MySQLハイスコア-ユーザーの個人ランク：誤った値を引き起こす重複したエントリ

これを試してください：

SET @rank = 0;
SELECT rank FROM (
    SELECT @rank:[email protected] + 1 AS rank, uid FROM playerstats ORDER BY overall DESC, 
    overallxp  DESC
) as tmp WHERE uid = 14950

avoveクエリはuser14950のランクを返します

このクエリは、すべてのユーザーとそのランクを一覧表示します

SET @rank=0;
SELECT rank, uid, overall, overallxp FROM (
SELECT @rank:[email protected] + 1 AS rank, uid, overall, overallxp FROM playerstats ORDER BY overall DESC, overallxp DESC
) as tmp