Ajax not getting only json output data (it print whole loaded view code.).? codeigntier

Question

Here is my little script code I want to get data from codeingiter controller. I get json data from controller to view ajax, but It print with html page code. any one can help me here, How can I solve this.

I only want to get json data ans a variable data to my page.

this is output that I am getting but this is comming with html code and I don't want html code.

[{"id":"1","p_name":"t_t11","p_type":"t_t1","paid_type":"0"},{"id":"2","p_name":"t_t12","p_type":"t_t1","paid_type":"1"},{"id":"3","p_name":"t_t1","p_type":"t_t1","paid_type":"0"}]

I have follow some question answers but can't et success, because that question's answers not related to me.

Link 1
Link 2 and many more...

<script>
   $("a.tablinks").on('click',function(e){
        e.preventDefault();     
        var p_name = $(this).attr('value');
        alert(p_name);
    $.ajax({            
            url:"<?php echo base_url(); ?>teq/gettabdata",                  
            dataType:'text',
            type: "POST",          
            data:{p_name : p_name},        
           success : function(data){
                    alert(data);
                if(data !=""){

                    var obj = JSON.parse(data);
                    alert(obj.id);
                    /*$.each(obj, function(key,val){
                         console.log(key);
                         console.log(val); //depending on your data, you might call val.url or whatever you may have
                    });*/

                }else{
                    alert(data+ '1');
                }
           },
           error : function(data){

                //var da = JSON.parse(data);
                alert(data+ '2');
                //alert(da+ '2 da ');
           }
        });
    });
</script>

Here is controller code.

  public function gettabdata(){

                    $p_name = $this->input->post('p_name');
                    //echo $p_name." this is paper name.!";

                    $tabs_data['res1'] = $this->db->distinct()->select('p_type')->from('t_name')->get()->result();      

                    //$p_name = $data;
                    $query['res'] = $this->db->select('*')->from('t_name')->where('p_type',$p_name)->get()->result();

                    echo json_encode($query['res']);
                    $this->load->view('teq', $tabs_data);


        }

The javascript is in the view teq?

Jacky Shek
– Jacky Shek

2017-04-18 09:36:38 +00:00
Commented Apr 18, 2017 at 9:36 — Jacky Shek
– Jacky Shek, Commented Apr 18, 2017 at 9:36

Sunny Khatri · Accepted Answer · 2017-04-18 07:34:04Z

2

You added view at the end of your function that return view's code.

Remove line:

$this->load->view('teq', $tabs_data);

answered Apr 18, 2017 at 7:34

Sunny Khatri

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3 Comments

Mahi Over a year ago

I have only one controller to load page so i must need it

Sunny Khatri Over a year ago

so you are getting data through gettabdata controller and set also with that using ajax?

Sunny Khatri Over a year ago

you can't do that with single controller because getting data and load view at same time so function return view also

Ankit Jain · Accepted Answer · 2017-04-18 16:55:14Z

1

You can either use 
if ($this->input->is_ajax_request()) {
       echo json_encode($data_set);
    }else{
//Procced with your load view

}

Or if you're avoiding ajax request check then please pass any extra paramter from your ajax call then then check for its existence at your controller and on behalf of it proceed your conditional statement . it will solve your problem

answered Apr 18, 2017 at 16:55

Ankit Jain

637 bronze badges

Comments

Prima Putra · Accepted Answer · 2017-04-18 09:23:28Z

0

Change your controller method like this:

public function gettabdata(){
    $p_name = $this->input->post('p_name');
    //echo $p_name." this is paper name.!";
    $tabs_data['res1'] = $this->db->distinct()->select('p_type')->from('t_name')->get()->result();      
    //$p_name = $data;
    $query['res'] = $this->db->select('*')->from('t_name')->where('p_type',$p_name)->get()->result();
    // if ajax request
    if ($this->input->is_ajax_request()) {
       echo json_encode($query['res']);
       return; // exit function
    }

    $this->load->view('teq', $tabs_data);
}

answered Apr 18, 2017 at 9:23

Prima Putra

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Comments

Deepak M · Accepted Answer · 2017-04-19 04:29:25Z

In your ajax code chage dataType: to json

$.ajax({            
    url:"<?php echo base_url(); ?>teq/gettabdata",                  
    dataType:'json',
    type: "POST",          
    data:{p_name : p_name},        
    success : function(res)
    {
        if(res !=""){
            alert(res.id);
        }else{
            alert(res+ '1');
        }
   }
});

And in your controller

public function gettabdata()
{
    if($this->input->post('p_name'))
    {
        $p_name = $this->input->post('p_name');
        $query['res'] = $this->db->select('*')->from('t_name')->where('p_type',$p_name)->get()->result();
        if($query['res'])
        {
            $resp = $query['res'];
        }
        else
        {
            $resp = array('status' => FALSE,'msg' => 'Failed');
        }
        echo json_encode($resp);
    }
    else
    {
        $tabs_data['res1'] = $this->db->distinct()->select('p_type')->from('t_name')->get()->result();
        $this->load->view('teq', $tabs_data);
    }
}

Hope this helps :)

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