ThinkPHP5.1 where多条件查询 2019-05-09T09:34:50 if ($keyword) { if ($sotype == "id") { $where[$stype] = $keyword; } else { $where = [ ['title', 'like', "%" . $keyword . "%"], ]; } } $where['level'] = 1; 当前页面是本站的「Baidu MIP」版。发表评论请点击:完整版 »