我想有过滤器下拉菜单在我的CMS符号 我的模型看起来像笨查询在条件字符串
public function load($sort,$order,$key,$value)
{ // $key='listening'; // $value="1";
//configure pagination
$config=array(
'base_url'=>base_url().'/index.php/companies/index',
'total_rows'=>$this->db->get('company')->num_rows(),
'per_page'=>$this->settings_model->get_per_page(),
'num_links'=>20
);
$this->pagination->initialize($config);
$this->db->select('company.id,
company.name,
company.logo,
company.status_id,
company.listening',FALSE);
$this->db->select('company_category.name as category,
company_category.id as category_id',FALSE);
$this->db->select('complain_status.cs_status as status',false);
$this->db->from('company');
$this->db->join('company_category','company_category.id = company.category_id');
$this->db->join('complain_stastus', 'complain_status.cs_id = company.status_id');
if(isset($_POST['key']))
{
$value= str_replace(' ', ' ', $_POST['value']);
var_dump($value);
if($value!='0')
$this->db->having ($_POST['key'], mysql_real_escape_string($value));
}
if($sort!='' || $sort!=NULL)
$this->db->order_by ($sort, $order);
$this->db->limit($config['per_page'], $this->uri->segment(3));
$result=$this->db->get();
if(!isset($_POST['key']))
$this->filter->set_filters_list($result->result_array());
return $result->result();
}
产生下面的查询
SELECT company.id, company.name, company.logo, company.status_id, company.listening, company_category.name as category, company_category.id as category_id, complain_status.cs_status as status
FROM (`company`)
JOIN `company_category` ON `company_category`.`id` = `company`.`category_id`
JOIN `complain_status` ON `complain_status`.`cs_id` = `company`.`status_id`
HAVING `category` = 'Health & recreation'
LIMIT 20
,你可以在这里看到的是当类别等于一些字符串与特殊字符如Health & recreation
它的问题,它失败,即使我试图查询生成的查询它通常在MYSQL上正常工作,并得到我的结果
注:我替换m的空间$value= str_replace(' ', ' ', $_POST['value']);
因为这个数据来自于失败时,它有选择的空间,所以我不得不来解析,并在后端代码提前
由于选择HTML元素
你怎么知道这就是生成的查询?你使用'echo $ this-> db-> last_query()'来确保这是正在生成的查询吗? – Catfish
没有其实我只是拼错在查询中的任何东西,如错误的表名,我从ajax响应生成的错误得到它.. –