load php file using ajax in codeigniter

Question

I'm new in web development. How can i load my php file that contains html in order to change the span content. Im using codeigniter by the way.

I always get an error.

here is my js file

$(document).ready(init);

function init(){
$('#home').click(doAjax);
}


function doAjax(){  
        $.ajax({
            data:{
                'n':1
            },
            type: 'get',
            success: ajaxSuccess,
            error: ajaxError,
            url: 'http://localhost/index.php/main/ajax'
        });


        function ajaxSuccess(data, textStatus){
            alert("sucess");
            $('#main-span').load(data);         
        }

        function ajaxError(data, textStatus){
            alert("error");
            $('#main-span').html('error');
        }





    }

controller

public function ajax(){
        $input = $this->input->get('n');
        $input = '/application/views/student_main.php';
            echo $input;
}

maincontent - this is where i want the span to update

<div class="container-fluid">
<div class="row">

<div class="col-sm-3 col-md-2 sidebar" id ="nav-sidebar">
<?php include('/application/views/navigation.php');?>
</div>

<span id="main-span">
        <?php include('/application/views/evaluation.php');?>
</span>


</div>
</div>

thanks in advance for those who can answer my problem.

Kumar V · Accepted Answer · 2014-02-22 05:53:49Z

1

Try this code:

$.ajax({
        url: "http://localhost/index.php/main/ajax",
        type: "get",
        data: {'n':1},

        success: function(data,txtstatus){
            alert("success");
            $('#main-span').load(data.next_url);   
        },
        error:function(){
            alert("failure");
            $('#main-span').html('error');
        }
    });

In your controller:

public function ajax(){
        $input = $this->input->get('n');
        $input = '/application/views/student_main.php'; // you can't access view directly. call this view from a controller.
        echo json_encode(array("next_url"=>$input));
}

edited Feb 22, 2014 at 5:53

answered Feb 22, 2014 at 5:40

Kumar V

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3 Comments

user2516654 Over a year ago

still, i got an error. i don't know what it is but when i check the error it says undefined

Kumar V Over a year ago

what is undefined? data?

user2516654 Over a year ago

sorry, i don't know but i just get the responsetext in the function ajaxError and all i got i undefined

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