2016-10-03 137 views
-2

这个问题真的让我很难过。我有4个文件,一个是DB的配置,其他为登录页,第三进行登录,处理和LSAT一个成功登录后重定向) 这是我的DB-connect.php:会话开始不“开始”

<?php 

     $servername = "localhost"; 
     $username = "dbroot"; 
     $password = "dbpassword"; 
     $dbname = "dbname"; 

     // Create connection 
     $conn = new mysqli($servername, $username, $password, $dbname); 
     // Check connection 
     if (!$conn) { 
      die("Connection failed: " . mysqli_connect_error()); 
     } 

     ?> 

形式在我的用户的login.php:

<form action="includes/login-process.php" id="login" class="formoid-metro-black" style="background-color:transparent;font-size:14px;font-family:'Open Sans','Helvetica Neue','Helvetica',Arial,Verdana,sans-serif;color:#FFFFFF;max-width:400px;min-width:150px; float:right;" method="post" enctype="multipart/form-data"><div class="title"><h2>Log In</h2></div> 
      <div class="element-input"><label class="title">Username<span class="required">*</span></label><input class="large" type="text" name="username" required="required"/></div> 
      <div class="element-password"><label class="title">Password<span class="required">*</span></label><input class="large" type="password" name="password" value="" required="required"/></div> 
      <div class="submit"><input type="submit" name="submit" value="Log In"/></div></form><script type="text/javascript" src="forms/sign-up-form_files/formoid1/formoid-metro-black.js"></script> 

登录-process.php:

<?php include ("db-connect.php"); 

if(isset($_POST['submit'])){ 

    $username = mysqli_real_escape_string($conn, $_POST['username']); 
    $password = mysqli_real_escape_string($conn, $_POST['password']); 

    $encrypt = md5($password); 

    $userquery = "SELECT * FROM user WHERE `user-username`='$username' AND `user-password`='$password'"; 

    $run = mysqli_query($conn,$userquery); 

    if(mysqli_num_rows($run)>0){ 

    $_SESSION['username']=$username; 
    ?> 
    <script>alert('Login successful.')</script> 
    <script>window.open('../user-profile.php','_self')</script> 
    <?php 
    } 
    else { 
    ?> 
    <script>alert('Username or Password Incorrect.','_self')</script> 
    <script>window.open('../user-login.php','_self')</script> 
    <?php 

    } 
} 


?> 

而在我的登录页面后,我成功登录,通过脚本警报然而证明是成功的,如同user-profile.php打开(我把session_start()放在最上面),它直接运行if语句中的条件,忽略了我已经成功登录的事实。 这是我的user-profile.php:

<!DOCTYPE html> 
     <?php 
     session_start(); 
     if(!isset($_SESSION['username'])){ 
     ?> 
     <script> alert('Please Log-in first.','_self')</script> 
     <?php include_once('user-login.php'); 
     } 
     else { 
     $first_name= $_SESSION['fname']; 
     $first_name= ucfirst($first_name); 
     ?> 
    //rest of my code 

上面的代码应该运行else语句中的内容,但是我不断获得首先登录的警报。 (只需忽略$ firstname代码,我想在下面输出它)。你们有没有关于此事的任何想法?如果是这样,pleeeease帮助我。谢谢!

+0

您是否设置了'$ _SESSION ['username']'的值? –

+0

是参考我的login-process.php代码 – IlanaWexler

+0

您确定要向公众公开您的真实数据库用户名和密码吗? – Shadow

回答

1

在您的login-process.php中,您必须先将值设置为session_start(),然后再将值设置为会话变量。

+0

谢谢!有效! – IlanaWexler

1

您正在直接访问会话而无需声明会话。用户session_start()访问会话变量之前,您可以使用session_close()访问会话变量。