database.php:
$db['default']['hostname'] = "192.168.2.104";
$db['default']['username'] = "webuser";
$db['default']['password'] = "----";
$db['default']['database'] = "vad";
$db['default']['dbdriver'] = "mysql";
$db['default']['dbprefix'] = "";
$db['default']['pconnect'] = TRUE;
$db['default']['db_debug'] = TRUE;
$db['default']['cache_on'] = FALSE;
$db['default']['cachedir'] = "";
$db['default']['char_set'] = "utf8";
$db['default']['dbcollat'] = "utf8_general_ci";
$db['stats']['hostname'] = "192.168.2.104";
$db['stats']['username'] = "webuser";
$db['stats']['password'] = "---";
$db['stats']['database'] = "vad_stats";
$db['stats']['dbdriver'] = "mysql";
$db['stats']['dbprefix'] = "";
$db['stats']['pconnect'] = TRUE;
$db['stats']['db_debug'] = TRUE;
$db['stats']['cache_on'] = FALSE;
$db['stats']['cachedir'] = "";
$db['stats']['char_set'] = "utf8";
$db['stats']['dbcollat'] = "utf8_general_ci";
问题是我只能在配置中定义一个
$DB2 = $this->load->database('stats', TRUE);
通过这种方式,我连接到第二个数据库,但我失go 了与第一个数据库的连接.有没有人知道如何加载这两个数据库,而不必在所有模型构造函数中执行以下操作?
$database1 = $this->load->database('database1', TRUE);
$database2 = $this->load->database('database2', TRUE);
向您致敬,
佩德罗