Question

我有点陷入了php / Mysql查询。我有两张桌子：

  table_users              table_ranks
----------------    ------------------------
| id  | points |    | name | points_needed |
----------------    ------------------------
| 1   |   2    |    | lvl0 |      0        |
| 2   |   10   |    | lvl1 |      10       |
| 3   |   21   |    | lvl2 |      20       |
| 4   |   29   |    | lvl3 |      30       |
----------------    ------------------------

我需要这样的输出：

User_1 = lvl0（因为用户有2分）
User_2 = lvl1（因为用户刚刚达到10分）
...
User_4 = lvl2（因为用户尚未达到30分）

想想你:)。

问候。

Answer 1

你可以这样做

SELECT
  tu.id,
  tr.name,
  tu.points
FROM table_ranks as tr
  LEFT JOIN (SELECT * FROM table_ranks LIMIT 1,69596585953484) as l
    ON l.points_needed = (SELECT MIN(points_needed) FROM table_ranks WHERE points_needed > tr.points_needed limit 1)
  LEFT OUTER JOIN table_users AS tu ON tu.points >= tr.points_needed AND tu.points < l.points_needed
WHERE tu.id IS NOT NULL
group by tu.id

Fiddle

输出

-------------------------
| id  | points   | name |
-------------------------
| 1   |   lvl0   | 2    |
| 2   |   lvl1   | 10   |
| 3   |   lvl2   | 21   |
| 4   |   lvl2   | 29   |
-------------------------

Answer 2

尝试一下，由于桌面设计有点乱，

SELECT  u.*, r.name
FROM    table_users u
        INNER JOIN
        (
            SELECT x.name, 
                   x.points_needed start_point, 
                   COALESCE(y.points_needed - 1, 2000000) end_point
            FROM
            (
              SELECT name, points_needed, @rank:=@rank+1 ranks
              FROM table_ranks a, (SELECT @rank:=0) b
              ORDER BY points_needed
            ) x
            LEFT JOIN 
            (
              SELECT name, points_needed, @rank1:=@rank1+1 ranks
              FROM table_ranks a, (SELECT @rank1:=0) b
              ORDER BY points_needed
            ) y ON x.ranks+1 = y.ranks
        ) r ON u.points BETWEEN r.start_point AND r.end_point

SQLFiddle Demo

用户的MYSQL SELECT排名（超过x＆amp;小于y）

2 个答案: