Question

我有一个商店桌子和一个项目桌子的家伙。所以我想从商店表中删除商店。如果商店中没有商品，则应将其删除；如果商店中没有商品，则不能将其删除。此处编写的代码不会删除商店，无论是否有物品。这是我的控制器：

public function deleteShop()
    {
        if ( ! $this->ion_auth->logged_in() OR ! $this->ion_auth->is_admin())
        {
            redirect('auth/login', 'refresh');
        }
        else
        {    
            $id = 0;        
            $shop_id = $_GET['shop_id']; 
            $results = $this->shop_model->getShopIdFromItem($shop_id);


            if($results) {
                    foreach ($results as $key) {
                        $id = $key->shop_id;
                    }
                    if($id == $shop_id) {
                        $this->session->set_flashdata('message', 'There is an item with this Shop. You cannot delete this!');
                        redirect(base_url() . 'admin/shop','refresh');
                    }
                } else {
                    $result = true;
                    if($result) {
                        redirect(base_url() . 'admin/shop', 'refresh');
                    } else {
                        echo "Something went wrong!";
                    }
                }
        }

这是我的模特

function getShopIdFromItem($shop_id) {
        $this->db->select('category_id');
        $this->db->from('item');
        $this->db->where('shop_id', $shop_id);
        $query = $this->db->get();
        if($query->num_rows() > 0) {
            return $query->result();
        } else {
            return false;
        }
    }

    function delete($shop_id)
    {
        $this->db->where('shop_id', $shop_id);
        $this->db->delete('shop'); 
        return $this->db->affected_rows();
    }

Answer 1

function delete($shop_id)
{
$this->db->from('item');
$this->db->where('shop_id', $shop_id);
$query = $this->db->get();
if($query->num_rows() == 0) {
    $this->db->where('shop_id', $shop_id);
    $this->db->delete('shop'); 
    return $this->db->affected_rows();
} else {
    return false;
}

更改您的删除方法...仅在没有可用项目的情况下才会删除..

如果在codeigniter中没有与其相关的子记录，则删除记录

1 个答案: