MySQL 中的排名函数

发布于 2024-09-11 05:54:26 字数 240 浏览 7 评论 0原文

我需要找出客户的等级。这里我根据我的要求添加相应的 ANSI 标准 SQL 查询。请帮我将其转换为 MySQL 。

SELECT RANK() OVER (PARTITION BY Gender ORDER BY Age) AS [Partition by Gender], 
  FirstName, 
  Age,
  Gender 
FROM Person

MySQL有没有查询排名的函数？

原文

I need to find out rank of customers. Here I am adding the corresponding ANSI standard SQL query for my requirement. Please help me to convert it to MySQL .

SELECT RANK() OVER (PARTITION BY Gender ORDER BY Age) AS [Partition by Gender], 
  FirstName, 
  Age,
  Gender 
FROM Person

Is there any function to find out rank in MySQL?

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同尘 2024-09-18 05:54:26

一种选择是使用排名变量，如下所示：

SELECT    first_name,
          age,
          gender,
          @curRank := @curRank + 1 AS rank
FROM      person p, (SELECT @curRank := 0) r
ORDER BY  age;

(SELECT @curRank := 0) 部分允许变量初始化，而无需单独的 SET 命令。

测试用例：

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');

结果：

+------------+------+--------+------+
| first_name | age  | gender | rank |
+------------+------+--------+------+
| Kathy      |   18 | F      |    1 |
| Jane       |   20 | F      |    2 |
| Nick       |   22 | M      |    3 |
| Bob        |   25 | M      |    4 |
| Anne       |   25 | F      |    5 |
| Jack       |   30 | M      |    6 |
| Bill       |   32 | M      |    7 |
| Steve      |   36 | M      |    8 |
+------------+------+--------+------+
8 rows in set (0.02 sec)

One option is to use a ranking variable, such as the following:

SELECT    first_name,
          age,
          gender,
          @curRank := @curRank + 1 AS rank
FROM      person p, (SELECT @curRank := 0) r
ORDER BY  age;

The (SELECT @curRank := 0) part allows the variable initialization without requiring a separate SET command.

Test case:

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');

Result:

+------------+------+--------+------+
| first_name | age  | gender | rank |
+------------+------+--------+------+
| Kathy      |   18 | F      |    1 |
| Jane       |   20 | F      |    2 |
| Nick       |   22 | M      |    3 |
| Bob        |   25 | M      |    4 |
| Anne       |   25 | F      |    5 |
| Jack       |   30 | M      |    6 |
| Bill       |   32 | M      |    7 |
| Steve      |   36 | M      |    8 |
+------------+------+--------+------+
8 rows in set (0.02 sec)

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呆橘 2024-09-18 05:54:26

这是一个通用的解决方案，它将分区上的密集等级分配给行。它使用用户变量：

CREATE TABLE person (
    id INT NOT NULL PRIMARY KEY,
    firstname VARCHAR(10),
    gender VARCHAR(1),
    age INT
);

INSERT INTO person (id, firstname, gender, age) VALUES
(1,  'Adams',  'M', 33),
(2,  'Matt',   'M', 31),
(3,  'Grace',  'F', 25),
(4,  'Harry',  'M', 20),
(5,  'Scott',  'M', 30),
(6,  'Sarah',  'F', 30),
(7,  'Tony',   'M', 30),
(8,  'Lucy',   'F', 27),
(9,  'Zoe',    'F', 30),
(10, 'Megan',  'F', 26),
(11, 'Emily',  'F', 20),
(12, 'Peter',  'M', 20),
(13, 'John',   'M', 21),
(14, 'Kate',   'F', 35),
(15, 'James',  'M', 32),
(16, 'Cole',   'M', 25),
(17, 'Dennis', 'M', 27),
(18, 'Smith',  'M', 35),
(19, 'Zack',   'M', 35),
(20, 'Jill',   'F', 25);

SELECT person.*, @rank := CASE
    WHEN @partval = gender AND @rankval = age THEN @rank
    WHEN @partval = gender AND (@rankval := age) IS NOT NULL THEN @rank + 1
    WHEN (@partval := gender) IS NOT NULL AND (@rankval := age) IS NOT NULL THEN 1
END AS rnk
FROM person, (SELECT @rank := NULL, @partval := NULL, @rankval := NULL) AS x
ORDER BY gender, age;

