所以我有一张表如下：
ID_STUDENT | ID_CLASS | GRADE
-----------------------------
1 | 1 | 90
1 | 2 | 80
2 | 1 | 99
3 | 1 | 80
4 | 1 | 70
5 | 2 | 78
6 | 2 | 90
6 | 3 | 50
7 | 3 | 90
我需要对它们进行分组，排序和排序，以便：
ID_STUDENT | ID_CLASS | GRADE | RANK
------------------------------------
2 | 1 | 99 | 1
1 | 1 | 90 | 2
3 | 1 | 80 | 3
4 | 1 | 70 | 4
6 | 2 | 90 | 1
1 | 2 | 80 | 2
5 | 2 | 78 | 3
7 | 3 | 90 | 1
6 | 3 | 50 | 2
现在我知道您可以使用临时变量排名like here，但我该如何为分组设置呢？感谢您的任何见解！
答案 0 :(得分：31)
SELECT id_student, id_class, grade,
@student:=CASE WHEN @class <> id_class THEN 0 ELSE @student+1 END AS rn,
@class:=id_class AS clset
FROM
(SELECT @student:= -1) s,
(SELECT @class:= -1) c,
(SELECT *
FROM mytable
ORDER BY id_class, id_student
) t
这非常简单：
id_class排序，id_student秒。@student和@class已初始化为-1 @class用于测试是否输入了下一组。如果id_class的先前值（存储在@class中）不等于当前值（存储在id_class中），则@student将归零。否则是递增。@class被赋予新值id_class，并将在下一行的第3步中用于测试。答案 1 :(得分：12)
Quassnoi的解决方案存在问题（标记为最佳答案）。
我有同样的问题（即模拟MySQL中的SQL窗口函数），我曾经使用用户定义的变量存储前一行值来实现Quassnoi的解决方案...
但是，也许在MySQL升级之后或者其他什么，我的查询不再起作用了。这是因为无法保证SELECT中字段的评估顺序。 @class赋值可以在@student赋值之前进行求值，即使它被放在SELECT中。
这在MySQL文档中提到如下：
作为一般规则，您不应该为用户变量赋值 并在同一语句中读取值。你可能会得到 你期望的结果，但这不能保证。的顺序 涉及用户变量的表达式的评估是未定义的 可能会根据给定声明中包含的元素进行更改; 另外，这个顺序不保证是相同的 发布MySQL服务器。
来源：http://dev.mysql.com/doc/refman/5.5/en/user-variables.html
最后我使用了这样的技巧，确保在阅读后分配@class：
SELECT id_student, id_class, grade,
@student:=CASE WHEN @class <> id_class THEN concat(left(@class:=id_class, 0), 0) ELSE @student+1 END AS rn
FROM
(SELECT @student:= -1) s,
(SELECT @class:= -1) c,
(SELECT *
FROM mytable
ORDER BY id_class, grade desc
) t
使用left（）函数只是用来设置@class变量。然后，将left（）（等于NULL）的结果连接到预期的结果是透明的。
不是很优雅，但它有效！
答案 2 :(得分：5)
从上面修改，这可行，但它比我认为的更复杂：
SELECT ID_STUDENT, ID_CLASS, GRADE, RANK
FROM
(SELECT ID_STUDENT, ID_CLASS, GRADE,
@student:=CASE WHEN @class <> id_class THEN 1 ELSE @student+1 END AS RANK,
@class:=id_class AS CLASS
FROM
(SELECT @student:= 0) AS s,
(SELECT @class:= 0) AS c,
(SELECT *
FROM Students
ORDER BY ID_CLASS, GRADE DESC
) AS temp
) AS temp2
答案 3 :(得分：4)
SELECT g1.student_id
, g1.class_id
, g1.grade
, COUNT(*) AS rank
FROM grades AS g1
JOIN grades AS g2
ON (g2.grade, g2.student_id) >= (g1.grade, g1.student_id)
AND g1.class_id = g2.class_id
GROUP BY g1.student_id
, g1.class_id
, g1.grade
ORDER BY g1.class_id
, rank
;
结果：
+------------+----------+-------+------+
| student_id | class_id | grade | rank |
+------------+----------+-------+------+
| 2 | 1 | 99 | 1 |
| 1 | 1 | 90 | 2 |
| 3 | 1 | 80 | 3 |
| 4 | 1 | 70 | 4 |
| 6 | 2 | 90 | 1 |
| 1 | 2 | 80 | 2 |
