codeigniter - dependent dropdown with jquery and a

问题:

view : learning_view.php

Here is the first dropdown which I am populating from database.

    <select name = 'category' id = 'category'>
        <option value="">-- Select Category --</option>
        <?php foreach($category as $item){ ?>
        <option value="<?php echo $item->id_cat; ?>"><?php echo $item->name; ?></option>
        <?php } ?>
    </select>
    <br><br>

What I want is to populate another dropdown which is dependent on the first dropdown. For that I have used the jQuery ajax post.

second dropdown:

    <select name = 'type' id = 'type'>
        <option value="">-- Select Type --</option>
        <?php foreach($type as $item){ ?>
        <option value="<?php echo $item->id_type; ?>"><?php echo $item->name; ?></option>
        <?php } ?>
    </select>
    <br><br>

ajax post:

    jQuery(document).ready(function(){
      $("#category").change(function() {
        var category_id = {"category_id" : $('#category').val()};
        console.log(category_id);

        $.ajax({
          type: "POST",
          data: category_id,
          url: "<?= base_url() ?>learning/dependent_dropdown",

          success: function(data){

            $.each(data, function(i, data){
            console.log(data.name);
            console.log(data.id_type)
            });
           }
         });
       });
     });

controller : learning.php

   public function dependent_dropdown()
   {
       if(isset($_POST['category_id']))
       {
           $this->output
           ->set_content_type("application/json")
           ->set_output(json_encode($this->learning_model->getType($_POST['category_id'])));
       }
   }

The data is coming from the database after ajax post which I checked by

    console.log(data.name);
    console.log(data.id_type)

in the console.

But couldn't able to figure out how to use the data in the second dropdown of my view.

I mean how can i populate the second dropdown with the data i have received after ajax post.

回答1:

I found a solution to my problem by modifying the success function of the ajax post:

success: function(data) {
    $.each(data, function(i, data) {
        $('#type').append("<option value='" + data.id_type + "'>" + data.name + "</option>");
    });
}

Which append the value into the drop down.

<select name="type" id="type">
    <option value="">-- Select Type --</option>
</select>

I just gave the id of the select block into the success function of the ajax post and appended the value. It works but there is a problem which is when I change the selection of the first dropdown new value appears but the values which were showing for the previous selection doesn't go away.

回答2:

Here is some modification of your answer

success: function(data)
{
    $('#type').html('<option value="">-- Select Type --</option>');

        $.each(data, function(i, data){
        $('#type').append("<option value='"+data.id_type+"'>"+data.name+"</option>");
        });
}

It will show only new items.

回答3:

view code:

<select id="category" name="prod_cat" class="form-control col-md-7 col-xs-12 category">
                    <?php if($category)
                        foreach ($category as $cat)
                        {
                    ?>
                            <option value="<?php echo $cat->cat_id;?>"><?php echo $cat->cat_title; ?></option>
                    <?php
                        }
                    ?>
                </select>

ajax code:

`

$(document).ready(function() {
    $('.category').change(function(){
      $.ajax({
        type: "POST",
        url: "<?php echo base_url();?>Product_admin/getSub",
        data:{id:$(this).val()}, 
        beforeSend :function(){
      $(".subcat option:gt(0)").remove();  
      $('.subcat').find("option:eq(0)").html("Please wait..");
        },                         
        success: function (data) {
          /*get response as json */
           $('.subcat').find("option:eq(0)").html("Select Subcategory");
          var obj=jQuery.parseJSON(data);
          $(obj).each(function()
          {
           var option = $('<option />');
           option.attr('value', this.value).text(this.label);           
           $('.subcat').append(option);
         });  
          /*ends */
        }
      });
    });
  });

`

Controller:

public function getSub()
{
      $result=$this->db->where('cat_id',$_POST['id'])
                    ->get('tbl_subcategory')
                    ->result();
    $data=array();
    foreach($result as $r)
    {
        $data['value']=$r->subcat_id;
        $data['label']=$r->subcat_title;
        $json[]=$data;


    }
    echo json_encode($json);

I too had same issue.. This is what I did..Worked well https://github.com/eboominathan/Dependent-Dropdown-in-Codeigniter-3.0.3

回答4:

i have done the same thing today. when i call the url via ajax i simply do the load-view procedure and load all the data in another small view file holding only select and option statment and a query to display data , and under the main view i create a div with id #divid and simply on success of ajax i did

$('#divid').html(data);

and that view will be displayed on the main page.

hopefully i make u understand.

回答5:

On success function write down this code. It will append your data in second dropdown:

$.ajax({
    type: "POST",
    data: category_id,
    url: "<?= base_url() ?>learning/dependent_dropdown",
    success: function(data) {
        $.each(data, function(i, data) {
            var opt = $('<option />'); 
            opt.val(data.id_type);
            opt.text(data.name);
            $('#your_second_dropdown_id_name').append(opt);
        });
    }
});