我想在codeigniter中搜索并查看我搜索的结果

我想在codeigniter中搜索并查看我搜索的结果,codeigniter,search,Codeigniter,Search,这是我的模型: function get_search() { $match = $this->input->post('search'); $this->db->like('p_category',$match); $this->db->or_like('p_place',$match); $this->db->or_like('p_price',$match); $query = $this->db->get('

这是我的模型:

function get_search() {
  $match = $this->input->post('search');
  $this->db->like('p_category',$match);
  $this->db->or_like('p_place',$match);
  $this->db->or_like('p_price',$match);
  $query = $this->db->get('tbl_property');
  if($query->num_rows() > 0)
         return $query->result();
    else
         return FALSE;
}   
以下是我的看法:

<?php 
        foreach($query as $pack_details): 
    ?>
    <table class="table table-hover table-bordered">
            <?php echo $pack_details->p_title?>
        </table>
    <?php 
    endforeach;?>


要搜索,将只显示我搜索的结果。

这是我的旧搜索结果,但您可以根据自己的搜索结果进行调整。您的表单将post发送给控制器。变量$art是搜索项。变量$num是结果数。jquery将表单发送到文章类搜索结果

控制器

public function seek()
{
    // this is the artist search display
  $art = html_escape(trim($this->input->post('art')));
  $this->db->like('artist', $art);
  $this->db->select('artist, song, album, mix_name');
  $query = $this->db->get('podcasts');
  $num = $query->num_rows();
  echo "<h3>We found $num $art song's</h3>";
  if($query->num_rows() > 0) {
    foreach ($query->result() as $row) {
         echo "<li class='search_list'>Song: $row->song <br> Album: $row->album <br> Podcast: $row->mix_name</li> ";
     }
  }else {
      echo "You searched for $art, Please check your spelling, or ensure the artists name is complete";
    }
}
观点很简单。请参阅jquery中的最后一行

<article class="search_results"></article>

<article class="search_results"></article>