PHP Fatal error: Call to undefined method resultSet::free() - php

Question

I am attempting to get a function working on an old php based application, autonomous lan party. It is based on php4 but it does work with what I want it to do on a php5 server (it is also only used in an intranet environment). I'm adding an accounting add-on from here. I can understand php, but still somewhat rusty.
Below is a small sample of code I'm stuck with...
<FORM ACTION="<?php echo $_SERVER['PHP_SELF']; ?>" METHOD="post" NAME="accounting" ID="accounting">
<strong>user: </strong><br>
<SELECT NAME="userid[]" SIZE="5" class="formcolors" TABINDEX="1" MULTIPLE>
<?php
$data = $dbc->query('SELECT username,userid FROM users ORDER BY username');
while ( $row = $data->fetchRow()) {
$option = '
<OPTION VALUE="%s" class="formcolors">%s</OPTION>';
printf($option, $row['userid'], $row['username']);
}
$data->free();
unset($option, $data);
?>
</SELECT>
The problem is once it reaches te $data->free(); line, the script stops executing and supplies the error mentioned in the topic. If I comment it out (twice, there is a similar line) the script runs but once i submit the data I get another error in the logfile...
PHP Warning: mysql_escape_string() expects parameter 1 to be string, array given
I believe that error is because I've commented out $data->free(); so it's not able to get the correct result.
I've pasted the full code from the file at pastebin here (didnt want to fill this page up with all the other code).
Any help or assistance would be much appreciated. Everything else on the application works as expected.

Answer 1

If I understand correctly, $data is a mysql result.
To free the result from memory, try using mysqli_free_result() instead.
$data = $dbc->query('SELECT username,userid FROM users ORDER BY username');
while ( $row = $data->fetchRow()) {
$option = '
<OPTION VALUE="%s" class="formcolors">%s</OPTION>';
printf($option, $row['userid'], $row['username']);
}
mysqli_free_result($data->free());
unset($option, $data);