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？

问题描述：

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.

我有一个视图 strong>，它有一个到控制器的AJAX POST strong> list_products_of_company code>： p>

  $（'select.product_dimension'）。change（function（）{
 var a =“a”; \  n var b =“b”; 
 $ .post（“＆lt;？php echo base_url（）;？＆gt; home / list_products_of_company”，
 {
a：a，b：b 
} 
）\  n}）; 
  code>  pre> 
 
  控制器 strong>： p> 
 
 
  function list_products_of_company（）{\  n $ a = $ this-＆gt; input-＆gt; post（'a'）; 
 $ b = $ this-＆gt; input-＆gt; post（'b'）; 
 $ this-＆gt; data [  'products_of_company'] = $ this-＆gt; Home_model-＆gt; get_products_of_company（$ a，$ b）; 
 $ this-＆gt; load-＆gt; view（'product_list'，$ this-＆gt; data）; 
  } 
  code>  pre> 
 
 如何使用从数据库中获取的数据加载视图？   p> 
 
 
 编辑： strong>  p> 
 
 
我不想在div中加载ajax响应。 我需要在窗口中加载整个视图页面。 p> 
  div>

答

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);
}

答

$('select.product_dimension').change(function () {
  $.ajax({
    type: 'POST',
    url: "<?php echo base_url(); ?>home/list_products_of_company",
    dataType: 'text',
    success:
      function(data){
        //change the id to your div id. where you want to load the view
        $('#yourdivid').html(data);
      }
  });
});

I think this will help

答

You need to update your html as per response. say for example you want to response set with in div with id container. you can do as follows

<div id="container"></div>

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

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？

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