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我對開發很陌生,不理會我的錯誤並幫助我。 如果證書匹配與否,我將重定向相同的頁面,並且錯誤代碼也不會顯示在登錄頁面上。 (實際上我希望使用html和mysql數據庫克里特登錄頁面,如果憑據匹配,它應該重定向到登錄頁面,如果憑據不匹配,它應該重定向到同一頁面並顯示無效的用戶名和密碼MSG)當憑據不匹配時如何重定向相同的頁面?
的login.php
<?php
include("process.php");
?>
<!DOCTYPE html>
<html lang="en-US">
<head>
<div class="line s-8 content text">
<div id="content" class="left-align contact-page" style="padding:0 !important">
<div class="margin">
<div class="s-12 l-6">
<div class="right"><br>
<p> User ID</p>
<p> Password</p>
</div>
</div>
<div class="s-11 l-6"><br>
<br>
<br>
<br>
<h2>User Login</h2>
<div class="customform" >
<div id="mail-status"></div>
<div class="block">
</div>
<div id="frm">
<form id="form" method="POST" action="process.php">
<div class="s-12 l-7">
<input type="email" id="email" name="email"required/></div>
<div class="s-12 l-7">
<input type="password" id="password" name="password" class="masked" required/></div>
<div class="s-3 l-7">
<button type="Login" id="btn" ; onClick="check(this.form)">Login</button></div>
</forms>
<span><?php echo $error; ?></span>
</body>
</html>
process.php
<?php
$error=""; // variabale store error message
if(isset($_POST['login'])){
if(empty($_POST['email']) || empty($_POST['password'])){
$error ="Username or password is invalid";
}
else
{
//define user and password
$email=$_POST['email'];
$password=$_POST['password'];
$conn=mysqli_connect("localhost", "root","");
//selecting database
$db=mysqli_select_db($conn, "ibricks");
$query =mysqli_query($conn, "SELECT * FROM user_master WHERE password='$password' AND email='$email'");
$rows=mysqli_num_rows($query);
if($rows==1){
header("Location: lpage.html");
}
else{
$error= "User ID or Password is Invalid";
}
mysqli_close($conn);
}
}
?>
我把它實在不行,現在,是否正確? –
*「忽略我的錯誤並幫助我」* - 就像你忽略了評論/回答。你的代碼有一個明顯的錯誤,但沒有告訴我們這個錯誤是什麼。 –