1
我初中笨的,我想要做的是連接多個數據庫檢索數據庫中的數據,但它不是爲我工作,保持返回我的404錯誤頁面爲什麼我的Codeigniter無法連接多個數據庫?
這裏是我的代碼
配置/ database.php中
$active_group = 'qm';
$active_record = TRUE;
$db['qm']['hostname'] = '192.168.0.128';
$db['qm']['username'] = 'callcenter';
$db['qm']['password'] = 'ca11c3nt3r';
$db['qm']['database'] = 'qm';
$db['qm']['dbdriver'] = 'mysql';
$db['qm']['dbprefix'] = '';
$db['qm']['pconnect'] = TRUE;
$db['qm']['db_debug'] = TRUE;
$db['qm']['cache_on'] = FALSE;
$db['qm']['cachedir'] = '';
$db['qm']['char_set'] = 'utf8';
$db['qm']['dbcollat'] = 'utf8_general_ci';
$db['qm']['swap_pre'] = '';
$db['qm']['autoinit'] = TRUE;
$db['qm']['stricton'] = FALSE;
/* call contact detail table */
$active_group = "reportcallcenter";
$active_record = TRUE;
$db['reportcallcenter']['hostname'] = '192.168.0.128';
$db['reportcallcenter']['username'] = 'callcenter';
$db['reportcallcenter']['password'] = 'ca11c3nt3r';
$db['reportcallcenter']['database'] = 'reportcallcenter';
$db['reportcallcenter']['dbdriver'] = 'mysql';
$db['reportcallcenter']['dbprefix'] = "";
$db['reportcallcenter']['pconnect'] = TRUE;
$db['reportcallcenter']['db_debug'] = TRUE;
$db['reportcallcenter']['cache_on'] = FALSE;
$db['reportcallcenter']['cachedir'] = "";
$db['reportcallcenter']['char_set'] = "utf8";
$db['reportcallcenter']['dbcollat'] = "utf8_general_ci";
$db['reportcallcenter']['swap_pre'] = '';
$db['reportcallcenter']['autoinit'] = TRUE;
$db['reportcallcenter']['stricton'] = FALSE;
控制器
<?php if (! defined('BASEPATH')) exit('No direct script access allowed');
class Qm extends CI_Controller {
public function __construct()
{
parent::__construct();
$this->permission->is_logged_in();
//load model
$this->load->helper('url');
$this->load->model('callcontactsdetails_model');
$this->load->database('qm', TRUE);
$this->load->database('reportcallcenter', TRUE);
}
function qm_form()
{
$data = array();
$data['page'] = 'qm_form';
if($query = $this->callcontactsdetails_model->get_all())
{
$data['recordings_record'] = $query;
}
$data['main'] = 'qm/qm_form';
$data['js_function'] = array('qm');
$this->load->view('template/template',$data);
}
}//end of class
?>
模型(我用My_model)
<?php
class Callcontactsdetails_model extends MY_Model {
protected $_table = 'callcontactsdetails';
protected $primary_key = 'id';
}
?>
我的屏幕返回結果 任何想法如何解決我的問題或任何錯誤,我做了?
感謝@ Gautam3164的答覆後,我加了你的負擔,但它還是回到了我同樣的錯誤顯示服務器錯誤 – Oscar 2013-05-09 07:23:54
你有沒有設置$ active_record = TRUE;爲這兩個連接.. ?? – Gautam3164 2013-05-09 07:25:33
雅,兩者也如真,我上傳了新的錯誤,比我以前的錯誤現在好,任何想法如何解決錯誤?我認爲它幾乎在那裏:) – Oscar 2013-05-09 07:29:27