查询未正确设置用户的排名，取决于用户的总积分

Question

我正在尝试获取每个用户的所有总积分，以降序对其进行排序并打印其排名。我的查询工作正常，但它不能正确设置用户的排名。

SET @rank := 0;
SELECT (@rank := @rank + 1) AS rank, u.username, SUM(s.score) AS totalScore
FROM solution AS s
INNER JOIN users u ON u.id = s.author_id
GROUP BY u.username
ORDER BY totalScore DESC

用户架构：

|---------------------|------------------|
|         id          |     username     |
|---------------------|------------------|
|          1          |       test1      |
|---------------------|------------------|
|          2          |       test2      |
|---------------------|------------------|
|          3          |       test3      |
|---------------------|------------------|
|          8          |       test4      |
|---------------------|------------------|

解决方案架构：

|---------------------|------------------|-----------------|
|         id          |     author_id    |      score      |
|---------------------|------------------|-----------------|
|          1          |       1          |       55        |
|---------------------|------------------|-----------------|
|          2          |       2          |       5         |
|---------------------|------------------|-----------------|
|          3          |       3          |       22        |
|---------------------|------------------|-----------------|
|          4          |       8          |       43        |
|---------------------|------------------|-----------------|
|          5          |       8          |       43        |
|---------------------|------------------|-----------------|

结果是：

|---------------------|------------------|-----------------|
|         rank        |     username     |    totalScore   |
|---------------------|------------------|-----------------|
|          4          |       test4      |       86        |
|---------------------|------------------|-----------------|
|          1          |       test1      |       55        |
|---------------------|------------------|-----------------|
|          3          |       test3      |       22        |
|---------------------|------------------|-----------------|
|          2          |       test2      |       5         |
|---------------------|------------------|-----------------|

为什么会这样？

预期结果应该是：

|---------------------|------------------|-----------------|
|         rank        |     username     |    totalScore   |
|---------------------|------------------|-----------------|
|          1          |       test4      |       86        |
|---------------------|------------------|-----------------|
|          2          |       test1      |       55        |
|---------------------|------------------|-----------------|
|          3          |       test3      |       22        |
|---------------------|------------------|-----------------|
|          4          |       test2      |       5         |
|---------------------|------------------|-----------------|

Answer 1

这将从MySql运行：

Select username, totalScore, Rank() over (order by totalScore desc) as Rank 
from (
SELECT  u.username, SUM(s.score) AS totalScore
FROM solution AS s
INNER JOIN users u ON u.id = s.author_id
GROUP BY u.username
ORDER BY totalScore DESC
)t

Answer 2

正如我在评论中所说。您首先要对分数进行排名才能对其进行排名。

SELECT 
    (@rank:=@rank + 1) rank, username, totalscore
FROM
    (SELECT 
        u.username, SUM(s.score) AS totalScore
    FROM
        solution AS s
    INNER JOIN users u ON u.id = s.author_id
    GROUP BY u.username
    ORDER BY totalScore DESC) t1,
    (SELECT @rank:=0) r1

请参见示例，第二个查询是我的

模式（MySQL v5.7）

CREATE TABLE users (
  `id` INTEGER,
  `username` VARCHAR(5)
);

INSERT INTO users
  (`id`, `username`)
VALUES
  ('1', 'test1'),
  ('2', 'test2'),
  ('3', 'test3'),
  ('8', 'test4');

CREATE TABLE solution (
  `id` INTEGER,
  `author_id` INTEGER,
  `score` INTEGER
);

INSERT INTO solution
  (`id`, `author_id`, `score`)
VALUES
  ('1', '1', '55'),
  ('2', '2', '5'),
  ('3', '3', '22'),
  ('4', '8', '43'),
  ('5', '8', '43');

---

查询＃1

SET @rank := 0;

没有要显示的结果。

---

查询＃2

SELECT (@rank := @rank + 1) AS rank, u.username, SUM(s.score) AS totalScore
FROM solution AS s
INNER JOIN users u ON u.id = s.author_id
GROUP BY u.username
ORDER BY totalScore DESC;

| rank | totalScore | username |
| ---- | ---------- | -------- |
| 4    | 86         | test4    |
| 1    | 55         | test1    |
| 3    | 22         | test3    |
| 2    | 5          | test2    |

---

查询＃3

SELECT
(@rank := @rank +1) rank
,username
,totalscore
FROM
(SELECT
u.username,
SUM(s.score) AS totalScore
FROM solution AS s
INNER JOIN users u ON u.id = s.author_id
GROUP BY u.username
ORDER BY totalScore DESC) t1,(SELECT @rank := 0) r1;

| rank | username | totalscore |
| ---- | -------- | ---------- |
| 1    | test4    | 86         |
| 2    | test1    | 55         |
| 3    | test3    | 22         |
| 4    | test2    | 5          |

---

View on DB Fiddle