请注意，变量赋值放置在 CASE 表达式内。这（理论上）解决了评估顺序问题。添加 IS NOT NULL 是为了处理数据类型转换和短路问题。

PS：通过删除所有检查平局的条件，可以轻松地将其转换为分区上的行号。

| id | firstname | gender | age | rank |
|----|-----------|--------|-----|------|
| 11 | Emily     | F      | 20  | 1    |
| 20 | Jill      | F      | 25  | 2    |
| 3  | Grace     | F      | 25  | 2    |
| 10 | Megan     | F      | 26  | 3    |
| 8  | Lucy      | F      | 27  | 4    |
| 6  | Sarah     | F      | 30  | 5    |
| 9  | Zoe       | F      | 30  | 5    |
| 14 | Kate      | F      | 35  | 6    |
| 4  | Harry     | M      | 20  | 1    |
| 12 | Peter     | M      | 20  | 1    |
| 13 | John      | M      | 21  | 2    |
| 16 | Cole      | M      | 25  | 3    |
| 17 | Dennis    | M      | 27  | 4    |
| 7  | Tony      | M      | 30  | 5    |
| 5  | Scott     | M      | 30  | 5    |
| 2  | Matt      | M      | 31  | 6    |
| 15 | James     | M      | 32  | 7    |
| 1  | Adams     | M      | 33  | 8    |
| 18 | Smith     | M      | 35  | 9    |
| 19 | Zack      | M      | 35  | 9    |

db<>fiddle 演示

Here is a generic solution that assigns dense rank over partition to rows. It uses user variables:

CREATE TABLE person (
    id INT NOT NULL PRIMARY KEY,
    firstname VARCHAR(10),
    gender VARCHAR(1),
    age INT
);

INSERT INTO person (id, firstname, gender, age) VALUES
(1,  'Adams',  'M', 33),
(2,  'Matt',   'M', 31),
(3,  'Grace',  'F', 25),
(4,  'Harry',  'M', 20),
(5,  'Scott',  'M', 30),
(6,  'Sarah',  'F', 30),
(7,  'Tony',   'M', 30),
(8,  'Lucy',   'F', 27),
(9,  'Zoe',    'F', 30),
(10, 'Megan',  'F', 26),
(11, 'Emily',  'F', 20),
(12, 'Peter',  'M', 20),
(13, 'John',   'M', 21),
(14, 'Kate',   'F', 35),
(15, 'James',  'M', 32),
(16, 'Cole',   'M', 25),
(17, 'Dennis', 'M', 27),
(18, 'Smith',  'M', 35),
(19, 'Zack',   'M', 35),
(20, 'Jill',   'F', 25);

SELECT person.*, @rank := CASE
    WHEN @partval = gender AND @rankval = age THEN @rank
    WHEN @partval = gender AND (@rankval := age) IS NOT NULL THEN @rank + 1
    WHEN (@partval := gender) IS NOT NULL AND (@rankval := age) IS NOT NULL THEN 1
END AS rnk
FROM person, (SELECT @rank := NULL, @partval := NULL, @rankval := NULL) AS x
ORDER BY gender, age;

Notice that the variable assignments are placed inside the CASE expression. This (in theory) takes care of order of evaluation issue. The IS NOT NULL is added to handle datatype conversion and short circuiting issues.

PS: It can easily be converted to row number over partition by by removing all conditions that check for tie.

| id | firstname | gender | age | rank |
|----|-----------|--------|-----|------|
| 11 | Emily     | F      | 20  | 1    |
| 20 | Jill      | F      | 25  | 2    |
| 3  | Grace     | F      | 25  | 2    |
| 10 | Megan     | F      | 26  | 3    |
| 8  | Lucy      | F      | 27  | 4    |
| 6  | Sarah     | F      | 30  | 5    |
| 9  | Zoe       | F      | 30  | 5    |
| 14 | Kate      | F      | 35  | 6    |
| 4  | Harry     | M      | 20  | 1    |
| 12 | Peter     | M      | 20  | 1    |
| 13 | John      | M      | 21  | 2    |
| 16 | Cole      | M      | 25  | 3    |
| 17 | Dennis    | M      | 27  | 4    |
| 7  | Tony      | M      | 30  | 5    |
| 5  | Scott     | M      | 30  | 5    |
| 2  | Matt      | M      | 31  | 6    |
| 15 | James     | M      | 32  | 7    |
| 1  | Adams     | M      | 33  | 8    |
| 18 | Smith     | M      | 35  | 9    |
| 19 | Zack      | M      | 35  | 9    |

