如何在oracle-apex中访问报表列值？

Question

I need the "elabora prenotazione" button to be shown only when the column "stato prenotazione" is "IN SOSPESO" I tried to set a condition but I don't know how to pick the column value

Answer 1

If I understand your question, this could help:

For the source of your report:

SELECT
    CASE
        WHEN stato_prenotazione = 'in sospeso' THEN 'elabora prenotazione'
        ELSE ''
    END name_of_column,
prenotazione,
username,
nome,
cognome,
viaggio,
stato_viaggio,
data_prenota,
stato_prenotazione,
numero_ospiti
FROM your_table

Then set the column "name_of_column" to type: link, and fill the target.

Answer 2

Wrap a condition around a call to apex_page.get_url within your SQL, so it will only produce a link when relevant

Example of function in use, sans condition: https://azureoutput.wordpress.com/2017/10/18/custom-hyperlink-in-interactive-report/

Use this to make the button look prettier, and maybe get some other ideas https://www.grassroots-oracle.com/2015/12/tutorial-include-action-button-in-report.html

See this in regard to escaping special characters, otherwise you'll just see HTML in your report https://www.grassroots-oracle.com/2017/01/escape-special-characters-apex-demo.html

Answer 3

This could be resolved by a simple case statement as @sara mentioned. something like:

select (case stato_prenotazione when 'in sospeso' then 'elabora prenotazione' end) as your_column

I would not keep else condition so that the column will simply contain null value if the condition is not met.