php - 登录后连接到不同的数据库？

Question

如何根据用户的登录凭据连接到不同的数据库？

db_helper.php

<?php

function setDb($company_name, $branch_name) {

    return array(
        'dsn'   => '',
        'hostname' => 'localhost',
        'username' => 'root',
        'password' => '',
        'database' => $company_name.'_'.$branch_name,
        'dbdriver' => 'mysqli',
        'dbprefix' => '',
        'pconnect' => FALSE,
        'db_debug' => (ENVIRONMENT !== 'production'),
        'cache_on' => FALSE,
        'cachedir' => '',
        'char_set' => 'utf8',
        'dbcollat' => 'utf8_general_ci',
        'swap_pre' => '',
        'encrypt' => FALSE,
        'compress' => FALSE,
        'stricton' => FALSE,
        'failover' => array(),
        'save_queries' => TRUE
    );

}
?>

Select_db.php (/libraries) --> 包含在 autoload.php 中

<?php

class Select_db {

    public function thisdb($db_settings) {

        return $this->load->database($db_settings, TRUE);
    }
}

?>

model.php(我这里好像报错了，但是打印不出来)

      if ($result_num == 1) {

            $first_row = $query->row();
            $stored_password = $first_row->password;


           if (crypt($password, $stored_password) == $stored_password) {
                //Successful login
                $sql = "SELECT
                            company.name As company_name,
                            branch.name As branch_name,
                            branch.id As branch_id
                        FROM 
                            account
                        INNER JOIN
                            branch
                        ON
                            account.branch_id = branch.id
                        INNER JOIN
                            company
                        ON 
                            branch.company_id = company.id
                        WHERE
                            account.username = ?
                ";
                $query = $this->db->query($sql, $username);

                $data = array();
                $data['company_name'] = $query->row()->company_name;
                $data['branch_id'] = $query->row()->branch_id;
                $branch_name = $query->row()->branch_name;

                $this->load->helper('db_helper');
                $db_settings = setDb($data['company_name'], $branch_name);
                $dbname = $data['company_name'].'_'.$branch_name;
                //$dsn = 'dbdriver://root:sprinthr123@localhost/'.$dbname;
                //$this->branch_db = $this->load->database($dsn);



                //I'm guessing the error is somewhere around here 
                //because when I comment it out I don't get the 500 error 
                //code anymore and I'm able to log in to the homepage
                $db = $this->Select_db->thisdb($db_settings); 
            } else {
                return FALSE;
            }
        } else {

            return 0;
        }

当通过 AJAX 请求登录时调用此模型。我在控制台中收到一个 500 错误代码，所以肯定有一些错误，但我不知道错误在哪里？

它必须是动态的，所以我不能简单地在配置中填充 database.php 文件。

Answer 1

在config目录下的database.php中添加这一行

/********* For Other Database***********************/
    $db['Other']['hostname'] = '192.168.1.191';
    $db['Other']['username'] = 'user589';
    $db['Other']['password'] = 'pass12345';
    $db['Other']['database'] = 'other_db';
    $db['Other']['dbdriver'] = 'mysql';
    $db['Other']['dbprefix'] = '';
    $db['Other']['pconnect'] = TRUE;
    $db['Other']['db_debug'] = TRUE;
    $db['Other']['cache_on'] = FALSE;
    $db['Other']['cachedir'] = '';
    $db['Other']['char_set'] = 'utf8';
    $db['Other']['dbcollat'] = 'utf8_general_ci';
    $db['Other']['swap_pre'] = '';
    $db['Other']['autoinit'] = TRUE;
    $db['Other']['stricton'] = FALSE;

在模型中使用“其他”数据库配置

public function get_a_queue() {
    $otherdb = $this->load->database('Other', TRUE); //loaad second databse
    $qr = $otherdb ->query("SELECT * from users");
    return $qr->result();
}

希望对你有帮助