Demo on db<>fiddle

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倾城泪 2024-09-18 05:54:26

虽然投票最多的答案排名，但它不会分区，您可以进行自连接来对整个内容进行分区：

SELECT    a.first_name,
      a.age,
      a.gender,
        count(b.age)+1 as rank
FROM  person a left join person b on a.age>b.age and a.gender=b.gender 
group by  a.first_name,
      a.age,
      a.gender

用例

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');

答案：

Bill    32  M   4
Bob     25  M   2
Jack    30  M   3
Nick    22  M   1
Steve   36  M   5
Anne    25  F   3
Jane    20  F   2
Kathy   18  F   1

While the most upvoted answer ranks, it doesn't partition, You can do a self Join to get the whole thing partitioned also:

SELECT    a.first_name,
      a.age,
      a.gender,
        count(b.age)+1 as rank
FROM  person a left join person b on a.age>b.age and a.gender=b.gender 
group by  a.first_name,
      a.age,
      a.gender

Use Case

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');

Answer:

Bill    32  M   4
Bob     25  M   2
Jack    30  M   3
Nick    22  M   1
Steve   36  M   5
Anne    25  F   3
Jane    20  F   2
Kathy   18  F   1

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鱼忆七猫命九 2024-09-18 05:54:26

对丹尼尔版本的调整，用于计算百分位数和排名。另外，两个分数相同的人将获得相同的排名。

set @totalStudents = 0;
select count(*) into @totalStudents from marksheets;
SELECT id, score, @curRank := IF(@prevVal=score, @curRank, @studentNumber) AS rank, 
@percentile := IF(@prevVal=score, @percentile, (@totalStudents - @studentNumber + 1)/(@totalStudents)*100),
@studentNumber := @studentNumber + 1 as studentNumber, 
@prevVal:=score
FROM marksheets, (
SELECT @curRank :=0, @prevVal:=null, @studentNumber:=1, @percentile:=100
) r
ORDER BY score DESC

示例数据的查询结果 -

+----+-------+------+---------------+---------------+-----------------+
| id | score | rank | percentile    | studentNumber | @prevVal:=score |
+----+-------+------+---------------+---------------+-----------------+
| 10 |    98 |    1 | 100.000000000 |             2 |              98 |
|  5 |    95 |    2 |  90.000000000 |             3 |              95 |
|  6 |    91 |    3 |  80.000000000 |             4 |              91 |
|  2 |    91 |    3 |  80.000000000 |             5 |              91 |
|  8 |    90 |    5 |  60.000000000 |             6 |              90 |
|  1 |    90 |    5 |  60.000000000 |             7 |              90 |
|  9 |    84 |    7 |  40.000000000 |             8 |              84 |
|  3 |    83 |    8 |  30.000000000 |             9 |              83 |
|  4 |    72 |    9 |  20.000000000 |            10 |              72 |
|  7 |    60 |   10 |  10.000000000 |            11 |              60 |
+----+-------+------+---------------+---------------+-----------------+

A tweak of Daniel's version to calculate percentile along with rank. Also two people with same marks will get the same rank.

set @totalStudents = 0;
select count(*) into @totalStudents from marksheets;
SELECT id, score, @curRank := IF(@prevVal=score, @curRank, @studentNumber) AS rank, 
@percentile := IF(@prevVal=score, @percentile, (@totalStudents - @studentNumber + 1)/(@totalStudents)*100),
@studentNumber := @studentNumber + 1 as studentNumber, 
@prevVal:=score
FROM marksheets, (
SELECT @curRank :=0, @prevVal:=null, @studentNumber:=1, @percentile:=100
) r
ORDER BY score DESC

Results of the query for a sample data -

+----+-------+------+---------------+---------------+-----------------+
| id | score | rank | percentile    | studentNumber | @prevVal:=score |
+----+-------+------+---------------+---------------+-----------------+
| 10 |    98 |    1 | 100.000000000 |             2 |              98 |
|  5 |    95 |    2 |  90.000000000 |             3 |              95 |
|  6 |    91 |    3 |  80.000000000 |             4 |              91 |
|  2 |    91 |    3 |  80.000000000 |             5 |              91 |
|  8 |    90 |    5 |  60.000000000 |             6 |              90 |
|  1 |    90 |    5 |  60.000000000 |             7 |              90 |
|  9 |    84 |    7 |  40.000000000 |             8 |              84 |
|  3 |    83 |    8 |  30.000000000 |             9 |              83 |
|  4 |    72 |    9 |  20.000000000 |            10 |              72 |
|  7 |    60 |   10 |  10.000000000 |            11 |              60 |
+----+-------+------+---------------+---------------+-----------------+

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屋顶上的小猫咪 2024-09-18 05:54:26

丹尼尔和萨尔曼的答案的结合。然而，排名不会给出，因为存在并列的连续序列。相反，它会跳过排名到下一个。所以最大总是达到行数。

    SELECT    first_name,
              age,
              gender,
              IF(age=@_last_age,@curRank:=@curRank,@curRank:=@_sequence) AS rank,
              @_sequence:=@_sequence+1,@_last_age:=age
    FROM      person p, (SELECT @curRank := 1, @_sequence:=1, @_last_age:=0) r
    ORDER BY  age;

架构和测试用例：

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');
INSERT INTO person VALUES (9, 'Kamal', 25, 'M');
INSERT INTO person VALUES (10, 'Saman', 32, 'M');

输出：

+------------+------+--------+------+--------------------------+-----------------+
| first_name | age  | gender | rank | @_sequence:=@_sequence+1 | @_last_age:=age |
+------------+------+--------+------+--------------------------+-----------------+
| Kathy      |   18 | F      |    1 |                        2 |              18 |
| Jane       |   20 | F      |    2 |                        3 |              20 |
| Nick       |   22 | M      |    3 |                        4 |              22 |
| Kamal      |   25 | M      |    4 |                        5 |              25 |
| Anne       |   25 | F      |    4 |                        6 |              25 |
| Bob        |   25 | M      |    4 |                        7 |              25 |
| Jack       |   30 | M      |    7 |                        8 |              30 |
| Bill       |   32 | M      |    8 |                        9 |              32 |
| Saman      |   32 | M      |    8 |                       10 |              32 |
| Steve      |   36 | M      |   10 |                       11 |              36 |
+------------+------+--------+------+--------------------------+-----------------+

Combination of Daniel's and Salman's answer. However the rank will not give as continues sequence with ties exists . Instead it skips the rank to next. So maximum always reach row count.

    SELECT    first_name,
              age,
              gender,
              IF(age=@_last_age,@curRank:=@curRank,@curRank:=@_sequence) AS rank,
              @_sequence:=@_sequence+1,@_last_age:=age
    FROM      person p, (SELECT @curRank := 1, @_sequence:=1, @_last_age:=0) r
    ORDER BY  age;

Schema and Test Case:

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');
INSERT INTO person VALUES (9, 'Kamal', 25, 'M');
INSERT INTO person VALUES (10, 'Saman', 32, 'M');

Output:

+------------+------+--------+------+--------------------------+-----------------+
| first_name | age  | gender | rank | @_sequence:=@_sequence+1 | @_last_age:=age |
+------------+------+--------+------+--------------------------+-----------------+
| Kathy      |   18 | F      |    1 |                        2 |              18 |
| Jane       |   20 | F      |    2 |                        3 |              20 |
| Nick       |   22 | M      |    3 |                        4 |              22 |
| Kamal      |   25 | M      |    4 |                        5 |              25 |
| Anne       |   25 | F      |    4 |                        6 |              25 |
| Bob        |   25 | M      |    4 |                        7 |              25 |
| Jack       |   30 | M      |    7 |                        8 |              30 |
| Bill       |   32 | M      |    8 |                        9 |              32 |
| Saman      |   32 | M      |    8 |                       10 |              32 |
| Steve      |   36 | M      |   10 |                       11 |              36 |
+------------+------+--------+------+--------------------------+-----------------+

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他不在意 2024-09-18 05:54:26

MySQL 5.7

在 MySQL 5.7 中，您可以使用 JSON 和局部变量，如下所示，以模拟 RANK() OVER (PARTITION BY ..)：

SELECT
  FirstName, Age, Gender,
  coalesce(
    json_extract(
      @rn := json_set(@rn, 
        @rnpath := concat('$."rn-', Gender, '"'), 
        @currn := coalesce(json_extract(@rn, @rnpath), 0) + 1,
        @prevpath := concat('$."pre-v-', Gender, '"'),
        Age,
        @prernpath := concat('$."pre-rn-', Gender, '"'),
        IF (json_extract(@rn, @prevpath) = Age, 
          coalesce(json_extract(@rn, @prernpath), @currn) div 1,
          @currn
        )
      ), 
      @prernpath
    ), 
    @currn
  ) AS rn2
FROM Person, (SELECT @rn := '{}') r
ORDER BY Age DESC;

每个分区键，这存储 ROW_NUMBER()< /code> ("rn-x")，根据排序键的上一个值 ("pre-vx")，以及上一个 ROW_NUMBER( ) ("pre-rn-x") 用于绑定行，以模拟 RANK() 行为。此处使用 JSON 的好处是 PARTITION BY 子句不会影响查询的顺序。

MySQL 8.0

从 MySQL 8 开始，您终于可以在 MySQL 中使用窗口函数：
https://dev.mysql.com/doc/refman/ 8.0/en/window-functions.html

您的查询可以以完全相同的方式编写：

SELECT RANK() OVER (PARTITION BY Gender ORDER BY Age) AS `Partition by Gender`, 
  FirstName, 
  Age,
  Gender 
FROM Person

MySQL 5.7

In MySQL 5.7, you can use JSON and local variables as follows, to emulate RANK() OVER (PARTITION BY ..):

SELECT
  FirstName, Age, Gender,
  coalesce(
    json_extract(
      @rn := json_set(@rn, 
        @rnpath := concat('$."rn-', Gender, '"'), 
        @currn := coalesce(json_extract(@rn, @rnpath), 0) + 1,
        @prevpath := concat('$."pre-v-', Gender, '"'),
        Age,
        @prernpath := concat('$."pre-rn-', Gender, '"'),
        IF (json_extract(@rn, @prevpath) = Age, 
          coalesce(json_extract(@rn, @prernpath), @currn) div 1,
          @currn
        )
      ), 
      @prernpath
    ), 
    @currn
  ) AS rn2
FROM Person, (SELECT @rn := '{}') r
ORDER BY Age DESC;

Per partition key, this stores the ROW_NUMBER() ("rn-x"), previous value according to the sort key ("pre-v-x"), as well as the previous ROW_NUMBER() ("pre-rn-x") for tied rows, to emulate RANK() behaviour. The benefit of using JSON here is that the PARTITION BY clause does not influence the ordering of the query.

MySQL 8.0

Starting with MySQL 8, you can finally use window functions also in MySQL:
https://dev.mysql.com/doc/refman/8.0/en/window-functions.html

Your query can be written exactly the same way:

SELECT RANK() OVER (PARTITION BY Gender ORDER BY Age) AS `Partition by Gender`, 
  FirstName, 
  Age,
  Gender 
FROM Person

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孤者何惧 2024-09-18 05:54:26

@Sam，你的观点在概念上非常好，但我认为你误解了MySQL文档在引用页面上所说的内容——或者我误解了:-)——我只是想添加这一点，以便如果有人对@感到不舒服丹尼尔的回答他们会更放心，或者至少会更深入地挖掘。

您会看到 SELECT 内的 "@curRank := @curRank + 1 ASrank" 不是“一个语句”，它是语句的一个“原子”部分，因此它应该是安全的。

您引用的文档继续显示了语句的 2 个（原子）部分中相同的用户定义变量的示例，例如，“SELECT @curRank, @curRank := @curRank + 1 AS rating”.

有人可能会说 @Daniel 的答案中使用了两次 @curRank：(1) "@curRank := @curRank + 1 AS rating" 和 (2) < code>"(SELECT @curRank := 0) r" 但由于第二次用法是 FROM 子句的一部分，我很确定它一定会首先被评估；本质上使它成为第二个也是前面的声明。

事实上，在您引用的同一个 MySQL 文档页面上，您会在注释中看到相同的解决方案——这可能是 @Daniel 从中获得的；是的，我知道这是评论，但这是官方文档页面上的评论，确实有一定的分量。

