Question

我尝试将一个表单插入到不同的一个表名是匹马，其他表名是 filename_img 查询意味着我在一个表中有不同的字段而我想要的其他字段将图像插入到其他表但我的一个表单插入字段和图像这是我的模型，请确认我的查询我错了，因为显示查询显示查询错误谢谢进展

模型

function horses_insert($data){
                    $postcode = addslashes($data['postcode']);
                    $height = addslashes($data['height']);
                    $breed_id = addslashes($data['breed_id']);
                    $colour_id = addslashes($data['colour_id']);
                    $age = addslashes($data['age']);
                    $gender = addslashes($data['gender']);
                    $name = addslashes($data['name']);
                    $price = addslashes($data['price']);
                    $description = addslashes($data['description']);
                    // $image = addslashes($data['image']);
                    $status = addslashes($data['status']);
                    $price_currency = addslashes($data['price_currency']);
                    $height_unit = addslashes($data['height_unit']);
                    $is_urgent = addslashes($data['is_urgent']);
                    $is_featured = addslashes($data['is_featured']);
                    $is_spotlight = addslashes($data['is_spotlight']);
                    $is_sale = addslashes($data['is_sale']);
                    if($is_urgent=="on"){
                        $is_urgent="1";
                    }else{
                        $is_urgent="0";
                    }

                    if($is_featured=="on"){
                        $is_featured="1";
                    }else{
                        $is_featured="0";
                    }

                    if($is_spotlight=="on"){
                        $is_spotlight="1";
                    }else{
                        $is_spotlight="0";
                    }

                    if($is_sale=="on"){
                        $is_sale="1";
                    }else{
                        $is_sale="0";
                    }




                    $result= $this->db->query("INSERT INTO horses (postcode,height,breed_id,colour_id,age,gender,name,price, description,is_urgent,is_featured,is_spotlight,is_sale,status,price_currency,height_unit) VALUES ('$postcode','$height','$breed_id','$colour_id','$age','$gender','$name','$price','$description','$is_urgent','$is_featured','$is_spotlight','$is_sale','$status','$price_currency','$height_unit')");
                    $id=$this->db->insert_id();

                    //return $id;
                    if ($result) {
            $userfile = addslashes($data['userfile']);

                        $query= $this->db->query("INSERT INTO filename_img (userfile,$id) VALUES ('$userfile','')");
                        $id=$this->db->insert_id();
                        return $id;
                    }

            }

Answer 1

试试这个：

$userfile = addslashes($data['userfile']); 

$commaSeprartedFilesName = implode(",",$userfile);

$query= $this->db->query("INSERT INTO filename_img (userfile,id) VALUES ('$commaSeprartedFilesName',$id)");

$id=$this->db->insert_id();

return $id;

UserFile应该是html中的数组，如

在codeginter中以一种形式发出两个不同的插入查询

1 个答案: