PHP, mySQL. saving on more than one different table using php action script

 $link = mysqli_connect("host", "username", "password", "database");
 $sql =  "INSERT INTO job_spec_contact (contact_info_key, datee,company_name,Physical_Address, recruitment_person,email,Telephone)
         VALUES('null','$datee','$company','$PAddress','$recruiter','$email','$telephone') ;";

 $sql . = "INSERT INTO job_company_infor (info_key, company_specialization,no_of_agents,no_of_resumes, org_structure)
         VALUES('null','$company','$agents','$resumes','$structure')";
 mysqli_multi_query($link, $sql);

Sorry for late reply.

If you are using mysql: (Not recommended due to unsecure)

$conn = mysql_connect('localhost','username','password', true, 65536) or die("cannot connect");
mysql_select_db('YourDBName') or die("cannot use database");

if(isset($_POST['Submit'])){
     $datee = $_POST['date'];
     $company = $_POST['company'];
     $PAddress = $_POST['PAddress'];
     $recruiter = $_POST['recruiter'];
     $email = $_POST['email'];
     $telephone = $_POST['telephone'];
     $company = $_POST['company'];
     $agents = $_POST['agents'];
     $resumes = $_POST['resumes'];
     $structure = $_POST['structure'];
}

$result = mysql_query("
    INSERT INTO job_spec_contact (contact_info_key, datee, company_name, Physical_Address, recruitment_person,email,Telephone)
                         VALUES('null','$datee','$company','$PAddress','$recruiter','$email','$telephone');

    INSERT INTO job_company_infor (info_key, company_specialization,no_of_agents,no_of_resumes, org_structure)
                         VALUES('null','$company','$agents','$resumes','$structure');

");

if (!$result) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}


while ($row = mysql_fetch_assoc($result)) {
    echo $row['datee'];
    echo $row['company_name'];
    ......
}

mysql_free_result($result);
?>

If you are using mysqli: (Recommended),

$conn = mysqli_connect('localhost','username','password') or die("cannot connect");
mysqli_select_db($conn, 'YourDBName') or die("cannot use database");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

if(isset($_POST['Submit'])){
     $datee = $_POST['date'];
     $company = $_POST['company'];
     $PAddress = $_POST['PAddress'];
     $recruiter = $_POST['recruiter'];
     $email = $_POST['email'];
     $telephone = $_POST['telephone'];
     $company = $_POST['company'];
     $agents = $_POST['agents'];
     $resumes = $_POST['resumes'];
     $structure = $_POST['structure'];
}

$query  = "INSERT INTO job_spec_contact (contact_info_key, datee, company_name, Physical_Address, recruitment_person, email, Telephone)
                         VALUES('null','$datee','$company','$PAddress','$recruiter','$email','$telephone')";
$query .= "INSERT INTO job_company_infor (info_key, company_specialization, no_of_agents, no_of_resumes, org_structure)
                         VALUES('null','$company','$agents','$resumes','$structure')";

if ($mysqli->multi_query($query)) {
    do {
        if ($result = $mysqli->store_result()) {
            while ($row = $result->fetch_row()) {
                printf("%s\n", $row[0]);//test here your values.
               //$datee = $row['datee'];
            }
            $result->free();
        }
    } while ($mysqli->next_result());
}

/* close connection */
$mysqli->close();

Hope this helps.

Note:

1. Check always POST is present or not by isset. (assume, input name Submit)

2. Use MySQLi/PDO instead of MySQL to avoid SQL injection.

3. Debug code by using echo, print_r, var_dump, etc.,

4. Try to use field names are in same pattern. For ex, Instead of Physical_Address, use physical_address like other fields. Telephone to telephone. datee to job_contact_date, etc.,

Just execute your queries one by one. And DO NOT write error_reporting(0); or the errors won't show. Plus where is your DB ?

 $sql1 =  "INSERT INTO job_spec_contact (contact_info_key, datee,company_name,Physical_Address, recruitment_person,email,Telephone)
             VALUES('null','$datee','$company','$PAddress','$recruiter','$email','$telephone')";

$sql2 = "INSERT INTO job_company_infor (info_key, company_specialization,no_of_agents,no_of_resumes, org_structure)
             VALUES('null','$company','$agents','$resumes','$structure')";

mysqli_query($db,$sql1) or die('error '.$sql1.'<br>'.mysqli_error($db));
mysqli_query($db,$sql2) or die('error '.$sql2.'<br>'.mysqli_error($db));