| 5 | 2 | 78 | 3 |
| 7 | 3 | 90 | 1 |
| 6 | 3 | 50 | 2 |
+------------+----------+-------+------+
答案 4 :(得分：1)
我做了一些搜索，找到this article来提出这个解决方案：
SELECT S2.*,
FIND_IN_SET(
S2.GRADE
, (
SELECT GROUP_CONCAT(GRADE ORDER BY GRADE DESC)
FROM Students S1
WHERE S1.ID_CLASS = S2.ID_CLASS
)
) AS RANK
FROM Students S2 ORDER BY ID_CLASS, GRADE DESC;
关于哪个更好的想法？
答案 5 :(得分：1)
SELECT ID_STUDENT, ID_CLASS, GRADE, RANK() OVER(
PARTITION BY ID_CLASS
ORDER BY GRADE ASC) AS 'Rank'
FROM table
ORDER BY ID_CLASS;
对于家庭作业，我遇到了类似的问题，发现MySQL（不能代表任何其他RDBMS）为其RANK（）方法具有分区参数。不知道为什么它不能解决这个问题。
答案 6 :(得分：1)
虽然我没有足够的声望来评论（有点幽默），但 MySQL 在最近几年取得了长足的进步。添加了窗口函数和 CTE（WITH 子句），这意味着现在支持 rank（和 row_number 等）。
我是同一个“Jon Armstrong - Xgc”，但那个帐户被旧电子邮件地址的风吹走了。
一个评论提出了一个关于 MySQL 是否支持排名窗口函数的问题。回答：是的。
我几年前的原始回复：
SELECT p1.student_id
, p1.class_id
, p1.grade
, COUNT(p2.student_id) AS rank
FROM grades AS p1
JOIN grades AS p2
ON (p2.grade, p2.student_id) >= (p1.grade, p1.student_id)
AND p1.class_id = p2.class_id
GROUP BY p1.student_id, p1.class_id
ORDER BY p1.class_id, rank
;
结果：
+------------+----------+-------+------+
| student_id | class_id | grade | rank |
+------------+----------+-------+------+
| 2 | 1 | 99 | 1 |
| 1 | 1 | 90 | 2 |
| 3 | 1 | 80 | 3 |
| 4 | 1 | 70 | 4 |
| 6 | 2 | 90 | 1 |
| 1 | 2 | 80 | 2 |
| 5 | 2 | 78 | 3 |
| 7 | 3 | 90 | 1 |
| 6 | 3 | 50 | 2 |
+------------+----------+-------+------+
9 rows in set (0.001 sec)
使用 ROW_NUMBER 窗口函数：
WITH cte1 AS (
SELECT student_id
, class_id
, grade
, ROW_NUMBER() OVER (PARTITION BY class_id ORDER BY grade DESC) AS rank
FROM grades
)
SELECT *
FROM cte1
ORDER BY class_id, r
;
结果：
+------------+----------+-------+------+
| student_id | class_id | grade | rank |
+------------+----------+-------+------+
| 2 | 1 | 99 | 1 |
| 1 | 1 | 90 | 2 |
| 3 | 1 | 80 | 3 |
| 4 | 1 | 70 | 4 |
| 6 | 2 | 90 | 1 |
| 1 | 2 | 80 | 2 |
| 5 | 2 | 78 | 3 |
| 7 | 3 | 90 | 1 |
| 6 | 3 | 50 | 2 |
+------------+----------+-------+------+
9 rows in set (0.002 sec)
使用RANK窗口函数：
WITH cte1 AS (
SELECT student_id
, class_id
, grade
, RANK() OVER (PARTITION BY class_id ORDER BY grade DESC) AS rank
FROM grades
)
SELECT *
FROM cte1
ORDER BY class_id, rank
;
结果：
+------------+----------+-------+------+
| student_id | class_id | grade | rank |
+------------+----------+-------+------+
| 2 | 1 | 99 | 1 |
| 1 | 1 | 90 | 2 |
| 3 | 1 | 80 | 3 |
| 4 | 1 | 70 | 4 |
| 6 | 2 | 90 | 1 |
| 1 | 2 | 80 | 2 |
| 5 | 2 | 78 | 3 |
| 7 | 3 | 90 | 1 |
| 6 | 3 | 50 | 2 |
+------------+----------+-------+------+
9 rows in set (0.000 sec)