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千年*琉璃梦 2024-09-18 05:54:26

确定给定值排名的最直接解决方案是计算其之前的值的数量。假设我们有以下值：

10 20 30 30 30 40

所有 30 值均被视为第三
所有 40 值均被视为第六（排名）或第四（密集排名）

现在回到原来的问题。以下是一些按 OP 中所述排序的示例数据（预期排名添加在右侧）：

+------+-----------+------+--------+    +------+------------+
| id   | firstname | age  | gender |    | rank | dense_rank |
+------+-----------+------+--------+    +------+------------+
|   11 | Emily     |   20 | F      |    |    1 |          1 |
|    3 | Grace     |   25 | F      |    |    2 |          2 |
|   20 | Jill      |   25 | F      |    |    2 |          2 |
|   10 | Megan     |   26 | F      |    |    4 |          3 |
|    8 | Lucy      |   27 | F      |    |    5 |          4 |
|    6 | Sarah     |   30 | F      |    |    6 |          5 |
|    9 | Zoe       |   30 | F      |    |    6 |          5 |
|   14 | Kate      |   35 | F      |    |    8 |          6 |
|    4 | Harry     |   20 | M      |    |    1 |          1 |
|   12 | Peter     |   20 | M      |    |    1 |          1 |
|   13 | John      |   21 | M      |    |    3 |          2 |
|   16 | Cole      |   25 | M      |    |    4 |          3 |
|   17 | Dennis    |   27 | M      |    |    5 |          4 |
|    5 | Scott     |   30 | M      |    |    6 |          5 |
|    7 | Tony      |   30 | M      |    |    6 |          5 |
|    2 | Matt      |   31 | M      |    |    8 |          6 |
|   15 | James     |   32 | M      |    |    9 |          7 |
|    1 | Adams     |   33 | M      |    |   10 |          8 |
|   18 | Smith     |   35 | M      |    |   11 |          9 |
|   19 | Zack      |   35 | M      |    |   11 |          9 |
+------+-----------+------+--------+    +------+------------+

计算 的 RANK() OVER (PARTITION BY Gender ORDER BY Age) Sarah，您可以使用以下查询：

SELECT COUNT(id) + 1 AS rank, COUNT(DISTINCT age) + 1 AS dense_rank
FROM testdata
WHERE gender = (SELECT gender FROM testdata WHERE id = 6)
AND age < (SELECT age FROM testdata WHERE id = 6)

+------+------------+
| rank | dense_rank |
+------+------------+
|    6 |          5 |
+------+------------+

计算<全部<的RANK() OVER (PARTITION BY Gender ORDER BY Age) /strong> 行，您可以使用此查询：

SELECT testdata.id, COUNT(lesser.id) + 1 AS rank, COUNT(DISTINCT lesser.age) + 1 AS dense_rank
FROM testdata
LEFT JOIN testdata AS lesser ON lesser.age < testdata.age AND lesser.gender = testdata.gender
GROUP BY testdata.id

这是结果（连接值添加在右侧）：

+------+------+------------+    +-----------+-----+--------+
| id   | rank | dense_rank |    | firstname | age | gender |
+------+------+------------+    +-----------+-----+--------+
|   11 |    1 |          1 |    | Emily     |  20 | F      |
|    3 |    2 |          2 |    | Grace     |  25 | F      |
|   20 |    2 |          2 |    | Jill      |  25 | F      |
|   10 |    4 |          3 |    | Megan     |  26 | F      |
|    8 |    5 |          4 |    | Lucy      |  27 | F      |
|    6 |    6 |          5 |    | Sarah     |  30 | F      |
|    9 |    6 |          5 |    | Zoe       |  30 | F      |
|   14 |    8 |          6 |    | Kate      |  35 | F      |
|    4 |    1 |          1 |    | Harry     |  20 | M      |
|   12 |    1 |          1 |    | Peter     |  20 | M      |
|   13 |    3 |          2 |    | John      |  21 | M      |
|   16 |    4 |          3 |    | Cole      |  25 | M      |
|   17 |    5 |          4 |    | Dennis    |  27 | M      |
|    5 |    6 |          5 |    | Scott     |  30 | M      |
|    7 |    6 |          5 |    | Tony      |  30 | M      |
|    2 |    8 |          6 |    | Matt      |  31 | M      |
|   15 |    9 |          7 |    | James     |  32 | M      |
|    1 |   10 |          8 |    | Adams     |  33 | M      |
|   18 |   11 |          9 |    | Smith     |  35 | M      |
|   19 |   11 |          9 |    | Zack      |  35 | M      |
+------+------+------------+    +-----------+-----+--------+

The most straight forward solution to determine the rank of a given value is to count the number of values before it. Suppose we have the following values:

10 20 30 30 30 40

All 30 values are considered 3rd
All 40 values are considered 6th (rank) or 4th (dense rank)

