如何从早期php代码中为表单分配变量以发布到下一个php页面

我正在制作一个通过提交表单收集用户评论的网站，然后将它们全部输出到可以从1-10评分的页面上。我在评分系统部分遇到了麻烦。

我有一个带有单选按钮的表单，用于我的评级，它通过'SendRating.php'发送值。我还想发送一个以前使用的变量，它是与特定评论和特定评级相关联的"id"。我试图使用隐藏的输入来做到这一点，但我似乎无法得到语法正确，无论我尝试我似乎得到一个错误。

我将展示我的两个php文件:

<?php
    $con=mysqli_connect("host","me","pwd","base");
    // Check connection
    if (mysqli_connect_errno($con))
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $query= "SELECT  id, comment, date_made, rating FROM table ORDER BY id DESC LIMIT 15";
    $result=mysqli_query($con, $query);
while ($row=mysqli_fetch_array($result))
    {
        $date=$row['date_made'];
        $id=$row['id'];
        $num_id=intVal($id);
        $current=$row['rating'];
        echo '<div id="text" onpaste="return false"; oncut="return false"; readonly>';
        echo '<div id="text2">';
        echo "Submitted: ";
        echo date('l jS F Y H:i:s ',strtotime($date));echo '&nbsp;'; echo '&nbsp;'; 
        echo '&nbsp;'; echo "No: ";  echo $row['id']; 
        echo '</div>';
        echo '<br>'; 
        echo nl2br($row['comment']);
        echo '<br>'; echo '<br>'; 
        echo '</a>';
        //FOOT
        echo '<div >'; 
        echo '<form id="radio" name="input_rating" action="SendRating.php"     method="post" >';
        echo '<input type="hidden" name="id" value="'$id'" >';
        echo '<input  id="radio" type="radio" name="rating"  onclick="this.form.submit()" value="1">1';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="2">2';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="3">3';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="4">4';
        echo '<input  id="radio" type="radio" name="rating"     onclick="this.form.submit()" value="5">5';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="6">6';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="7">7';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="8">8';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="9">9';
        echo '<input  id="radio" type="radio" name="rating" onclick="this.form.submit()" value="10">10';
        echo '&nbsp;'; echo '&nbsp;'; echo '&nbsp;'; echo  "Rating: ";
        echo '&nbsp';
        echo  "$current";
        echo '</form>';
        echo '</div>';
        //END FOOT
         echo '</div>'; 
         echo '<p></p>';
    }
?> 

表单操作由以下操作处理(或者如果一切正常的话):

<?php
    $con2=mysqli_connect("host","me","pwd","base");
    // Check connection
    if (mysqli_connect_errno($con2))
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
        $value=$_POST['rating'];
        $id=$_POST['id'];
    mysqli_query  ($con2, "UPDATE table SET totalraters = totalraters + 1 WHERE id='$id' ");
    mysqli_query  ($con2, "UPDATE table SET ratingsum = ratingsum + '$value' WHERE id='$id' ");
    $query2 = mysqli_query ($con2, "SELECT FROM table ratingsum, totalraters WHERE id = '$id' ");
    $result2 = mysqli_query($con2, $query2);
    $row2 =mysqli_fetch_array($result2);
    $ratingsum=$row2['ratingsum']; 
    $totalraters=$row2['totalraters'];
    $current = $ratingsum/$ratingtotal;
    mysqli_query ($con2, "INSERT INTO table rating VALUE '$current' WHERE id='$id' ");
    include ('/index.html')
?>   

include页应该将用户带回到更新评级的起始页。

ratingsum是给出的评分的总和，totalraters是进行评分的人数。$current应该是给出当前评级的平均值。

我现在遇到的麻烦是

echo '<input type="hidden" name="id" value="'$id'" >';

我似乎无法从一个地方获得id到另一个地方以在进一步的查询中使用。这一行出现了语法错误。感谢收到的帮助，相信我已经尝试了所有正确的解决方案!!请使用过程，因为我是新手，不能理解面向对象的所有…它让我头疼;-)