thinkphp6模型繼承問題
作者:網(wǎng)站建設(shè) | 發(fā)布日期:2020-12-30
首先看這部分代碼,這部分代碼是能夠正常返回?cái)?shù)據(jù)的:
use think\Model;/*** @mixin \think\Model*/class Menu extends Model{public function getMenuList($condition = [],$order='',$json = true){$where = array();if (!empty($condition['title'])){$condition['title'] = trim($condition['title']);$where[] = ['title','like', '%'.(string)$condition['title'].'%'];}if (isset($condition['pid'])){$where[] = ['pid','=',intval($condition['pid']) ];}if (isset($condition['hide'])){$where[] = ['hide','=',$condition['hide']];}if (isset($condition['status'])){$where[] = ['status','=',$condition['status']];}$list = $this->where($where)->order($order)->page($condition['page'],$condition['limit'])->select()->toArray();return $list;}
但是,如果我集成了另外一個(gè)非Model的公用模型的話,就會(huì)返回?cái)?shù)據(jù)是空的了,如:
namespace app\base\model;use think\Model;/*** @mixin \think\Model*/class Menu extends Base{

