duanfei8897 2012-11-13 07:32
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Codeigniter:如何在提交ajax表单后向用户返回成功/失败消息

I'm starting to get the hang of Codeigniter, but apparently am missing something here.

I'm not sure how to return data to the user after an ajax form submission. Here are the relevant bits of my code, with the question inside:

View:

if (confirm("Are you sure?"))
    {
        var form_data = $('form').serialize();

        $.post("<?php echo site_url();?>" + "/products/update_multiple", (form_data),
// this is what I don't understand: how to populate the variable result with data returned from the model/controller?
           function(result) {
                    $('#myDiv').html(result);
            }
            )
    }

Controller:

function update_multiple()
{

 $this->load->model('products_model');
 $this->products_model->updateMultiple();
// What goes here, to grab data from the model and send it back to the ajax caller?
}

Model:

function updateMultiple()
{
// I have a drop down list to select the field to update. The values of the drop down list look like this: "TableName-FieldName". Hence the explode.

    $table_field = explode("-",$this->input->post('field'));
    $table = $table_field[0];
    $field = $table_field[1];

    $data = $this->input->post('data');

    $rows = explode("
", $data);

    foreach($rows as $key => $row)
    {
        $values = explode(" ", $row);
        $arr[$key]["code"] = $values[0];
        $arr[$key][$field] = $values[1];
    }


   if ($this->db->update_batch($table, $arr, 'code'))
   {
       // Here is the problem. What should I return here, so that it would be sent back to the calling ajax, to populate the result variable?
   }

}
  • 写回答

1条回答 默认 最新

  • donoworuq450547191 2012-11-13 07:41
    关注

    Return json

       if ($this->db->update_batch($table, $arr, 'code'))
      {
       print json_encode(array("status"=>"success","message"=>"Your message here"));
      }
    

    jQuery inside your confirm function

    $.ajax({
    type: "POST",
    url: "<?php echo site_url();?>" + "/products/update_multiple",
    data:  $('form').serialize(),
        dataType: "json",
        success: function(content) {
        if (content.status == "success") {
           $('#myDiv').html(content.message);
        } 
            }
        });
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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