How open modal pages conditionally?

Question

I am using Oracle APEX 21.1. I have a table visits with a column visit_type. visit_type has only values 1 or 2. There are 2 dialog pages(1 and 2). I need to create a query that returns a link that opens page 1 when visit_type = 1 and opens page 2 when visit_type = 2. Should I select a string with <a href=""</a> tag or use APEX_PAGE.GET_URL API. Either way, kindly, give me an example.

Answer 1

Here is a typical format

  apex_string.format(
      '<a class="t-Button t-Button--hot t-Button--simple t-Button--stretch" href="%s">%s</a>'
    , apex_page.get_url
       (p_application => 'YOUR_APP'
       ,p_page        => t.visit_type
       ,p_items       => 'p'||t.visit_type||'_id'
       ,p_values      => t.visit_id
       )
    , 'Link text'
    ) as link_target

Don't forget to not escape special characters in your column definition.