Question

所以这是我的问题，我正在我的学校项目中创建自己的upvote和downvote功能。我想要发生的是当我点击向上箭头图像时，它会将记录插入数据库。它会在我的数据库中为upvote表添加一条记录。

<script type="text/javascript">

$( document ).ready(function() {



   $("#try").click(function(){
        var username = $("#username").val();
        var thread_id = $("#thread_id").val();
        var vote = $("#vote").val();

        $.ajax(
            {
                type:"post",
                url: "<?php echo site_url();?>/Users/add_vote/",
                data:{ username:username, thread_id:thread_id, vote:vote},
                success:function(data)
                {
                   alert("POST SUBMITTED");

                }


            });



});
   });

这是我的表格

<div id = "active_upvote_div">
                <form>
                    <input type="hidden" name = "username" id="username" value= "try">
                    <input type="hidden" name="thread_id" id="thread_id" value= "1">
                    <input type="hidden" name="vote" id="vote" value= "1">
                <button type="submit" id="try"><img src="<?php echo base_url();?>/public/images/upvote.png"   height:"25px" width="25px"/> </button>
                </form>
            </div>

我的控制器只调用模型

这是我的模特

 public function ajax()
      {
         $add_user = array(
        'username'     => $this->input->post('username'),
        'thread_id' => $this->input->post('thread_id'),
        'vote' => $this->input->post('vote'),

             );

       $this->db->insert('thread_upvote', $add_user);
      }

希望立即做出回应:(

Answer 1

在codeigniter中$this->input->post()相当于$_POST，这会使发布的名称数组为key/value对。

在controller's add_vote()功能中执行此操作...

public function add_vote()
{
$data = $this->input->post();
$this->load->model('model_namel');//loads model
$this->model_name->ajax($data);//call model's function with data
}

然后在model's ajax()函数

public function ajax($data=array())
{
   $this->db->insert('thread_upvote', $data);// inserts into database
}

Answer 2

在你的Ajax中：

$( document ).ready(function() {

   $("#active_upvote_div").click(function(){
       var username = $("#username").val();
       var thread_id = $("#thread_id").val();
       var vote = $("#vote").val();

       $.ajax({
           type:"POST",
           dataType: 'json', //add this to call json
           url: "<?php echo site_url();?>/Users/add_vote/",
           data:{ username:username, thread_id:thread_id, vote:vote},
           success:function(data)
           {
               if(data.success == true){
                    alert("POST SUBMITTED");
               } else{
                    alert("Failed to Post");
               }
           }
       });
   });
});

在您的控制器中：

Class Users extends CI_Controller{
   public function add_vote()
   {
       $result = $this->name_of_model->ajax(); //call the ajax function in you model
       echo json_encode($result); //check the return if true or false and pass to json
   }
}

在你的模特中：

public function ajax()
{
   $add_user = array(
      'username'     => $this->input->post('username'),
      'thread_id' => $this->input->post('thread_id'),
      'vote' => $this->input->post('vote'),
   );

   if($this->db->insert('thread_upvote', $add_user))
   {
       return true;
   } else{
       return false;
   }
}

如何在单击按钮时插入数据库记录 - Codeigniter，jquery，ajax

2 个答案: