sql ORDER BY multiple values in specific order?

Question

Ok I have a table with a indexed key and a non indexed field. I need to find all records with a certain value and return the row. I would like to know if I can order by multiple values.

Example:

id     x_field
--     -----
123    a
124    a
125    a
126    b
127    f
128    b
129    a
130    x
131    x
132    b
133    p
134    p
135    i

pseudo: would like the results to be ordered like this, where ORDER BY x_field = 'f', 'p', 'i', 'a'

SELECT *
FROM table
WHERE id NOT IN (126)
ORDER BY x_field 'f', 'p', 'i', 'a'

So the results would be:

id     x_field
--     -----
127    f
133    p
134    p
135    i
123    a
124    a
125    a
129    a

The syntax is valid but when I execute the query it never returns any results, even if I limit it to 1 record. Is there another way to go about this?

Think of the x_field as test results and I need to validate all the records that fall in the condition. I wanted to order the test results by failed values, passed values. So I could validate the failed values first and then the passed values using the ORDER BY.

What I can't do:

• GROUP BY, as I need to return the specific record values
• WHERE x_field IN('f', 'p', 'i', 'a'), I need all the values as I'm trying to use one query for several validation tests. And x_field values are not in DESC/ASC order

After writing this question I'm starting to think that I need to rethink this, LOL!

Answer 1

...
WHERE
   x_field IN ('f', 'p', 'i', 'a') ...
ORDER BY
   CASE x_field
      WHEN 'f' THEN 1
      WHEN 'p' THEN 2
      WHEN 'i' THEN 3
      WHEN 'a' THEN 4
      ELSE 5 --needed only is no IN clause above. eg when = 'b'
   END, id