PHP MYSQL連接變量不存在？當它

-4

注意：未定義的變量：P中CONN：\ XAMPP \ htdocs中\ CraftedLogin \上線authenticate.php 13

警告：請求mysql_query（）預計參數2是資源，NULL給出 P中：\ XAMPP \ htdocs中\ CraftedLogin \ authenticate.php上線13

警告：mysql_num_rows（）預計參數1是資源，空給出在第16行的P：\ xampp \ htdocs \ CraftedLogin \ authenticate.php中錯誤的用戶名或密碼！無法登錄，GFG

這是我的代碼：

<?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>

而且，這裏是我的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"); 
?>

誰能幫助我？

注意：我已經嘗試將$ connection更改爲$ conn，但這次仍然會給出$ num變量的錯誤，但我認爲我可以很快對此進行排序。

NEW CODE：

<?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"); 
?>

來源

2013-12-21 MCRocks999

更改'$ connection = mysql_connect（$ host，$ user，$ pass）;'到'$ conn = mysql_connect（$ host，$ user，$ pass）;' –

變化

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

到

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

這就是爲什麼它不工作。您正在使用錯誤的變量名稱。

編輯：那很可能是你想要什麼：

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

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

$numr = mysql_num_rows($result);

來源

2013-12-21 11:59:28 aksu

嘗試仍然不起作用 – MCRocks999

是錯誤，你收到與你的問題相同？ – aksu

編號它給出了：警告：mysql_num_rows（）期望參數1是資源，null在第11行P：\ xampp \ htdocs \ CraftedLogin \ authenticate.php中給出在上面的問題中查看新代碼 – MCRocks999

你的連接是$connection當你使用$conn其他任何地方。

來源

2013-12-21 11:59:17

在這段代碼：

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

你傳入$康恩，但你的連接被定義爲$連接。將其更改爲：

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

將解決此問題。

來源

2013-12-21 11:59:54

修正：

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

來源

2013-12-21 12:00:06 floww

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

變化到

$康恩=的mysql_connect（$主機，$用戶，$通）;

來源

2013-12-21 12:06:59 shafi

PHP MYSQL連接變量不存在？當它

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