2016-07-05 117 views
-3

我正在使用PHP和Mysqli數據庫創建登錄頁面,我寫了查詢,但是mysqli_num_rows()在給定值時給了我一個錯誤。 PS:我在查詢這樣做在面向對象的格式mysqli_row_nums在查詢時不起作用

<?php 



$host = "localhost"; 
$user = "root"; 
$pass = ""; 
$db = "customerdb"; 


$connection = @new mysqli($host, $user, $pass, $db); 
if ($connection->connect_errno) { 
    die("Connection failed!"); 
    exit(); 
} 


if (isset($_POST['submit'])) { 
    $username = $connection->real_escape_string($_POST['username']); 
    $password = $connection->real_escape_string($_POST['password']); 
    $script = "SELECT * FROM customer_management WHERE 
    customer_management.Username='".$username."'AND WHERE customer_management_Password='".$password."'"; 
    $result = $connection->query($script, MYSQLI_USE_RESULT); 
    $check = $result->num_rows; 




    if ($check >= 1){ 
     echo "Welcome to this website"; 

     } 

    else{ 

     echo"Sorry but your input is incorrect!"; 

    } 


} 

?> 

<!DCOTYPE html> 
<html lang='en'> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title>Login Page</title> 
</head> 

<body> 
<h1>Login Page</h1> 
<form method="post" action="login.php"> 
<input type="text" placeholder = 'username' name="username" /><br /><br /> 
<input type="password" placeholder="password" name="password" /><br /><br /> 
<input type="submit" name="submit" value="Log In" /> 
</form> 
</body> 
</html> 
+0

旁註:這是不正確的<!DCOTYPE html>'如果這是你真正的代碼。 –

+1

在查詢中有多個WHERE子句。刪除第二個。 – Marcus

+0

檢查了這個錯誤........哦,別介意。只要RTM將雅? http://dev.mysql.com/doc/refman/5.7/en/where-optimizations.html並且不要使用純文本密碼;你肯定會完成的。 –

回答

0

變化customer_management_Passwordcustomer_management.Password

+0

這不是唯一的錯誤與他們的代碼 –

+0

堅持在這裏牛仔..'customer_management_Password 「你怎麼不是他們的專欄名稱? –

+0

這是一個假設,我覺得這是一個受過教育的假設。 – segFault