How can I create a insert/update form for multiple table in Oracle Apex 5.1.4 - oracle

I have craeted two tables. I have also created a page and shown all data as classic report by inner joining between two tables. Now I want to know how can I create a form to update both table at a time?
For single table it can be done easily by creating form and report page. But In case of multiple table what will be way?
Thanks

Create new page by the new page creation wizard using the Form > Single Page Master Detail template.
It will guide you.

Here's one option: create a view, then create instead-of trigger(s) upon that view. Use the same view as a data source on the Apex page.
I know for sure that it works in Oracle Forms; never tried it in Apex, though, but I believe that it should be just fine.

Related

Unable to retrieve report data using Dynamic Action in Oracle APEX

I have two tables: Application (PK: Application_ID) and App_support_domain (column: team_name, FK: Application_ID referencing Application table)
I am trying to display the data present in the Application table based on the selection of team names.
I have created a LOV popup (P2_NEW) to select the team name.
After selecting the team name I want to show the application data in the form of a report, so I have created a dynamic action that will show the body region which has the application data.
But the SQL query that I have written does not return any data. I think it is not taking the P3_NEW LOV value and displaying the application data.
Please let me know if you need more info/clarification.
Could you please take a look and help me out?
A "show" dynamic action will not resubmit the report. Add 2 actions to the dynamic action. One to refresh and an 2nd one to show. That will force the report to to refresh. Make sure you put P2_NEW in the "Page Items to submit" for the report.

Oracle Apex 19.1 (how to add, edit, delete data through a form)

I'm new to oracle apex so this might be simple.
I have an application that i'm currently building. I have an 'employee' table in the SQL workshop. When i attempt to create a form linked to the table there is no option to edit, delete, or add entries into the table ,once the form is completed?
This is essentially what i need help with. I need to be able to manipulate the 'employee' table through the form created rather than through anything within the sql workshop. Just for context i am not the workspace admin, however i am a contributor.
I would appreciate if anyone could provide me with a quick step by step guide into creating this desired form accurately.
I'd suggest you to create (using the Wizard, of course) Report with Form on Table. It will create
interactive report you'd use to view data stored in that table
form which will be used to insert/update/delete data
the same form will be called when you push
the "Create" button in order to create new rows, or
icon at the beginning of every line in the report in order to update/delete rows
This combination (report + form) works nicely for ages, so ... try it. I hope you'll find it useful.

JSP page to immediately display whenever a new row is added in table?

I have oracle dbTable letssay testDbTable having column columnA and columnB.Whenever a new row is added in database table,it should display content of that row on JSP page in following manner.
columnA(value):----- and then columnB(value):........
lestsay after some time a new row is added in testDbTable.
then output should be in two line. Added new row in testDbTable should be aove on previous row that was displayed earlier on JSP page.
columnA(value):----- and then columnB(value):........
columnA(value):----- and then columnB(value):........
But such action should be at same time whenever a new row is added.
The simplest way i could think of that check on database again and again for some predefined lenth of time,but this solution seems to be wasteful and suboptimal to me, so i was wondering is there a better way.
i think there should be a trigger on table,when a new row is added in db then trigger should be fired and display row content on JSP page. I know "how to create trigger on table" but after that i am blank,how will i display just added row on JSP. After trigger on db table how will i approach? Can Ajax solve my problem like auto refresh or watch on db table or sothing else?
is there anyway i can listen to a trigger on my database from a .jsp script and force my client page to be refreshed ? Can you provide me any example link related to ajax,oracle dbtriggers and JSP?
In this case you should try reverse ajax technology like Comet.

Oracle Apex Apps. Sending data to two tables on submit

I have two tables, HELP_PROBLEMS and HELP_SOLUTIONS.
HELP_PROBLEMS has a number of different fields, including PROBLEM_ID, SOLUTION_ID, PROBLEM_DESC etc.
HELP_SOLUTIONS has just SOLUTION_ID and SOLUTION_DESC.
I'm trying to build form that when you view the edit form of a problem, you can add the solution in a form below it.
In my mind, when I hit the button Add Solution (submit), it creates the new row in HELP_SOLUTIONS, and adds the SOLUTION_ID to the current row in HELP_PROBLEMS. (and update a few other things but trying to keep this as basic as possible).
Is this at all possible? I thought the 'master form' might work but I couldn't seem to sort it out as planned.
Thanks,
C.
You can create a pl/sql block that inserts a value in two different tables.
create a page process and select pl/sql
then try something like
begin
--insert into help_problems
--insert into help_solutions
end;

Oracle APEX form based on multiple tables

Can I create a form in APEX which creates/updates records in multiple tables? For example two tables with a one-to-one relationship.
If you are using Automated Row Fetch mechanism, you can use two different ARF for two tables. If you want to have more control, you can define your INSERT/UPDATE statements yourself usign Page Processes.
A simple way would be to create a view that returns the joined tables and base the Apex form on that. You may need to add INSTEAD OF triggers to the view to be able to insert and update successfully.

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