php - Display saved checkbox value -

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i have checkbox options save in db. able view , select multiple options , save them in db. issue want display saved information don't know how that.

<form action="save_comp.php" method="post"> <?php //display   include ('mysql_connect.php');  $sql = mysql_query("select * competency "); //$row = mysql_fetch_array($sql);  while($row = mysql_fetch_array($sql)) {  echo"<input type='checkbox' name='comp[]' value= ".$row['id']." /> ".$row['competency']." <br />";   }  ?> <input name="submit" type="submit" value="submit" /> </form>

save db

<?php session_start(); $id = $_session['user_id']; //$id = 3;  include ('mysql_connect.php');  $insstr = '';  foreach($_post['comp'] $val){ $insstr .=$val.","; } mysql_query("insert competency_result (user_id,result) values ( '$id', '$insstr' )") or die(mysql_error());   echo'<script>alert("inserted successfully")</script>';  ?>

all want display saved information in table format. tried doing showed me saved id

<?php $res= mysql_query("select * competency_result user_id = '$user'")or die(mysql_error());  while($row = mysql_fetch_array($res)) {     echo"<tr>";     echo"<td>&nbsp;$row[result]</td>";                 ?>  <?php       echo"</tr>"; } ?>                               

<form action="save_comp.php" method="post"> <?php //display   include ('mysql_connect.php');  $sql = mysql_query("select * competency "); //$row = mysql_fetch_array($sql);  while($row = mysql_fetch_array($sql)) {  echo"<input type='checkbox' name='comp[". $row['id']. "]' value='". $row['competency'] ."' /> ".$row['competency']." <br />";   }  ?> <input name="submit" type="submit" value="submit" /> </form>

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