Question

我在xampp的phpmyadmin中创建了我的数据库“MUSIC”。我正在尝试使用我存储在OML文件夹中的php文件连接到此数据库，而该文件又存在于htdocs文件夹中，我的代码是：

$db_name="MUSIC";
$db_user="root";
$db_pwd="ash123";
$db_host="localhost";

$connect = mysql_connect("localhost","root","ash123");
mysql_select_db("MUSIC");
echo "connection successful";

但是当我转到localhost并输入localhost/connect.php时，我收到此错误：

Object not found!
the requested URL was not found on this server.if u entered the URL manually please check your spelling and try again.If you think this is a server error,please contact webmaster.
Error 404
localhost
Apache/2.4.3 (Unix) OpenSSL/1.0.1c PHP/5.4.7

我已经搜索了但是我找到了解决这个问题的方法，请帮忙！

Answer 1

我正在尝试使用我拥有的php文件连接到此数据库将存储在OML文件夹中，而OML文件夹又存在于htdocs文件夹中

因此，URI将是/OML/connect.php，而不仅仅是/connect.php，对吧？

Answer 2

尝试在地址栏中输入：“localhost / OML / connect.php”

Answer 3

$connect = mysqli_connect("localhost","root","ash123","MUSIC");

在数据库连接文件中替换此代码。删除您显示的代码。

Answer 4

mysql_connect()为deprecated，您不正确地选择了数据库。

您的代码应为：

$db_name="MUSIC";
$db_user="root";
$db_pwd="ash123";
$db_host="localhost";

$link = mysqli_connect($db_host, $db_pwd, $db_name);
if (!$link->connect_error) {
  echo "connection successful";
} else {
  echo $link->connect_error
}

此时建立连接，并选择DB。

Answer 5

请尝试将其连接到您的数据库

SELECT bundle_id,asset_id ,
(CASE 
WHEN weekday_am IS NULL  or weekday_am = 0.00  where customer_id = 5 and price_id = 3 THEN weekday_am
WHEN weekday_am IS NULL  or weekday_am = 0.00 where price_id = 2 THEN weekday_am  
ELSE weekday_am IS NULL  or weekday_am = 0.00 where price_id = 1 THEN weekday_am  
END as 'weekday_am'),(CASE 
WHEN weekend_am IS NULL  or weekend_am = 0.00 where customer_id = 5 and price_id = 3 THEN weekend_am
WHEN weekend_am IS NULL  or weekend_am = 0.00 where price_id = 2 THEN weekend_am  
ELSE weekend_am IS NULL  or weekend_am = 0.00 where price_id = 1 THEN weekend_am  
END as 'weekend_am')
FROM 
bundle_prices where bundle_id = 1

使用php连接到mysql数据库时“找不到对象”

5 个答案: