我的代码在执行时中断 - 不期待 T_ELSE，期待 T_FUNCTION

Question

我已经在 lamp、wamp 和 xampp 上尝试过我的代码 - 它只适用于 lamp。然而，如果我将 php.ini 文件从 lamp 换成 xampp，它就可以工作 - 因此，我猜测我在 LAMP 上以一种草率的方式编码，而我的 php.ini 却令人恼火地允许。

目前，我的 LAMP php.ini 破坏了我的 xampp mysqli，在我看来，我的 lamp 代码在任何情况下都一定是脏的，所以我想知道你们是否能看到这里需要清理什么？

class datamanagement{


 protected $mysql_host = "localhost";
 protected $mysql_username = "root";
 protected $mysql_password = "";
 protected $mysql_database = "data";

 protected $security_table = "users";

 function __construct($security_level = 0)
 {
   $this->security($security_level);
 }

 protected function mysql_connect_func(){ 
          // ...standard mysql connect stuff
 }

 protected function security($security_level){


  session_start();

  if(isset($_GET['logout'])){

   session_unset();
  }


  if($security_level > 0)
  {


   if(!isset($_SESSION['initiated'])){

    if(!isset($_POST['username']) || empty($_POST['username']) || empty($_POST['password']))
    {

     if(isset($_GET['logout']))
     {
     $string = rtrim($_SERVER['PHP_SELF'], '?logout');
     }


     ?>

     <div class="main_container">
    <div class="form_container">

    <?php

    if(isset($_GET['logout']))
    {
     echo "<p>successfully logged out</p>";

    }else
    {

     echo "<p>Access to this section require logging in</p>";

    }
    ?>

       <form method="post" action="<?php echo $string ?>"><input
        type="hidden" name="login" value="true"></input>

       <div><label for="title">Username:</label> <input name="username"
        type="text" value="<?php echo $_POST['username']; ?>"></input><?php if(isset($_POST['username']) && $_POST['username'] == ''){echo "username required";}?><br>
       </div>

       <div><label for="post">Password:</label> <input name="password"
        type="password"></input><?php if(isset($_POST['username']) && $_POST['password'] == ''){echo "password required";}?><br>
       </div>

<input type="submit" value="Sign in" name="submit"></input></form>

     </div>
    </div>

    <?


     exit();


    } // end if - no username or password were posted
    else{

     $this->mysql_connect_func();

     $sql = "SELECT * FROM " . $this->security_table . " WHERE  username='" . $_POST['username'] . "'";


     $result = mysql_query($sql) or die(mysql_error());
     $row = mysql_fetch_assoc($result);


     if(mysql_num_rows($result) != 0)
     {


      if(sha1($_POST['password']) == $row['password'])
      {
      session_regenerate_id();
      $_SESSION['initiated'] = "true";
      $_SESSION['username'] = $row['username'];
      $_SESSION['authority'] = $row['authority'];

      } // end if sha1 of $_POST password == $row password
      else {
       ?>

        <div class="main_container">
        <div class="form_container">


       <?php
       echo 'Incorrect password<br><a href="' . $_SERVER['PHP_SELF'] . '">Please try again</a>';
       ?>
       </div>
       </div>


       <?php

       exit();


      }  // if password is wrong


     } // end if no rows with username returned
     else{
?>

        <div class="main_container">
        <div class="form_container">


       <?php
       echo 'Incorrect username <br><a href="' . $_SERVER['PHP_SELF'] . '">Please try again</a>';
       ?>
       </div>
       </div>


       <?php

       exit();

     } // if username not found


    } // end else - no username or password were posted

   } // end if - check the session !initiated

   else {   //*THIS IS THE LINE THAT THROWS THE ERROR IN XAMPP AND WAMP*

    if($_SESSION['authority'] < $security_level)
    {

     die("security clearance insufficient");
    }


   } // end else - check the session !initiated


  } // end if $security_level <= 0

 } // end of function security()

}

我很抱歉，这是一大堆代码，我想不出一种合乎逻辑的方法来分解它，而不损害某人帮助我找到错误的能力。如果你们对将野兽的大小缩小到更易读的程度有任何建议，请说出来！

Answer 1

我不确定这是否对您有帮助 - 但是在 第 67 行中，您在 $string 变量之后缺少终端 ; 。

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