PHP登錄腳本'if（isset（$ _ POST ['submit']））''不工作

我的網站有一個PHP登錄腳本，但if(isset($_POST['submit']))似乎不工作。它始終以回聲"Unable to login. Please contact administrator.";返回。該代碼是

<?php 

error_reporting(E_ALL); 
ini_set('display_errors', 1); 

session_start(); 

require "dbc.php"; 
if(isset($_POST['submit'])) 
{ 
    $username = mysql_real_escape_string($_POST['username']); 
    $password = mysql_real_escape_string($_POST['password']); 
    $enc_password = md5($password); 

    if($username && $password) 
    { 
     $query = mysql_query("SELECT * FROM users WHERE username='$username'"); 
     $numrow = mysql_num_rows($query); 

     if($numrow!=0) 
     { 
      while($row = mysql_fetch_assoc($query)) 
      { 
       $db_username = $row['username']; 
       $db_password = $row['password']; 
      } 

      if($username==$db_username&&$enc_password==$db_password) 
      { 
       echo "Logged in <a href='membersarea.php'>Click here to enter the members area</a>"; 
       $_SESSION['username']=$db_username; 
       header("location: membersarea.php"); 
      } 
      else 
      { 
       echo "Incorrect Username/Password"; 
      } 
     } 
     else 
     { 
      echo "Incorrect Username/Password"; 
     } 
    } 
    else 
    { 
     echo "All fields are required"; 
    } 
} 
else{ 
    echo "Unable to login. Please contact administrator."; 
} 

?> 
<!--V0.3.5 Alpha-->

這是你在那裏輸入信息

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 

<html xmlns="http://www.w3.org/1999/xhtml"> 
    <head> 
     <meta http-equiv="Content-Type" content="text/html;charset=utf-8" /> 
     <meta name="description" content="Home to and its game downloads."/> 
     <meta name="keywords" content="Games, Downloads, Free, Fun, Game Development, Roll Ball, "/> 
     <meta name="author" /> 
     <link rel="stylesheet" type="text/css" href="stylesheet.css"/> 
     <title> 

     </title> 
    </head> 
    <body> 
     <div id="validxhtml"> 
      <p> 
       <a href="http://validator.w3.org/check?uri=referer"><img 
       src="http://www.w3.org/Icons/valid-xhtml10" alt="Valid XHTML 1.0 Transitional" height="31" width="88" /></a> 
      </p> 
     </div> 
     <div id="validcss"> 
     <p> 
      <a href="http://jigsaw.w3.org/css-validator/check/referer"> 
       <img style="border:0;width:88px;height:31px" 
        src="http://jigsaw.w3.org/css-validator/images/vcss-blue" 
        alt="Valid CSS!" /> 
      </a> 
     </p> 

    </div> 
     <div id="title"> 
      <h1 class="hcenter"></h1> 
     </div> 
     <div id="subtitle"><h2 class="hcenter">Login</h2></div> 
     <div id="links"> 
      <ul> 
       <li><a href="index.html" title="Home">Home</a></li> 
       <li><a href="game.html" title="Current Games">Games</a></li> 
       <li>Login</li> 
       <li><a href="contact.html" title="Contact Me">Contact Me</a></li> 
      </ul> 
     </div> 
     <div id="body"><h3>Please Login</h3> 
     <?php if(isset($_GET['error'])){echo $_GET['error'];echo "<br />";} ?> 
      <form action="login.php" method="POST"> 
        Username: <input type="text" name="username"><br> 
        Password: <input type="password" name="password"><p> 
       <input type="submit" value="Login"> 
      </form> 

      <a href="register.php">Don't have an account? Register here.</a> 
     </div> 
     <div id="policy" class="hcenter"> 
      <a target="_blank" href="http://privacypolicies.com/privacy/view/PUlI8q">Privacy policy</a> 
     </div> 
      <script type="text/javascript"> 
    (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){ 
    (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), 
    m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m) 
    })(window,document,'script','//www.google-analytics.com/analytics.js','ga'); 

    ga('create', 'UA-73371388-2', 'auto'); 
    ga('send', 'pageview'); 

</script> 
    </body> 
</html> 
<!--V0.3.5 Alpha-->

我也有最後一兩件事要說，第二個腳本它，你在你的登錄信息輸入一個，當頁面你點擊'登錄'它將它指向第一個。如果您有任何問題，請隨時詢問我。

來源

2016-03-04 SkrillexNukehulk

您正在使用' mysql_ *'函數在你的代碼中。請看看這個問題，爲什麼你應該遠離他們：http://stackoverflow.com/q/12859942/4464702 – RAnders00

你不必在你的HTML代碼name="submit"任何輸入，所以這就是爲什麼如果條款是skiped

變化

<input type="submit" value="Login">

要：

<input type="submit" value="Login" name="submit">

來源

2016-03-04 22:41:34 Rhopercy

非常感謝你！它的工作原理，我不能相信這個錯誤是多麼愚蠢。 – SkrillexNukehulk

沒問題，不要忘記選擇最佳答案：D（答覆方這個答案） – Rhopercy

我必須再等9分鐘，因爲網站不會讓我。 – SkrillexNukehulk

PHP登錄腳本'if（isset（$ _ POST ['submit']））''不工作

回答

相關問題