How can I get the rank of rows relative to total number of rows based on a field?

I have a scores table that has two fields:

• user_id
• score

I'm fetching specific rows that match a list of user_id's. How can I determine a rank for each row relative to the total number of rows, based on score? The rows in the result set are not necessarily sequential (the scores will vary widely from one row to the next). I'm not sure if this matters, but user_id is a unique field.

Edit

@Greelmo

I'm already ordering the rows. If I fetch 15 rows, I don't want the rank to be 1-15. I need it to be the position of that row compared against the entire table by the score property. So if I have 200 rows, one row's rank may be 3 and another may be 179 (these are arbitrary #'s for example only).

Edit 2

I'm having some luck with this query, but I actually want to avoid ties

    SELECT
       s.score
       , s.created_at
       , u.name
       , u.location
       , u.icon_id
       , u.photo
       , (SELECT COUNT(*) + 1 FROM scores WHERE score > s.score) AS rank
    FROM
        scores s
    LEFT JOIN
        users u ON u.uID = s.user_id
    ORDER BY
        s.score DESC
        , 开发者_开发知识库s.created_at DESC
    LIMIT 15

If two or more rows have the same score, I want the latest one (or earliest - I don't care) to be ranked higher. I tried modifying the subquery with AND id > s.id but that ended up giving me an unexpected result set and different ties.

Select S.score, S.created_at, U.name
   , U.location, U.icon_id, U.photo
   , (Select Count(*) + 1
        From scores S2
        Where S2.score > S.score
            Or (S2.score = S.Score And S2.created_at > S.created_at)
        ) AS rank
From scores S
    Left Join users U 
        On U.uID = S.user_id
Order By S.score DESC, S.created_at DESC
LIMIT 15

Of course, if it is possible for two scores to have the same created_at date, then you will still get ties and need to determine a third tie-breaker.

You could order the data in your query.

SELECT user_id, score FROM table ORDER BY score ASC;

This will give you your data from lowest score to highest.

If this doesn't answer your question, then I don't understand what you're asking.

EDIT

To get the position, while iterating through the database results, just keep a counter.