CSS popup登錄表單和PHP登錄腳本

Question

我有一個CSS彈出窗口，顯示一個登錄表單。當我嘗試登錄時，只是重新加載頁面，彈出窗口再次出現。

下面是打開彈出

<a href="#login_form" class="btn signInBtn">Sign In</a> 

鏈接這裏是我的彈出登錄表單的HTML。

<a href="#x" class="overlay" id="login_form"></a>  
    <div class="popup"> 
     <h2 class="modal-header">Sign In</h2> 
     <div class="modal-body"> 
      <form class="signIn-form"> 
       <div class="errorMessage alert alert-error" id="errorMsg"> 
        <h4>Whoops</h4> 
        <p> 
         <?php echo $errorMsg; ?> 
        </p> 
       </div> 
       <fieldset class="l-formMain"> 
        <ul> 
         <li> 
          <label class="applyForm-label" for="email">Email</label> 
          <input class="applyForm-input required" type="email" id="login_email" placeholder="name@example.com" tabindex="1" /> 
         </li> 
         <li> 
          <label class="applyForm-label" for="password">Password</label> 
          <input class="applyForm-input required" type="password" id="login_password" tabindex="2" /> 
          <span class="forgotPassLi"> 
           <a href="#forgotpass" class="v-secondary" style="position:relative; left:-150px;">Forgot your Password?</a> 
          </span> 
         </li> 
        </ul> 
       </fieldset> 
       <div class="modal-buttonHolder"> 
        <input type="submit" class="btn btn-large" id="login" value="Sign In" tabindex="3" /> 
       </div>      
      </form> 
     </div> 
     <a href="#close" class="modal-closeBtn">×</a>  
    </div> 

這是我的PHP：

if(isset($_POST['login'])){ 
    $login_email = $_POST['login_email']; 
    $login_password = $_POST['login_password']; 

    // error handling conditional checks go here 
    if ((!$login_email) || (!$login_password)) { 
    $errorMsg = 'Please fill in both fields'; 
    } else { // Error handling is complete so process the info if no errors 
    include 'scripts/connect_to_mysql.php'; // Connect to the database 
    $email = mysql_real_escape_string($login_email); // After we connect, we secure the string before adding to query  
    $pass = md5($login_password); // Add MD5 Hash to the password variable they supplied after filtering it  
    // Make the SQL query 
    $sql = "SELECT * FROM members WHERE email='$email' AND password='$password' AND email_activated='1'"; 
      $result = mysql_query($sql); 
    if($result){ 
     $login_check = mysql_num_rows($result); 
    } 
    // If login check number is greater than 0 (meaning they do exist and are activated) 
    if($login_check > 0){ 
      while($row = mysql_fetch_array($result)){ 

       $id = $row["ID"]; 
       $_SESSION['ID'] = $id; 
       // Create the idx session var 
       $_SESSION['idx'] = base64_encode("xxxxxxxxxxxxxxxxxxx$id"); 
       // Create session var for their username 
       $email = $row["email"]; 
       $_SESSION['email'] = $email; 
       $_SESSION['userId'] = $row["ID"]; 

       mysql_query("UPDATE members SET last_log_date=now() WHERE ID='$id' LIMIT 1"); 

      } // close while 
      // All good they are logged in, send them to homepage then exit script 
      if (isset($_SESSION["email"]) || count($_SESSION["email"]) > 0) { 
       header("Location: http://localhost/dashboard.php"); 
      }    
    } else { // Run this code if login_check is equal to 0 meaning they do not exist 
     $errorMsg = "Either the email or password (or both) are incorrect. Make sure that you've typed them correctly and try again"; 
    } 
    }// Close else after error checks 
    } 
?> 

Answer 1

你的形式不張貼到任何地方。所以默認情況下它會轉到當前頁面。

嘗試增加的動作和方法屬性的形式：

<form method="post" action="http://url.of.PHP.page/containing/login/code">