Question

我有一个小问题：我创建了一个函数，其中包含一个连接多个表的查询，该函数返回查询结果。我的问题是：我如何查询该函数的结果？到目前为止，搜索到的解决方案都没有为我工作。这是代码：

public  function manytomany(){
$this->db->select(*)
        ->from('tbl_users')
        ->join('tbl_users_cars','tbl_users.user_id = tbl_users_cars.user_id')
        ->join('tbl_cars','tbl_users_cars.car_id = tbl_cars.id')
        ->join('tbl_car_model','tbl_cars.car_model_id = tbl_car_model.car_model_id')
        ->order_by("user_id", "asc")
        ->group_by("user_id");
return $query = $this->db->get()->result();

如何在同一函数中查询结果？但功能之外？谢谢！

Answer 1

我不确定你想要它，但试试这个：

public  function manytomany() {

$query = $this->db->select(*)
        ->from('tbl_users')
        ->join('tbl_users_cars','tbl_users.user_id = tbl_users_cars.user_id')
        ->join('tbl_cars','tbl_users_cars.car_id = tbl_cars.id')
        ->join('tbl_car_model','tbl_cars.car_model_id = tbl_car_model.car_model_id')
        ->order_by("user_id", "asc")
        ->group_by("user_id");

$query = $this->db->get()->result();

return $query;
}
public function otherfunction()
{
$result_query = $this->manytomany();
}

Answer 2

试试这个：

public  function manytomany(){
        $this->db->select(*)
                ->from('tbl_users')
                ->join('tbl_users_cars','tbl_users.user_id = tbl_users_cars.user_id')
                ->join('tbl_cars','tbl_users_cars.car_id = tbl_cars.id')
                ->join('tbl_car_model','tbl_cars.car_model_id = tbl_car_model.car_model_id')
                ->order_by("user_id", "asc")
                ->group_by("user_id");
        $sql = $this->db->get_compiled_select();
        $this->db->query('SELECT * FROM ($sql) as table WHERE ...');
    ...
    }

如何在Codeigniter中查询另一个查询的结果

2 个答案: