Can you add form data from a single form into more than one mySQL table?

Can the script (referenced below) be modified to post form data into more than one table?

How to send submissions to your MySQL database using PHP
https://www.jotform.com/help/126-How-to-send-Submissions-to-Your-MySQL-Database-Using-PHP

Yes, you can modify the query here, you can write join table queries to insert values in multiple tables.

$query = "SELECT * FROM `table_name` WHERE `submission_id` = '$submission_id'";

Let us know if you have further questions.

Thanks.

roneet provided a solution which if I were more skilled in mySQL, could probably have ran with and solved my issue.  Given I'm not a skilled mySQL developer, I discovered another means for posting form data from a single form into multiple tables within my mySQL database.

Using the sample php file jotform provides for posting data to your personal mySQL database, I modified the following section of this form (Ref: https://www.jotform.com/help/126-How-to-send-Submissions-to-Your-MySQL-Database-Using-PHP).

I simply replicated the "search submission ID" section for each table I wanted to post form data into.  In my case, I had a single form collecting information relevant to two tables.  Hopefully the following makes sense.  Perhaps any mySQL developers who may run across this can hang a little meat on this solution to whether this is a good or bad approach.

// search submission ID

$query = "SELECT * FROM `table_name1` WHERE `submission_id` = '$submission_id'";

$sqlsearch = mysql_query($query);

$resultcount = mysql_numrows($sqlsearch);

if ($resultcount > 0) {

    mysql_query("UPDATE `table_name1` SET

                                `tbl1item1` = '$tbl1item1',

                                `tbl1item2` = '$tbl1item2',

                                `tbl1item3` = '$tbl1item3',

                                WHERE `submission_id` = '$submission_id'")

     or die(mysql_error());

} else {

mysql_query("INSERT INTO `table_name1` (submission_id, formID, IP, tbl1item1, tbl1item2, tbl1item3)

                               VALUES ('$submission_id', '$formID', '$ip', '$tbl1item1', '$tbl1item2', '$tbl1item3') ")

    or die(mysql_error()); 

}

// This section for the 2nd table 

$query = "SELECT * FROM `table_name2` WHERE `submission_id` = '$submission_id'";

$sqlsearch = mysql_query($query);

$resultcount = mysql_numrows($sqlsearch);

if ($resultcount > 0) {

    mysql_query("UPDATE `table_name2` SET

                                `tbl2item1` = '$tbl2item1',

                                `tbl2item1` = '$tbl2item1',

                                WHERE `submission_id` = '$submission_id'")

     or die(mysql_error());

} else {

mysql_query("INSERT INTO `table_name2` (submission_id, tbl2item1, tbl2item1)

                               VALUES ('$submission_id', '$tbl2item1', '$tbl2item1') ")

    or die(mysql_error()); 

}