Question

我有一张桌子给玩家：

set.seed(1234)

a <- factor(c("ID1","ID2","ID3","ID4","ID5"))
b <- runif(5)
b1 <- runif(5)
b2 <- runif(5)
b3 <- runif(5)
c1 <- runif(5)
c2 <- runif(5)
c3 <- runif(5)
df <- data.frame(a,b1,b2,b3,c1,c2,c3)

我正在尝试根据player_id分组的出现次数对它们进行排名。

这就是我的尝试：

+------------+-----------+
| player_id  |   date    |
+------------+-----------+
|          1 | 27/1/2018 |
|          1 | 27/1/2018 |
|          1 | 27/1/2018 |
|          2 | 27/1/2018 |
|          2 | 27/1/2018 |
|          3 | 27/1/2018 |
+------------+-----------+

结果我得到了：

SELECT total2.player_id, COUNT(*) + 1 AS rank FROM (
    SELECT player_id, COUNT(*) AS total
    FROM players
    GROUP BY player_id
) total2
INNER JOIN (
    SELECT player_id, COUNT(*) AS total
    FROM players
    GROUP BY player_id
) total3 ON total3.total > total2.total
GROUP BY total2.player_id

结果我想要：

+------------+------+
| player_id  | rank |
+------------+------+
|          2 |    2 |
|          3 |    3 |
+------------+------+

Answer 1

试试这个

这里是小提琴

SELECT a.player_id, @row_number:=@row_number+1 AS row_number 
FROM (SELECT player_id, COUNT(*) AS total
    FROM players
    GROUP BY player_id) a,
(SELECT @row_number:=0) AS t
ORDER BY TOTAL desc

排名行跳过排名1

1 个答案: