这是我的代码
$RestaurantId=14;
$data['status']=0;
$this->db->set('a.status', $data['status']);
$this->db->set('b.status', $data['status']);
$this->db->where('a.id', $RestaurantId);
$this->db->join('b','a.id = b.business_location_id');
$result =$this->db->update('b_locations_info as a,b_images as b');
错误:
表kpbaiger.b_locations_info as a,b_images
不存在
在SQL上:
更新
b_locations_info as a,b_images
为b
设置a
。status
='0',b
。status
='0' 在a
。id
='14'