2016-02-26 59 views
0

您好我试图将用户名,他们键入到logattempt表 但它不承认我已经宣布

<?php 
$servername = "localhost"; 
$user = "root"; 
$dbpassword=""; 
$dbname = "hrms"; 
$username=$_POST['username']; 
$password=$_POST['password']; 
$conn = new mysqli($servername, $user, $dbpassword, $dbname); 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 

$sql = "SELECT * FROM users WHERE username='$username' AND password='$password'"; 
$result = $conn->query($sql); 

if ($result->num_rows > 0) { 

header("location: main.php"); 
} else { 
$sql = "INSERT INTO logattempt (id, username, timest, failedatt) 
VALUES ('John', '$username', '[email protected]','asdfasdf')"; 
$conn->query($sql); 
    header("location: index.php"); 
} 
$conn->close(); 
?> 
+0

打印您的$ sql,然后尝试在phpmyadmin控制台中查询,看看会发生什么 –

+0

您插入“约翰”对id字段,也许这就是产生错误。你可以打印insert sql并尝试在phpmyadmin中运行。它会显示你的SQL查询中的错误究竟在哪里。 –

+0

@Michael Dela Cruz:id字段是否自动增加? –

回答

0

$用户名假设u盘Id字段作为自动递增

变化从

$sql = "INSERT INTO logattempt (id, username, timest, failedatt) 
VALUES ('John', '$username', '[email protected]','asdfasdf')"; 

$sql = "INSERT INTO logattempt (id, username, timest, failedatt) 
VALUES (null, '$username', '[email protected]','asdfasdf')";