Question

每当我尝试登录我的网站时，它总是会发出：

注意：未定义的变量：第13行的P：\ xampp \ htdocs \ CraftedLogin \ authenticate.php中的conn

警告：mysql_query（）期望参数2为resource，给定null   在第13行的P：\ xampp \ htdocs \ CraftedLogin \ authenticate.php

警告：mysql_num_rows（）期望参数1为resource，null   在第16行的P：\ xampp \ htdocs \ CraftedLogin \ authenticate.php中给出   错误的用户名或密码！无法以gfg身份登录

这是我的代码：

<?php

$username = $_POST['username'];
$password = $_POST['password'];
include("dbConfig.php");

$sql = "SELECT * FROM `users` WHERE `username` = '$username' AND `password` = '$password'";

$rs = mysql_query($sql,$conn);

$num = mysql_num_rows( $rs );

if( $num != 0 )
{ 
  $msg = "<h3>Welcome $username - your log-in succeeded!</h3>";
}
else
{
  $msg = "Wrong username or password! Cannot login as $username";
}
?>

<html>

 <head>
  <title>Log-In</title>
  </head>

  <body>
   <?php echo( $msg ); ?>
  </body>

</html>

这也是我的dbConfig.php：

<?
$host = "localhost";
$user = "CraftedLogin";
$pass = "craftedlogin";
$db = "craftedlogin";
$self =     $_SERVER['PHP_SELF'];
$referer =  $_SERVER['HTTP_REFERER'];
$connection = mysql_connect($host, $user, $pass);
if ( !$conn )
{
echo "Error connecting to database.\n";
}
$rs = @mysql_select_db( $db, $conn ) 
        or die( "Could not select database" );
?>

任何人都可以帮助我吗？

注意：我已经尝试将$ connection更改为$ conn但是这次仍然使用$ num变量发出错误，但我想我可以快速解决这个问题。

新代码：

<?php

$username = $_POST['username'];
$password = $_POST['password'];
include("dbConfig.php");

$sql = "SELECT * FROM `users` WHERE `username` = '$username' AND `password` = '$password'";

$rs = @mysql_select_db( $db, $conn );

$numr = mysql_num_rows( $rs );

if( $numr != 0 )
{ 
  $msg = "<h3>Welcome $username - your log-in succeeded!</h3>";
}
else
{
  $msg = "Wrong username or password! Cannot login as $username";
}
?>

<html>

 <head>
  <title>Log-In</title>
  </head>

  <body>
   <?php echo( $msg ); ?>
  </body>

</html>

和dbConfig.php：

<?
$host = "localhost";
$user = "CraftedLogin";
$pass = "craftedlogin";
$db = "craftedlogin";
$self =     $_SERVER['PHP_SELF'];
$referer =  $_SERVER['HTTP_REFERER'];
$conn = mysql_connect($host, $user, $pass);
if ( !$conn )
{
echo "Error connecting to database.\n";
}
$rs = @mysql_select_db( $db, $conn ) 
        or die( "Could not select database" );
?>

Answer 1

当您在其他地方使用$connection时，您的联系为$conn。

Answer 2

更改

$connection = mysql_connect($host, $user, $pass);

到

$conn = mysql_connect($host, $user, $pass);

这就是为什么它不起作用。你使用了错误的变量名。

编辑：这可能就是你想要的：

$sql = "SELECT * FROM `users` WHERE `username` = '$username' AND `password` = '$password'";

$result = mysql_query( $sql, $conn );

$numr = mysql_num_rows( $result );

Answer 3

在这段代码中：

$rs = @mysql_select_db( $db, $conn ) or die( "Could not select database" );

您正在传递$ conn，但您的连接被定义为$ connection。将其更改为：

$rs = @mysql_select_db( $db, $connection) or die( "Could not select database" );

将解决此问题。

Answer 4

修复：

$conn = mysql_connect($host, $user, $pass);

Answer 5

$connection = mysql_connect($host, $user, $pass);

更改为

$ conn = mysql_connect（$ host，$ user，$ pass）;

PHP MYSQL连接变量不存在？什么时候呢

5 个答案: