Ordering SQL query by specific field values

Question

I've got a sql query (using Firebird as the RDBMS) in which I need to order the results by a field, EDITION. I need to order by the contents of the field, however. i.e. "NE" goes first, "OE" goes second, "OP" goes third, and blanks go last. Unfortunately, I don't have a clue how this could be accomplished. All I've ever done is ORDER BY [FIELD] ASC/DESC and nothing else.

Any suggestions?

Edit: I really should clarify: I was just hoping to learn more here. I have it now that I just have multiple select statements defining which to show first. The query is rather large and I was really hoping to learn possibly a more effecient way of doing this: example:

SELECT * FROM RETAIL WHERE MTITLE LIKE 'somethi%' AND EDITION='NE'
UNION 
SELECT * FROM RETAIL WHERE MTITLE LIKE 'somethi%' AND EDITION='OE'
UNION
SELECT * FROM RETAIL WHERE MTITLE LIKE 'somethi%' AND EDITION='OP'
UNION (etc...)

Answer 1

Add those values to another table with a numeric column for their rank:

Edition  Rank
NE       1
OE       2
OP       3

Join the tables, and sort on the RANK field.

Answer 2

SELECT /other fields/
CASE WHEN 'NE' THEN 1
WHEN "OE" THEN 2
WHEN "OP" THEN 3
ELSE 4 END AS OrderBy
FROM /Tables/
WHERE /conditions/
ORDER BY OrderBy, /other fields/

Answer 3

Order By Case Edition
    When 'NE' Then 1
    When 'OE' Then 2
    When 'OP' Then 3
    Else 4 End

Answer 4

Try:

select *
from MyTable
order by
case [FIELD] 
    when 'NE' then 1
    when 'OE' then 2
    when 'OP' then 3
    when '' then 4
    else 5
end

Answer 5

Try this:

ORDER BY FIND_IN_SET(EDITION, 'NE,OE,OP,')