php - How to insert multiple ID's in one linked table? -

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i want insert de employeeid , knowledgeid in knowledgedetail. creates employee nothing in knowledgedetail. i'm there no code, have tried many things have no idea.
first in addprofile.php make profile , @ least choose yoour knowledge. question if make profile , choose knowledge how de id's in knowledgedetail.

table 1
employees: employeeid, name, establishment, e-mail, phonenumber, photo, description
table 2
knowledge: knowledgeid, knowledge
table 3
knowledgedetail: knowledgedetailid, employeeid knowledgeid

employeeid out employees have relation employeeid out knowledgedetail and
knowledgeid out knowledge have relation knowledgeid out knowledegedetail

addprofile.php

<?php include("connection.php"); ?> <!doctype html> <html> <head>     <title>information system</title>      <link rel="stylesheet" type="text/css" href="css/test.css">     <meta charset ="utf-8">     <link rel='stylesheet' href='http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/smoothness/jquery-ui.css' type='text/css' media='screen' />       <link rel='stylesheet' href='css/ui.multiselect.css' type='text/css' media='screen' />     <script src="../informatiesysteem/js/jquery.min.js"></script>     <script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js'></script>     <script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js'></script>     <script type='text/javascript' src='../informatiesysteem/js/ui.multiselect.js'></script>     <script type='text/javascript'>             jquery(document).ready(function() {                 jquery("#selectitems").multiselect();             });     </script> </head>  <body>     <div id="container">         <div id="logo"></div>             <div id="header">             <h1>add profile</h1>             </div>                     <div id="menu">                     </div>                         <div id="content">                 <?php                 $result = mysql_query("select knowledgeid, knowledge knowledge");                 $items = array();                  $selected = array();                 while ($row = mysql_fetch_array($result)){                   $id [] = $row [ 'knowlegdeid' ];                 $items[] = $row[ 'knowledge' ];                   }                 //form processing                 if (isset($_post['selectitems'])) {                 $selected = $_post['selectitems'];                 }                        if (!empty($selected)) : print_r($selected); endif;                 ?>                 <form enctype="multipart/form-data" id="my form" method="post" action="addedprofile.php">                 name:            <input type="text" name="name" /></br>                 establishment:   <input type="text" name="establishment"/></br>                 e-mail:          <input type="email" name="email"/></br>                             phonenumber:     <input type="tel" name="phonenumber"/></br>                     photo:           <input type="file" name="photo"/></br>                 description:     <textarea rows="4" cols="50" name="description"></textarea></br>                 knowledge: <select name="selectitems[]" id="selectitems" multiple="multiple" style="width: 450px; height: 180px;">                 <?php //first add list of selected items if                 foreach ($selected $sel) { ?>                 <option value="<?php echo $sel; ?>" selected="selected"><?php echo $id[$sel]; $items[$sel];?></option>                 <?php } ?>                 <?php foreach ($items $i => $v) { //then insert items, skipping added above                 if (in_array($d, $i, $selected)) : continue; endif; //skip ?>                 <option value="<?php echo $d, $i; ?>"><?php echo $v; ?></option>                 <?php } ?>                 </select>                                                     </br></br></br></br>                 <input type="submit" name="add_profile" value="add profile" />                                        </form>                 </div> </body> </html>

addedprofile.php

<!doctype html> <html> <meta http-equiv="refresh" content=";url=addprofile.php" /> </html> <?php include ("connection.php"); $name = $_post['name']; $establishment = $_post['establishment']; $email = $_post['email']; $phonenumber = $_post['phonenumber']; $photo = $_post['photo']; $description = $_post['description'];   $sql = "insert employees           (          name,           establishment,          email,          phonenumber,          photo,          description          )          values ('". $name ."', '". $establishment ."', '". $email ."', '". $phonenumber ."', '". $photo ."', '". $description ."')";  $sqldetail = "insert knowledgedetail          (         employeeid,         knowledgeid              )         values .......................";                   $add = mysql_query($sql);                           if ($add === false){                    echo 'profile not created';              }             else {                                 echo "profile created";                  }              echo '</br>';    $knowledge = mysql_query($sqldetail);                                                 if ($add === false){                echo 'knowledge not added';          }         else {                             echo "knowledge added";              }              echo '</br>';   ?>

here's 1 thing that's wrong code:

$knowledge = mysql_query($sqldetail);                                         if ($add === false){     echo 'knowledge not added'; } else {                    echo "knowledge added"; }

in if statement, should compare $knowledge , not $add. so, should be:

$knowledge = mysql_query($sqldetail);                                         if ($knowledge === false){     echo 'knowledge not added'; } else {                    echo "knowledge added"; }

also, add call mysql_error() every time mysql_query() fails:

echo "mysql error: sql = $sql -- error=".mysql_error()";

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