I don't think you could express it in the SELECT directly (someone else
will probably build a statement refuting that) but you could do:

SELECT r.RDB$FIELD_NAME FROM RDB$RELATION_FIELDS r where
r.RDB$RELATION_NAME=<your table name here>

and then process that result set to build your appropriate SELECT for
the target table. So I suppose you could build a statement with a
quantity of derived sets using IIF and WHERE EXISTS - the question is do
you need to express this as single SELECT, or can you accomplish your
goal either via your calling program or a stored procedure?
--
Daniel


On 10/9/2017 1:36:20 AM, "Elmar Haneke elmar@...
[firebird-support]" <firebird-support@yahoogroups.com> wrote:

>
>>Is there a way to include a column in a SELECT but substitute a value
>>if
>>it doesn't? I need my code to work with different versions of my db
>>schema.
>>
>>eg I want to SELECT A, B, C, ... but C might not exist.
>>
>>SELECT * would work of course, except it will fetch a bunch of columns
>>I
>>don't need.
>
>You can read the list of fields available in advance and modify your
>query ommitting missing columns.
>
>
>Elmar
>
>
>
>------------------------------------
>Posted by: Elmar Haneke <elmar@...>
>------------------------------------
>
>++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>
>Visit http://www.firebirdsql.org and click the Documentation item
>on the main (top) menu. Try FAQ and other links from the left-side
>menu there.
>
>Also search the knowledgebases at
>http://www.ibphoenix.com/resources/documents/
>
>++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>------------------------------------
>
>Yahoo Groups Links
>
>
>