Question

我正在创建一个网页。网页有一个表单，表单数据通过.php页面发送，保存在MySQL数据库中。我正在使用XAMPP 这是我的剧本：

  $dbc = mysqli_connect('http://localhost:8080/127.0.0.1','root',' ','aliendatabase');
  $query = "INSERT INTO aliens_abduction (first_name,last_name,when_it_happened, how_long, " .
  "how_many, alien_description, what_they_did, fang_spotted, other, email) " .
  "VALUES ('Nadreen', 'Muhammad Ali', 'since birth', '21 years', 'all of them', 'sad cretures', " .
  "'ruined me', 'yes', 'I may have seen your dog. Contact me.', " .
  "'sally@gregs-list.net')";
  $result = mysqli_query($dbc,$query)
  or die('Error querying database.');
  mysqli_close($dbc);

我一直在收到这些错误：

Warning: mysqli_connect(): php_network_getaddresses: getaddrinfo failed:
The requested name is valid, but no data of the requested type was found

和

mysqli_connect(): (HY000/2002): php_network_getaddresses: getaddrinfo failed:
The requested name is valid, but no data of the requested type was found

Answer 1

而不是：

$dbc = mysqli_connect('http://localhost:8080/127.0.0.1','root',' ','aliendatabase');

使用此：

$dbc = mysqli_connect('localhost','root','','aliendatabase');

XAMPP / MySQL：在数据库连接期间出错

1 个答案: