因爲未定義的變量而從DB中檢索數據問題

Question

我的意圖是從數據庫檢索數據並創建了以下功能但收到了未定義的函數，成績，錯誤消息。代碼如下：

 <form action="processor.php" method="post"> 
     <h4>Question # 1</h4> 
     <p>What grade are you in?</p> 
     <label class="checkbox"><input type="checkbox" name="grade" value="1"> Freshmen</label> 
     <label class="checkbox"><input type="checkbox" name="grade" value="2"> Sophomore</label> 
     <label class="checkbox"><input type="checkbox" name="grade" value="3"> Junior</label> 
     <label class="checkbox"><input type="checkbox" name="grade" value="4"> Senior</label> 

     <h4>Question # 2</h4> 
     <p>What is your current GPA?</p> 
     <select name="gpa"> 
     <option value="4">3.5 or above</option> 
     <option value="3">3.0-3.4</option> 
     <option value="2">2.5-2.9</option> 
     <option value="1">2.0-2.4</option> 
     <option value="0">Lower</option> 
     </select> 

     <h4>Question # 3</h4> 
     <p>Where do you excel the most academically?</p> 
     <select name="school" multiple="multiple"> 
     <option value ="1">Mathematics</option> 
     <option value ="2">Literature</option> 
     <option value ="3">History</option> 
     <option value ="4">Humanities</option> 
     <option value ="5">Science</option> 
     </select> 

然後處理表單提交的PHP文件：

<?php 
require_once('processor.php'); 
class Processor { 
    #code 

function get_response($id) { 
$dsn = "mysql:host=localhost;dbname=user"; 
$username = "root"; // database username 
$password = "stewie12"; // database password 
try { 
    $enter = new PDO($dsn, $username, $password); 
    $sql = "SELECT response FROM recommendations WHERE id = ? "; 
    $new_item = $enter->prepare($sql); 
      $new_item->setFetchmode(PDO::FETCH_ASSOC); 
    $new_item->execute(array($id)); 
    foreach($new_item as $nw) { 
     return $nw['response']; 
    } 
} catch (Exception $e) { 
    echo $e->getMessage(); 
    exit; 
} 
return ""; 
    } 



function GetWeird() { 
    $santiago="1992"; 
if ($santiago="1992") { 
     $id = 1; 
    } else { 
     $id = 2; 
} 
     echo $this->get_response($id); 
    } 


function Grades() { 
$grade = $_POST['grade']; 
if ($grade =="1") { 
    echo "You're a freshmen"; 
} elseif ($grade == "2") { 
    echo "You're a sophomore"; 
} elseif ($grade == "3") { 
    echo "You're a junior."; 
} elseif ($grade == "4") { 
    echo "You're a senior."; 
} else { 
    echo "Something is wrong."; 
} 

} 

function Gpa() { 
$gpa = $_POST['gpa']; 
if ($gpa =="1") { 
    echo "You strongly need to up your GPA."; 
} elseif ($gpa == "2") { 
    echo "You're an average student."; 
} elseif ($gpa == "3") { 
    echo "You're an above average student."; 
} elseif ($gpa == "4") { 
    echo "You're an excellent sudent."; 
} else { 
    echo "Something is wrong."; 
    } 

    } 

function School() { 
$school = $_POST['school']; 
if ($school =="1") { 
echo "You're into Math"; 
} elseif ($school == "2") { 
    echo "You're into Lit"; 
    } elseif ($school == "3") { 
    echo "You're into history."; 
} elseif ($school == "4") { 
    echo "You're into humanities."; 
} elseif ($school == "5") { 
    echo "You're into science."; 
    } else { 
    echo "Something is wrong."; 
    } 

    } 

和視圖頁面：

 <?php $processor = new Processor(); ?> 
    <h4>Question # 1</h4> 
    <p><?php Grades($grade); ?></p> 
    <h4>Question # 2</h4> 
    <p><?php Gpa($gpa); ?></p> 
     <h4>Question # 3</h4> 
    <p><?php School($school); ?></p> 

Answer 1

要訪問以下代碼的類功能使用，

你必須這樣使用$class_variable->function_name()

<?php $processor = new Processor(); ?> 
<h4>Question # 1</h4> 
<p><?php $processor->Grades($grade); ?></p> 
<h4>Question # 2</h4> 
<p><?php $processor->Gpa($gpa); ?></p> 
    <h4>Question # 3</h4> 
<p><?php $processor->School($school); ?></p>