Question

这是我的模特：

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;
}

这是我的观点：

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

想要搜索并仅显示我搜索过的结果。

Answer 1

这是我的旧版本，但您可以将其改编为您的版本。您的表单将帖子发送给控制器。变量$ art是搜索项。变量$ num是结果的数量。 jquery将表单发送到文章类search_results

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";
    }
}

如果您希望搜索显示在同一页面上，请使用Jquery

$(function () {
"use strict";
$("#search").submit(function () {
    var data = $('#search').serialize();
    //alert(data); return false;
    $.ajax({
        url: "/display/seek",
        data: data,
        type: "POST",
        success: function (msg) {
            $('article.search_results').html(msg).show();
        }
    });
    return false;
 });
});

查看很简单。请参阅jquery中的最后一行

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

我想在codeigniter中搜索并查看我搜索过的结果

1 个答案: