我想在codeigniter中搜索并查看我搜索的结果
这是我的模型: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('
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>