2016-11-18 15 views
-1

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); 
    } 
    } 
    ?> 
+0

我把它實在不行,現在,是否正確? –

+0

*「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。 –

回答

0

您可以取代

$error= "User ID or Password is Invalid"; 

隨着

header("Location: /login.php?errorMessage=invalidLogin"); 

然後偵聽$_GET您的登錄頁面上

<?php 
if(isset($_GET['errorMessage'])){ 
    echo 'Credentials are invalid. Please try again'; 
} 
?> 
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