oracle select based on priority

Question

I'm trying to rewrite some old SQL queries that look particularly awful. I am wondering if there is a more efficient way to prioritize values in a where statement for order of precedence. Basically the table contains multiple email_code records per user but I want to prioritize based on what records are preferred. In this case if the email_code is WORK it should be selected. But if there is no WORK record then HOME should be selected, and so on. Here is an example of what I am working with. There has to be a more graceful way to do this...?

select 
    * 
from 
    email m
where
    status_ind='A'
    and decode(email_code, 'WORK',1,
                           'HOME',2,
                           'ALT1',3,
                           'ALT2',4,5) = (select 
                                          min(decode(email_code, 'WORK',1,
                                                                 'HOME',2,
                                                                 'ALT1',3,
                                                                 'ALT2',4,5)) 
                                        from 
                                          email
                                        where 
                                          email_uid = m.email_uid
                                          and status_ind='A');

Answer 1

Try:

SELECT * FROM (
   SELECT e.*,
          dense_rank() over (PARTITION BY user_id
            ORDER BY CASE email_code
               WHEN 'WORK' THEN 1
               WHEN 'HOME' THEN 2
               WHEN 'ALT1' THEN 3
               WHEN 'ALT2' THEN 4
               ELSE 5
          END ) As priority
   FROM emails e
   WHERE status_ind='A'
)
WHERE priority = 1