登錄後連接到其他數據庫？

如何根據用戶的登錄憑據連接到不同的數據庫？登錄後連接到其他數據庫？

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（/庫） - >包括在autoload.php

<?php 

class Select_db { 

    public function thisdb($db_settings) { 

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

?>

model.php（I似乎是在這裏得到一個錯誤，但我不能打印出來）

 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; 
       //$dsn = 'dbdriver://root:[email protected]localhost/'.$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; 
     }

通過AJAX請求登錄時調用此模型。會發生什麼是我在控制檯中得到一個500錯誤代碼，所以一定有一些錯誤，但我不知道在哪裏？

它必須是動態的，所以我不能只是簡單地在config中填充database.php文件。

來源

2017-02-16 herondale

請參閱此HTTPS：/ /codeigniter.com/user_guide/database/configuration.html – Naincy

它必須是動態的嗎？因爲有多個數據庫可以訪問。哪一個將被訪問取決於成功登錄的用戶名（已經涵蓋）。主要問題是建立連接。 @Naincy – herondale

你寫連接腳本並建立連接。成功登錄後關閉所有其他數據庫連接（PHP函數）並啓用只有你想要的。希望這會有所幫助 –

在database.php中添加這一行config目錄

/********* 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;

使用 '其他' 數據庫配置模型

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

希望這將幫助你

來源

2017-02-16 08:45:40

但是我不能預知哪個數據庫需要被訪問，因爲它取決於用戶的用戶名？ – herondale

登錄後連接到其他數據庫？

回答

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