Question

我一直收到这个错误：

Warning: mysql_query() [function.mysql-query]: Access denied for user 'mcabinet'@'localhost' (using password: NO) in /home/mcabinet/public_html/games/db_edit/airportmadness4.php on line 3

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/mcabinet/public_html/games/db_edit/airportmadness4.php on line 3
No Database found.

这是我的整个.php页面：

<?php
include("../templates/base/template2/mysql_connect.php");




$gametitle = "Airport Madness TEST";
$gamedescription = "DESCRIPtion testing a description. LOL !";
$image1url = "http://website-gamesite.com/games/images/lhfdsjk.jpg";
$categorycode = "adv";
$gametitle2 = "airportmadnesstest";


mysql_query("INSERT INTO Games VALUES (null,'$gametitle','$gamedescription','$image1url','$categorycode','0','$gametitle2'") or die ("Error Inserting Values into the Games Table");

?>

谢谢！我刚开始使用PHP和MySQL。

编辑：以下是包含文件的内容：

<?php

$db_host = "localhost";
$db_username = "mcabinet_admin";
$db_password = "4jf8ido9A";
$db_name = "mcabinet_games";

@mysql_connect($db_host,$db_username,$db_password) or die ('MySQL Not found // Could Not Connect.');
@mysql_select_db("$db_name") or die ("No Database found.");


?>

使用您的提示进行编辑后，我仍然收到错误。

编辑：我做了更多改动。这很奇怪，因为当我从数据库中获取内容时，内容显示正常，但仍然显示错误。当我运行单独的脚本时，尝试添加行，它甚至都不会连接。

Answer 1

尝试将include移到第一个查询行上方！喜欢这个

<?php
  include("../templates/base/template2/mysql_connect.php");
  $query = mysql_query("SELECT * FROM Games WHERE id = '1' ");

要发出MySQL请求，您需要先连接。

Answer 2

试试这个：

<?php
$db_host = "localhost";
$db_username = "mcabinet_admin";
$db_password = "4jf8ido9A";
$db_name = "mcabinet_games";

mysql_connect($db_host,$db_username,$db_password) or die ('MySQL Not found // Could Not Connect.');
mysql_select_db("$db_name") or die ("No Database found.");


$gametitle = "Airport Madness TEST";
$gamedescription = "DESCRIPtion testing a description. LOL !";
$image1url = "http://website-gamesite.com/games/images/lhfdsjk.jpg";
$categorycode = "adv";
$gametitle2 = "airportmadnesstest";


mysql_query("INSERT INTO Games VALUES (null,'$gametitle','$gamedescription','$image1url','$categorycode','0','$gametitle2'") or die ("Error Inserting Values into the Games Table");

?>

此外，您的用户似乎设置为不在phpmyadmin中使用密码，这是否正确？

Answer 3

问题在于：

mysqli_connect('$db_host','$db_username','$db_password')
    or die ('MySQL Not found // Could Not Connect.');

这应该是：

mysql_connect($db_host, $db_username, $db_password)
    or die ('MySQL Not found // Could Not Connect.');

由于您单引号引用了字符串，因此最终会传递'$db_host'（等）的字符串文字，而不是您想要的值。

此外，您似乎正在使用mysqli进行连接，但使用mysql运行查询。我猜它值得使用其中一个 - 它们是两个不同的库。

Answer 4

您应该在运行查询之前建立连接：

1) $link = mysqli_connect($host, $username, $password) 2) mysqli_select_db($link, $db_name); 3) $result = mysqli_query($link, $query);

Answer 5

分辨!!!修复：

简单改变：

mysql_query("INSERT INTO Games VALUES (null,'$gametitle','$gamedescription','$image1url','$categorycode','0','$gametitle2'") or die ("Error Inserting Values into the Games Table");

到：

mysql_query("INSERT INTO Games VALUES (null,'$gametitle','$gamedescription','$image1url','$categorycode','0','$gametitle2')") or die ("Error Inserting Values into the Games Table");

Answer 6

一个有趣的场景是

<h1>Some heading</h1>
<?php
    insert_data(a1,b1);
?>
<h2>other html stuff</h2>
<?php
    include('database_connnection.php'); <--- problem is this line

    function insert_data(a1,b1)
    {
         Query = "Insert INTO MYTABLE VALUES(a1,b1)";
    }
?>

上面的代码将产生错误，因为在代码进一步进行调用include（database ...）之前会先调用insert_data（）函数，因此会出现错误。解决方法是将include（database）移动到函数内部或在调用insert_data查询之前

<?php

    function insert_data(a1,b1)
    {
         include('database_connnection.php');
         Query = "Insert INTO MYTABLE VALUES(a1,b1)";
    }
?>

使用密码NO的用户拒绝MySQL错误访问

6 个答案: