2016-07-07 42 views
0

我目前正在使用用户和密码表访问mysql数据库的登录页面。每当我尝试运行我的代码,我得到以下错误无论怎样我的评论:在登录页面出现奇怪的PHP错误

Parse error: syntax error, unexpected '=', expecting ',' or ';' in /home/scs/licenta/an2/gr927/rpie1814/public_html/testphpathome/login.php on line 15

这里是HTML主页:

<!DOCTYPE html> 
<html> 
<head> 
    <title> 
     Gaming site 
    </title> 

</head> 
<body> 

<h1> 
    Log-In 
</h1> 

<form action="login.php" method="post"> 
    Username: <br> 
    <input type="text" name="user"> <br> 
    Password: <br> 
    <input type="text" name="pass"><br> 

    <input type="submit" value="Submit"> 
</form> 

</body> 
</html> 

这里是PHP代码:

<?php 

$user = $_POST["user"]; 
$pass = $_POST["pass"]; 


global $conn = mysqli_connect("localhost", "rpie1814", "rpie1814", "rpie1814"); 

if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 

$sql = "SELECT * FROM Users"; 
global $result = mysqli_query($conn, $sql); 

if(mysqli_num_rows($result) > 1) 
{ 
    header("Location: /game.php"); 
    exit; 
} 
else 
{ 
    echo "Unknown user"; 
} 

mysqli_close($conn); 
exit; 

?> 
+0

从连接中删除'global'并得到结果 – Saty

+0

@replace'$ connect'用'$ con' – Iceman

+0

嘿为什么你使用了全局单词。任何特定的原因????? – user1234

回答

1

你必须从连接($conn)和结果($result)删除global: -

$conn = mysqli_connect("localhost", "rpie1814", "rpie1814", "rpie1814"); 

替换:

$result = mysqli_query($connect, $sql); 

要: -

$result = mysqli_query($conn, $sql); // actually your connection variable is $conn not $connect 
0

替换:

global $result = mysqli_query($connect, $sql); 

人:

$result = mysqli_query($conn, $sql); 
+0

我替换了它。谢谢,但同样的错误依然存在。现在它说它是在线13. –

+0

你用'$ conn'替换了'$ connect'吗? –