Java conector oracle sequence for inserting

Hy there…
I’m trying to insert a new record into an oracle db using a form.
I looked at your code on
docs.dhtmlx.com/doku.php?id=dhtm … ex_queries

And, in php, you suggest:
$grid->sql->sequence(“EMPLOYEES_INC.nextVal”); //sequence name
$grid->render_table(“EMPLOYEES”,“EMPLOYEE_ID”,“FIRST_NAME,LAST_NAME”);

I’ve tried to translate that in Java, but the connector does NOT have a “sql.sequence” parameter,
i.e. the following gives compile errors on the “c.sql.sequence” line:

Connection conn = ( new DataBaseConnection()).getConnection();
FormConnector c = new FormConnector(conn, DBType.Oracle);
c.sql.sequence(“TIPOGIUNTA_SEQ.nextVal”);
c.render_table(“TIPI_GIUNTE”, “TIPOGIUNTA_ID”, “DEFINIZIONE”);

In java you can use

DBDataWrapper sql = (DBDataWrapper)c.sql;
sql.sequence("TIPOGIUNTA_SEQ.nextVal");

Thanks…
I tried that but I get an error inserting… the log says:

====================================
Log started, Thu Apr 11 17:48:34 CEST 2013

id parameter was missed
Done in : 10ms

But shouldn’t it be auto-generating the id parameter from the sequence ???

But shouldn’t it be auto-generating the id parameter from the sequence ???
It is the different issue, when loading data to the form it expects to receive the id parameter.

For example you can use

form.load("some.php")

you need to use

form.load("some.php?id=123"); 

which will select record with id = 123, and will load it in form

I understand, but I am generating a NEW record
So I do a form.clear();
Input data on the form
And then save … So I don’t have an ID!!!

… sorry, I wasn’t clear … I’m getting the error when I’m SAVING the new record.

The issue was confirmed and fixed.
Fixed jar will be published as part of oncoming update ( was sent to you through support ticket as well )