php - 当我登录到另一个表CODEIGNITER时如何插入外键ID

Question

我想将我的 user_ID 插入到基于用户登录的冒险表中。但问题是我从未从用户登录中获取值，该值始终为 NULL

这是错误页面:

Error Number: 1048

Column 'user_id' cannot be null

INSERT INTO adventure (name, category, place, state, user_id) VALUES ('asd', 'gunung', 'asd', 'asd', NULL)

Filename: C:/xampp/htdocs/login-ci/system/database/DB_driver.php

Line Number: 691

这是我的模型

public function getUserId()
    {
        return $this->session->users['user_id'];
    }

这是我的 Controller

public function addTrip(){
        $this->load->model('userModel');
        $user_id = $this->session->users['user_id'];

        $newTrip = ['name' => $this->input->post('name'),
                    'category' => $this->input->post('category'),
                    'place' => $this->input->post('place'),
                    'state' => $this->input->post('state'),
                    'user_id' => $this->input->$user_id
                    ];


        $this->db->insert('adventure',$newTrip);
        // $this->db->insert('adventure',$data);

        redirect('userController/profile');
    }

请有人告诉我，如果我在代码中遗漏了某些内容，非常感谢

Answer 1

您在创建插入数组时遇到问题。将您的代码更改为:

$user_id = $this->session->users['user_id'];
$newTrip = ['name' => $this->input->post('name'),
                    'category' => $this->input->post('category'),
                    'place' => $this->input->post('place'),
                    'state' => $this->input->post('state'),
                    'user_id' => $user_id // You already have $user_id from session
                    ];