Php 从mysql数据库授权用户_Php_Mysql

Php 从mysql数据库授权用户

php mysql

Php 从mysql数据库授权用户,php,mysql,Php,Mysql,我的php很烂，在这里找不到错误。该脚本从html中获取两个变量“username”和“password”，然后对照MySQL数据库检查它们。当我运行此命令时，会出现以下错误“查询为空” 秘密区域您正在将$sql传递给mysql\u query，但您实际将查询存储在$slq <? if ((!$_POST[username]) || (!$_POST[password])) { header("Location: show_login.html"); exit; } $

我的php很烂，在这里找不到错误。该脚本从html中获取两个变量“username”和“password”，然后对照MySQL数据库检查它们。当我运行此命令时，会出现以下错误“查询为空”


秘密区域

您正在将

$sql

传递给

mysql\u query

，但您实际将查询存储在

$slq

<?
if ((!$_POST[username]) || (!$_POST[password])) {
    header("Location: show_login.html");
    exit;
}
$db_name = "testDB";
$table_name = "auth_users";
$connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());
$slq = "SELECT * FROM $table_name WHERE username ='$_POST[username]' AND password = password('$_POST[password]')";
$result = @mysql_query($sql, $connection) or die(mysql_error());
$num = mysql_num_rows($result);
    if ($num != 0) {
        $msg = "<p>Congratulations, you're authorised!</p>";
    } else {
        header("Location: show_login.html");
        exit;
    }
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Secret Area</title>
</head>

<body>
<? echo "$msg"; ?>   

</body>
</html>