建立猜测动物游戏

Question

对于单一作业，我必须创建一个网页，用户可以回答是/否问题，系统会猜测他们正在考虑的动物。数据库必须是自引用表/二进制树。

我已经构建了似乎运行正常的数据库。

列是

nodeID，parentID，message，answerYesID，answerNoID。 parentID和两个answerId都是fk引用nodeID。

使用PHP我创建了按钮和一个检索第一个节点并询问第一个问题的函数。我已经到了系统要求下一个适当问题的地步。这是我被困的地方。我已经尝试了各种切换方式和if语句，但我正在努力正确遍历二叉树。

我创建了一个变量$ currentnode，我将用它来跟踪用户的位置，以便可以提出正确的问题，但是，在下一个if块中使用它会导致系统向前跳。

非常感谢任何帮助

$query = "SELECT `message`, `nodeID` FROM `creature`";//basic query that will be used at                    all times
$where = "";//empty where clause, will be altered 



//function created to be re used after an answer is given
function output($query,$where,$dbconn){
$result = mysqli_query($dbconn, $query.$where);
while($row = mysqli_fetch_assoc($result)) 
echo "{$row['message']}
    {$row['nodeID']}<br>";


 echo '<form action="play.php" method="POST">
<input type="radio" name="answer" value="1">yes
<input type="radio" name="answer" value="0">no
<input type = "submit" name = "submit" value = "submit">
<input type = "submit" name = "restart" value = "restart game">
</form>';}



?>



<form action="play.php" method="POST">
<input type="submit" name="start" value="Start Game">
</form>


<?php

if ( isset( $_POST['start'] ))
{ 
$where = "WHERE `nodeID` = '1'";
$currentnode = '1';
output($query,$where,$dbconn);
}



 if ($currentnode = 1){
if ( isset( $_POST['submit'] )){

            switch ($_POST['answer']) {

                case 1:     $where = "WHERE `nodeID` = '2'";
                                    output($query,$where,$dbconn);
                            $currentnode = 2; 
                            break;  

                case 0:     $where = "WHERE `nodeID` = '25'";
                              output($query,$where,$dbconn);
                            $currentnode = 25;
                            break;


                            }   

                    }           
}

if ($currentnode = 2){
if ( isset( $_POST['submit'] )){

            switch ($_POST['answer']) {

                case 1:     $where = "WHERE `nodeID` = '3'";
                             output($query,$where,$dbconn);
                            $currentnode = 3; 
                            break;  

                case 2:     $where = "WHERE `nodeID` = '4'";
                             output($query,$where,$dbconn);
                            $currentnode = 4;
                            break;


                            }   

                    }           
}   





if ( isset( $_POST['restart'] ))
{ 

$where = "WHERE `nodeID` = '1'";
output($query,$where, $dbconn);
}