MYSQL排序按W /计数 [英] MYSQL Order By W/Count

问题描述

我的表有用户名和用户点，我正在处理领导板列表。我想根据积分得到前十名玩家。我已将此查询表达为：

My table has usernames and points of user, and I am working on a leader boards list. I would like to get the top ten players based on points. I have formulated this query as:

SELECT
    users.username,
    users.points
FROM users
ORDER BY users.points DESC
LIMIT 10

我也想得到玩家站在根据1号播放器，而不创建一个新的列。是否有MYSQL查询，通过点获得 ORDER BY DESC ，以及从 COUNT > ORDER 找到列表中的用户？ （ WHERE username = ）。

However, I would also like to get where the player stands in accordance to the number 1 player, without creating a new column for it. Is there a MYSQL query to get the ORDER BY DESC by points as well as COUNT from that ORDER to find the user in the list? (WHERE username=).

编辑澄清：以上语句。 I.E. 1 - user1 - 1000点，2 - user2 - 750点..等等... N - currentPlayer - currentPoints。我不介意使用 JOIN 语句，但我不希望有一个新的玩家列按排名排序。

Edit for clarification: I would like the count of the users from desc by the above statement. I.E. 1 - user1 - 1000 points, 2 - user2 - 750 points.. etc etc... N - currentPlayer - currentPoints. I don't mind using a JOIN statement, but I don't want to have a new column of players sorted by ranking.

推荐答案

尝试：

SELECT * FROM (
    select @rownum:=@rownum+1 `rank`, u.username, u.points 
    from users u, (SELECT @rownum:=0) r
    order by points desc) AS X
WHERE username ="Bob";

示例数据：

CREATE TABLE users(
  id int auto_increment primary key,
  username varchar (30),
  points INT
  );

INSERT INTO users(username,points)
VALUES ('Bob', 1000), ('Jack',750), ('Joe', 500)

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