<?
require('./lib/db_mysql.inc');
require('./lib/query_sql.inc');
class db_user extends Query{
var $Host = "localhost";
var $Database = "test";
var $User = "root";
var $Password = "";
}
class db_xq extends Query{
var $Host = "localhost";
var $Database = "test1";
var $User = "root";
var $Password = "";
}
$db_user = new db_user; //连接数据库test
$db_xq = new db_xq; //连接数据库test1
if(!$db_user->query('select * from test')){
echo "数据库操作错误";
}
if( !$db_user->next_record() )
{
echo "没有数据";
exit(1);
}
do
{
echo $db_user->Record["id"]."<br>";
}while($db_user->next_record());
if(!$db_xq->query('select * from test1')){
echo "数据库操作错误";
}
if( !$db_xq->next_record() )
{
echo "没有数据";
exit(1);
}
if(!$db_user->query('delete from test where aid =100')){ //这里发现$db_user连接的数据库变成了test1,报错
echo "数据库操作错误";
}
?>
我是认为它既然是派生出了两个类,那么$db_user,$db_xq应该是互不干扰才对啊,要不这个定义两个类有什么用呢?麻烦高手帮忙指点指点


最新回复
require('./lib/query_sql.inc');
class db_user extends Query{
var $Host = "localhost";
var $Database = "test";
var $User = "root";
var $Password = "";
}
class db_xq extends Query{
var $Host = "localhost";
var $Database = "test1";
var $User = "root";
var $Password = "";
}
在这里 两个class不能使用相同的user名,如果user名相同的情况下,mysql只会维护最后一个linkid