如何让php连接到mysql？

Question

我已经尝试了一切使我的网站连接到数据库,但我总是得到他的错误:无法连接到主数据库我有2个文件

define('DBHOST','localhost');
define('DBUSER','root');
define('DBPASS','root');
define('DBNAME','test');
define('dbslave','test');  
define('dbsiteid','1');
define('dbprefix','_blog');

和connect.php

error_reporting(0);
$connect = mysql_connect("$DBHOST", "$DBUSER", "$DBPASS");
if ( ! $connect) {
    die('Could not connect to Database server');
}
$siteid  = "$dbsiteid";
$prefix  = "$dbprefix";
$dbmast  = "$DBNAME";
$dbslave = "$dbslave"; 
$cmast   = mysql_select_db("$DBNAME");
if ( ! $cmast) {
    die('Could not connect to Master Database');
}
$cslave = mysql_select_db("$dbslave");
if ( ! $cslave) {
    die('Could not connect to Slave Database');
}

如何通过连接解决此错误或我做错了什么？

Answer 1

您不会将常量放在引号中,它们不会以常量开头$,并且按惯例都是大写的.

 define('DBHOST','localhost');
define('DBUSER','root');
define('DBPASS','root');
define('DBNAME','test');
define('DBSLAVE','test');  
define('DBSITEID','1');
define('DBPREFIX','_blog');

$connect = mysql_connect(DBHOST, DBUSER, DBPASS);
if (!$connect) {die('Could not connect to Database server');}
$cmast = mysql_select_db(DBNAME);
if (!$cmast) {die('Could not connect to Master Database');}
$cslave = mysql_select_db(DBSLAVE);
if (!$cslave) {die('Could not connect to Slave Database');}

另外,定义常量只是为了将它们分配给变量是愚蠢的,浪费资源.并且在开发时不要关闭错误报告,因为它会隐藏您的错误.你想做相反的事情而他们就是这样.