2017-04-26 50 views
0

此表單是爲了從用戶&獲得登錄詳細信息而提供的登錄頁面中執行login.php文件的 詳細信息。未執行Html表單操作字段

<div id="log_btn"> 
<section id="form_before_launch"> 
<p style="font-family: Arial, Helvetica, sans-serif;">LogIn</p> 
<form action="login.php" id="form" method="POST" onsubmit="return false"> 
    <div> 
     <div> 
      **<!-- Fields of Login page -->** 
      <input type="text" id="username" name="uname" placeholder="Username" required /> 
     </div> 
     <div> 
      <input type="password" id="password" name="pass" placeholder="password" required /> 
     </div> 
     **<!-- Captcha is Displayed in id="ip" & users enters the captcha in id "op" on Submit login page is executed -->** 
     <div> 
      <input type="text" id="ip" name="ip" /> 
      <br> 
      <input type="text" id="op" name="op" /> 
     </div> 
     <br><br><br> 
     <div id="btn"> 
      <input type="submit" id="submit" value="Log In" /> 
     </div> 
</form> 
</div> 

的login.php的一部分,如果你需要看測試的形式發送真正的結果輸入到的login.php

<?php 

$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname  = "Train"; 
$conn  = mysqli_connect($servername, $username, $password, $dbname); 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 
session_start(); 
if ($_SERVER["REQUEST_METHOD"] == "POST") { 
    $ip_uid = $_POST['uname']; 
    $ip_pass = $_POST['pass']; 
    $ip  = $_POST['ip']; 
    $op  = $_POST['op']; 
} 
function input($data) 
{ 
    $data = trim($data); 
    $data = stripslashes($data); 
    $data = htmlspecialchars($data); 
    return $data; 
} 
$sql = "SELECT user_name,pass FROM `Reg_User` WHERE user_name='$uid' AND pass='$pass'; "; 
$result = mysqli_query($conn, $sql); 
if (mysqli_num_rows($result) > 0) { 
    while ($row = mysqli_fetch_assoc($result)) { 
     $user_name = $row["user_name"]; 
     $pass  = $row["pass"]; 
    } 
} 
if ($ip_uid == $user_name && $ip_pass == $pass) { 
    $_COOKIE['$user_name'] = $user_name; 
    echo " <script> alert('Successfully Logged IN');"; 
    header('Location: http://localhost/TrainProject/Landing_page.php'); 
} else { 
    echo " <script> alert('Wrong Credentials Login failed');"; 
    header('Location: http://localhost/TrainProject/new_home.html'); 
} 

?> 
+0

檢查我的答案更新將解決您的問題 –

+1

這是爲什麼標記爲PHP?沒有代碼;那它在哪裏?你如何訪問這個? –

+0

我已經添加了php代碼 –

回答

7

明確取消表單提交:

onsubmit="return false" 

如果您想要提交表單,請將其刪除。

+0

我刪除onsubmit =「返回false」仍然相同 –

+0

@TejasPatil:適用於我:https://jsfiddle.net/1qntm17e/你有代碼,可靠地演示了這個問題? – David

+0

@大衛可能是因爲他們缺少「php」。主只知道他們如何訪問該文件。 –

1

更改爲

<div id="log_btn"> 
    <section id="form_before_launch"> 
    <p style="font-family: Arial, Helvetica, sans-serif;">LogIn</p> 
    <form action="login.php" id="form" method="POST"> 
     <div> 
      <div> 
       **<!-- Fields of Login page -->** 
       <input type="text" id="username" name="uname" placeholder="Username" required /> 
      </div> 
      <div> 
       <input type="password" id="password" name="pass" placeholder="password" required /> 
      </div> 
      **<!-- Captcha is Displayed in id="ip" & users enters the captcha in id "op" on Submit login page is executed -->** 
      <div> 
       <input type="text" id="ip" name="ip" /> 
       <br> 
       <input type="text" id="op" name="op" /> 
      </div> 
      <br><br><br> 
      <div id="btn"> 
       <input type="submit" id="submit" value="Log In" /> 
      </div> 
    </form> 
    </div> 

您的問題在HTML有放慢參數onsubmit="return false"

+2

OP爲什麼要「改成」這個代碼?答案很好* **將永遠解釋所做的事情以及爲什麼這樣做,不僅是爲了OP,而且是爲了將來的訪問者SO。 –

+0

解釋,謝謝您的評論 –

+1

[Little Bobby](http: /bobby-tables.com/)說*** [你的腳本處於SQL注入攻擊風險。](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in -php)***瞭解[MySQLi]的[prepared](http://en.wikipedia.org/wiki/Prepared_statement)語句(http://php.net/manual/en/mysqli.quickstart.prepared- statement.php)。甚至[轉義字符串](http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string)是不安全的! –