How to get jquery ajax response data and display in codeigniter view html div id

Question

I am new in codeigniter.

I want to display data in codeigniter view div id from ajax on window load.

View.php

<script type='text/javascript' language='javascript'>
$(window).load(function() 
{
    $.ajax({
       type: "POST",
       url: "<?php echo base_url(); ?>colleges/index",       
       success: function(server_response
       {
           if(server_response == 'success')
           {
              $("#fillgrid").html(server_response); 
           } 
           else{
              alert('Not OKay');
           }
                      }
   });   //$.ajax ends here
});
</script>

What is the problem here? Are you stuck somewhere?

Shrikant Mavlankar
– Shrikant Mavlankar

2016-07-15 17:42:39 +00:00
Commented Jul 15, 2016 at 17:42 — Shrikant Mavlankar
– Shrikant Mavlankar, Commented Jul 15, 2016 at 17:42

Louys Patrice Bessette · Accepted Answer · 2016-07-15 18:22:53Z

0

I added explanations directly in code:

$(window).load(function() 
{
    $.ajax({
       type: "POST",
       url: "<?php echo base_url(); ?>colleges/index",       
       success: function(server_response)           // Added missing parenthesis
       {
           if(server_response == 'success')         // You already are in the success callback.
           {                                        
              $("#fillgrid").html(server_response); 
           } 
           else{
              alert('Not OKay');
           }
                      }
   });   //$.ajax ends here
});

Function simplifed:

$.ajax({
    type: "POST",
    url: "<?php echo base_url(); ?>colleges/index",       
    success: function(server_response){
        $("#fillgrid").html(server_response); 
    },
    error: function(){
        alert('Not Okay');
    }
});   //$.ajax ends here

Don't place an Ajax function inside a .load() method !
.load is a jQuery shorthand for .ajax().
So it's one or the other... But not both.

Using .load(), it would be:

$("#fillgrid").load("<?php echo base_url(); ?>colleges/index");

edited Jul 15, 2016 at 18:22

answered Jul 15, 2016 at 17:47

Louys Patrice Bessette

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5 Comments

Durgesh Pandey Over a year ago

In my model.php public function get_all_college() { $this->db->order_by("id","DESC"); $query = $this->db->get("tbl_colleges"); return $query->result_array(); } In my controller.php public function index() { $list['college'] = $this->Colleges_model->get_all_college(); $this->load->view('header'); $this->load->view('Colleges/all_colleges', $list); $this->load->view('footer'); } It is right above code.

Louys Patrice Bessette Over a year ago

Ajax server_response contains the html produced. I'm not really used to model/controller/view framework... You have to make something to echo the result.

Louys Patrice Bessette Over a year ago

«It is right above code.» It does'nt matter where the PHP script is versus the jQuery script. PHP executes server-side... And jQuery executes on client-side, "long" after the PHP. ;) So here, you produce a page that recalls PHP again... I hope you don't mke the common error : a page that calls itself... The ajax called content should be another PHP script than the one of the main page.

Louys Patrice Bessette Over a year ago

Here is a previous answer I made to "explain" things about a working Ajax example... May be usfefull to you. stackoverflow.com/questions/38187894/…

Louys Patrice Bessette Over a year ago

Hey!! I just saw that!!! You placed an Ajax call inside a .load() method ?!?! I edited my answer.

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