Now back to the original question. Here is some sample data which is sorted as described in OP (expected ranks are added on the right):

+------+-----------+------+--------+    +------+------------+
| id   | firstname | age  | gender |    | rank | dense_rank |
+------+-----------+------+--------+    +------+------------+
|   11 | Emily     |   20 | F      |    |    1 |          1 |
|    3 | Grace     |   25 | F      |    |    2 |          2 |
|   20 | Jill      |   25 | F      |    |    2 |          2 |
|   10 | Megan     |   26 | F      |    |    4 |          3 |
|    8 | Lucy      |   27 | F      |    |    5 |          4 |
|    6 | Sarah     |   30 | F      |    |    6 |          5 |
|    9 | Zoe       |   30 | F      |    |    6 |          5 |
|   14 | Kate      |   35 | F      |    |    8 |          6 |
|    4 | Harry     |   20 | M      |    |    1 |          1 |
|   12 | Peter     |   20 | M      |    |    1 |          1 |
|   13 | John      |   21 | M      |    |    3 |          2 |
|   16 | Cole      |   25 | M      |    |    4 |          3 |
|   17 | Dennis    |   27 | M      |    |    5 |          4 |
|    5 | Scott     |   30 | M      |    |    6 |          5 |
|    7 | Tony      |   30 | M      |    |    6 |          5 |
|    2 | Matt      |   31 | M      |    |    8 |          6 |
|   15 | James     |   32 | M      |    |    9 |          7 |
|    1 | Adams     |   33 | M      |    |   10 |          8 |
|   18 | Smith     |   35 | M      |    |   11 |          9 |
|   19 | Zack      |   35 | M      |    |   11 |          9 |
+------+-----------+------+--------+    +------+------------+

To calculate RANK() OVER (PARTITION BY Gender ORDER BY Age) for Sarah, you can use this query:

SELECT COUNT(id) + 1 AS rank, COUNT(DISTINCT age) + 1 AS dense_rank
FROM testdata
WHERE gender = (SELECT gender FROM testdata WHERE id = 6)
AND age < (SELECT age FROM testdata WHERE id = 6)

+------+------------+
| rank | dense_rank |
+------+------------+
|    6 |          5 |
+------+------------+

To calculate RANK() OVER (PARTITION BY Gender ORDER BY Age) for All rows you can use this query:

SELECT testdata.id, COUNT(lesser.id) + 1 AS rank, COUNT(DISTINCT lesser.age) + 1 AS dense_rank
FROM testdata
LEFT JOIN testdata AS lesser ON lesser.age < testdata.age AND lesser.gender = testdata.gender
GROUP BY testdata.id

And here is the result (joined values are added on right):

+------+------+------------+    +-----------+-----+--------+
| id   | rank | dense_rank |    | firstname | age | gender |
+------+------+------------+    +-----------+-----+--------+
|   11 |    1 |          1 |    | Emily     |  20 | F      |
|    3 |    2 |          2 |    | Grace     |  25 | F      |
|   20 |    2 |          2 |    | Jill      |  25 | F      |
|   10 |    4 |          3 |    | Megan     |  26 | F      |
|    8 |    5 |          4 |    | Lucy      |  27 | F      |
|    6 |    6 |          5 |    | Sarah     |  30 | F      |
|    9 |    6 |          5 |    | Zoe       |  30 | F      |
|   14 |    8 |          6 |    | Kate      |  35 | F      |
|    4 |    1 |          1 |    | Harry     |  20 | M      |
|   12 |    1 |          1 |    | Peter     |  20 | M      |
|   13 |    3 |          2 |    | John      |  21 | M      |
|   16 |    4 |          3 |    | Cole      |  25 | M      |
|   17 |    5 |          4 |    | Dennis    |  27 | M      |
|    5 |    6 |          5 |    | Scott     |  30 | M      |
|    7 |    6 |          5 |    | Tony      |  30 | M      |
|    2 |    8 |          6 |    | Matt      |  31 | M      |
|   15 |    9 |          7 |    | James     |  32 | M      |
|    1 |   10 |          8 |    | Adams     |  33 | M      |
|   18 |   11 |          9 |    | Smith     |  35 | M      |
|   19 |   11 |          9 |    | Zack      |  35 | M      |
+------+------+------------+    +-----------+-----+--------+

