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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");
}
$conn->close();
?>
打印您的$ sql,然后尝试在phpmyadmin控制台中查询,看看会发生什么 –
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。 –
@Michael Dela Cruz:id字段是否自动增加? –