Php MYSQL选择用户排名（大于x小于y）

我有点被php/Mysql查询卡住了。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：

• 用户_1=lvl0（因为用户有2分）
• 用户_2=lvl1（因为用户刚刚达到10分）
• 用户_4=lvl2（因为用户尚未达到30分）

我想你：）

---

问候。

试试看，因为桌子设计有点乱

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

---

试试看，由于桌子的设计有点凌乱

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

---

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

---

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

---

那么你尝试了什么？展示你的努力？你试过了吗？我发现了这个：但我对“AS”这个词有点迷茫。而且排名也不完全一样，顺便说一句：\那么你尝试了什么？展示你的努力？你试过了吗？我发现了这个：但我对“AS”这个词有点迷茫。而且排名也不完全一样，顺便说一句：\太棒了！做这项工作，而且是灵活的。非常感谢：）太棒了！做这项工作，而且是灵活的。非常感谢：）这似乎有效，但与@raheel shan的答案相比有点复杂。我认为你的BTWIt似乎有效，但与@raheel shan的答案相比有点复杂。顺便说一句，你觉得怎么样