年齡計算器不顯示計算的年齡

Question

此代碼是基於使用組合框的用戶的選定值來計算年齡，並且它必須在下面提供的文本框中顯示年齡。每次用戶更改組合框中的選定值時，都必須刷新年齡。但我目前的代碼並沒有顯示計算出來的年齡。

<?php 
       $month = date("m"); //without leading zero(o) 
       $year = date("Y"); //four digit format 
       $day = date("d"); 
       $st_year = "1950"; //Starting Year 
       $month_names = array("January", "February", "March","April", "May", "June", "July", "August", "September", "October", "November", "December"); 
      ?> 

      <form name="Month_Year" id="Month_Year" method="post"> 
      <select name="month" id="month"> 
      <?php 
      for ($i=1; $i<=12; $i++) { 
       echo "<option "; 
       if ($i == $month) { 
        echo "selected=\"selected\" "; 
       } 
       echo "value=\"$i\">", $month_names[$i-1], "</option>\n"; 
      } 
      ?> 
      </select> 
      <select name="year" id="year"> 
      <?php 
      for ($i=$st_year; $i<=$year; $i++) { 
       echo "<option "; 
       if ($i == $year) { 
        echo "selected=\"selected\" "; 
       } 
       echo "value=\"$i\">$i</option>\n"; 
      } 
      ?> 
      </select> 
      <select name="day" id="day"> 
      <?php 
      for ($i=1; $i<=31; $i++) { 
       echo "<option> "; 
       if ($i == $day) { 
        echo "selected=\"selected\" "; 
       } 
       echo "value=\"$i\">$i</option>\n"; 
      } 
      ?> 
      </select> 
      // I used this code to combine the selected value in the combo box. 
      <?php $Convertdays = $month."/".$day."/".$year; 

      echo $Convertdays; 
      ?> 
      <script type="text/javascript"> 
      var birth = new Date(<?php '$Convertdays'?>); 
      var check = new Date(); 
      var milliDay = 1000 * 60 * 60 * 24; // a day in milliseconds; 
      var ageInDays = (check - birth)/milliDay; 
      var ageInYears = Math.floor(ageInDays/365); 
      var age = ageInDays/365 ; 
      </script> 

我每次改變組合框的值，該文本框會顯示0 讓我們假設該文本框的值是1991年10月19日它應該顯示21