayout();
$this->_helper->viewRenderer->setNoRender(true);
echo Zend_Json::encode($your_json_data);
exit;

This will prevent rendering of your layout and view and will only output your json.

You could also use $.getJSON. This will treat your returned data as json.

使用AJAX将Zend PHP返回的数组转换为JQuery，而无需重新加载页面

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