2015-07-21 60 views
3

我正在致力于codeigniter应用程序,以允许用户添加编辑和修改数据库。动态地在codeigniter中添加数据库

我想通过循环重复此代码:

$db['db1']['hostname'] = 'localhost'; 
$db['db1']['username'] = 'ts4l_wp13'; 
$db['db1']['password'] = 'O&3D6c(0zD70.^9'; 
$db['db1']['database'] = 'ts4l_wp13'; 
$db['db1']['dbdriver'] = 'mysql'; 
$db['db1']['dbprefix'] = ''; 
$db['db1']['pconnect'] = FALSE; 
$db['db1']['db_debug'] = TRUE; 
$db['db1']['cache_on'] = FALSE; 
$db['db1']['cachedir'] = ''; 
$db['db1']['char_set'] = 'utf8'; 
$db['db1']['dbcollat'] = 'utf8_general_ci'; 
$db['db1']['swap_pre'] = ''; 
$db['db1']['autoinit'] = TRUE; 
$db['db1']['stricton'] = FALSE; 

我这样做:

@$get_data = mysql_query("SELECT * FROM `database_name`"); 
    while($getdata = mysql_fetch_assoc($get_data)){   
    $id='db'.$getdata['id']; 
    $iid="$id"; 
        $db["$iid"]['hostname'] = 'localhost'; 
        $db["$iid"]['username'] = $getdata['username']; 
        $db["$iid"]['password'] = $getdata['password']; 
        $db["$iid"]['database'] = $getdata['name']; 
        $db["$iid"]['dbdriver'] = 'mysql'; 
        $db["$iid"]['dbprefix'] = ''; 
        $db["$iid"]['pconnect'] = FALSE; 
        $db["$iid"]['db_debug'] = TRUE; 
        $db["$iid"]['cache_on'] = FALSE; 
        $db["$iid"]['cachedir'] = ''; 
        $db["$iid"]['char_set'] = 'utf8'; 
        $db["$iid"]['dbcollat'] = 'utf8_general_ci'; 
        $db["$iid"]['swap_pre'] = ''; 
        $db["$iid"]['autoinit'] = TRUE; 
        $db["$iid"]['stricton'] = FALSE; 
    } 

和该块。

// Loading db and running query. 
     $CI = &get_instance();  
     $this->db1 = $CI->load->database('db1', TRUE); 
     $this->db1->set($data); 
     $this->db1->insert($this->_table_name); 
     $id=$this->db1->insert_id(); 

我这样做:“你已经指定了一个无效的数据库连接组”

$this->load->model('mdb_m'); 
    $dbms =$this->mdb_m->get(); 

     if (count($dbms)) { 
      foreach ($dbms as $dbm) { 
       $mid=$dbm->id; 
       $dbc='db'.$mid; 
       $CI = &get_instance(); 
       $this->db.$mid = $CI->load->database("$dbc" , TRUE); 
       $this->db.$mid->set($data); 
       $this->db.$mid->insert($this->_table_name); 
      } 
     } 

但只在阵列中的第一个数据库中插入,然后停止和retrive这个错误

回答

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