drn1008 2015-06-19 06:02
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当从AJAX POST请求调用控制器时,如何从Codeigniter中的控制器加载视图?

I have a view which has an AJAX POST to a controller list_products_of_company :

$('select.product_dimension').change(function () {
  var a = "a";
  var b = "b";
  $.post("<?php echo base_url(); ?>home/list_products_of_company",
            {
                a : a, b : b
            }       
 )
});

Controller :

function list_products_of_company(){ 
        $a = $this->input->post('a');
        $b = $this->input->post('b');
        $this->data['products_of_company'] = $this->Home_model->get_products_of_company($a,$b);
        $this->load->view('product_list',$this->data);
}

How do I load the view with the data that I got from the database?

Edit:

I don't want to load the ajax response inside a div. I need the entire view page to be loaded in the window.

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3条回答 默认 最新

  • douji1999 2016-11-14 16:20
    关注

    If you want to load the view from your codeigniter controller don't use AJAX here instead I suggest you to modify your HTML like this:

    <form id="form" method="post" action="action.php">
        <select name="select" id="select">
            <option value="">Select</option>
            <option value="1">One</option>
            <option value="2">Two</option>
        </select>
        <input type="submit" style="visibility:hidden;" name="sub" value="submit" />
    </form>
    

    And your jquery code like this so that you can directly submit the form on change of the select box and the action goes to the CI controller:

    $(function () {
        $('#select').change(function () {
            $('#form').submit();
        });
    });
    

    And your CI home controller:

    function list_products_of_company(){ 
            $a = $this->input->post('a');
            $b = $this->input->post('b');
            $this->data['products_of_company'] = $this->Home_model->get_products_of_company($a,$b);
            $this->load->view('product_list',$this->data);
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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