How to open a page in the APEX after I click on the "submit" button?

Hi gurus,

Here's my question (I'm new to APEX so please bear with me and provide details - I'm not new to programming, however). Ok. I have open application, a menu and the option of a menu I want to be a wizard to create an object (signature).

Once this wizard is done (it contains 3 pages linked modeless dialog), the button "submit" I want:

1 / all data is committed to the DB

2 / all data to be loaded into a new page that contains the full record, which was completed in steps the wizard 3

.. and I think that's all...

I understand the page after sending else can't do anything, not even to be triggered if everything must be done before. So, I made a dynamic action on the button end like this:

But I will not be opened in a new page but as part of the application (below you can see that opens in a new page):

Can you guide me please? Something must be either very easy, I guess...

This is the first question.

Second question: how to fill this new page with the instructions I just submit - how can I find out the rowid of the folder has just presented. Or... is there an another way of modeling that? What I want to do is create a record in a database either by attending either directly. If I click on cancel I will go on the record in page 1 where I can see full fields, if I choose assistant and then once I'm done I want to see all the data even in Page1. Where I can possibly do some mod.

Thank you!

Gabriela

(if it is not clear that I can show in a short owc internally)

Point 1: Creating a new branch on the last page of the wizard, which will take place on the State of the button "FINISH" Give branch URL of page that you want to navigate.

Point 2: You can save the primary key of the row and go to the parent page using the direction of the point 1.

Probably if you can build an example on apex.oracle.com, we can help more.

Tags: Database

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