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别挽留 2024-09-18 05:54:26

如果你只想对一个人进行排名，你可以执行以下操作：

SELECT COUNT(Age) + 1
 FROM PERSON
WHERE(Age < age_to_rank)

此排名对应于 oracle RANK 函数（如果你有相同年龄的人，他们会获得相同的排名，并且之后的排名是不连续的）。

它比在子查询中使用上述解决方案之一并从中进行选择以获得一个人的排名要快一点。

这可以用来对每个人进行排名，但它比上述解决方案慢。

SELECT
  Age AS age_var,
(
  SELECT COUNT(Age) + 1
  FROM Person
  WHERE (Age < age_var)
 ) AS rank
 FROM Person

If you want to rank just one person you can do the following:

SELECT COUNT(Age) + 1
 FROM PERSON
WHERE(Age < age_to_rank)

This ranking corresponds to the oracle RANK function (Where if you have people with the same age they get the same rank, and the ranking after that is non-consecutive).

It's a little bit faster than using one of the above solutions in a subquery and selecting from that to get the ranking of one person.

This can be used to rank everyone but it's slower than the above solutions.

SELECT
  Age AS age_var,
(
  SELECT COUNT(Age) + 1
  FROM Person
  WHERE (Age < age_var)
 ) AS rank
 FROM Person

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め可乐爱微笑 2024-09-18 05:54:26

为了避免 Erandac 的答案中结合 Daniel 和 Salman 的答案中的“however”，可以使用以下“分区”之一解决方法”

SELECT customerID, myDate

  -- partition ranking works only with CTE / from MySQL 8.0 on
  , RANK() OVER (PARTITION BY customerID ORDER BY dateFrom) AS rank, 

  -- Erandac's method in combination of Daniel's and Salman's
  -- count all items in sequence, maximum reaches row count.
  , IF(customerID=@_lastRank, @_curRank:=@_curRank, @_curRank:=@_sequence+1) AS sequenceRank
  , @_sequence:=@_sequence+1 as sequenceOverAll

  -- Dense partition ranking, works also with MySQL 5.7
  -- remember to set offset values in from clause
  , IF(customerID=@_lastRank, @_nxtRank:=@_nxtRank, @_nxtRank:=@_nxtRank+1 ) AS partitionRank
  , IF(customerID=@_lastRank, @_overPart:=@_overPart+1, @_overPart:=1 ) AS partitionSequence

  , @_lastRank:=customerID
FROM myCustomers, 
  (SELECT @_curRank:=0, @_sequence:=0, @_lastRank:=0, @_nxtRank:=0, @_overPart:=0 ) r
ORDER BY customerID, myDate

此代码片段中第三个变体中的分区排名将返回连续的排名数字。这将产生类似于 rank() over partition by 结果的数据结构。作为示例，请参见下文。特别是，对于每个新的partitionRank，partitionSequence将始终从1开始，使用以下方法：

customerID    myDate   sequenceRank (Erandac)
                          |    sequenceOverAll
                          |     |   partitionRank
                          |     |     | partitionSequence
                          |     |     |    | lastRank
... lines ommitted for clarity
40    09.11.2016 11:19    1     44    1   44    40
40    09.12.2016 12:08    1     45    1   45    40
40    09.12.2016 12:08    1     46    1   46    40
40    09.12.2016 12:11    1     47    1   47    40
40    09.12.2016 12:12    1     48    1   48    40
40    13.10.2017 16:31    1     49    1   49    40
40    15.10.2017 11:00    1     50    1   50    40
76    01.07.2015 00:24    51    51    2    1    76
77    04.08.2014 13:35    52    52    3    1    77
79    15.04.2015 20:25    53    53    4    1    79
79    24.04.2018 11:44    53    54    4    2    79
79    08.10.2018 17:37    53    55    4    3    79
117   09.07.2014 18:21    56    56    5    1   117
119   26.06.2014 13:55    57    57    6    1   119
119   02.03.2015 10:23    57    58    6    2   119
119   12.10.2015 10:16    57    59    6    3   119
119   08.04.2016 09:32    57    60    6    4   119
119   05.10.2016 12:41    57    61    6    5   119
119   05.10.2016 12:42    57    62    6    6   119
...

To avoid the "however" in Erandac's answer in combination of Daniel's and Salman's answers, one may use one of the following "partition workarounds"

SELECT customerID, myDate

  -- partition ranking works only with CTE / from MySQL 8.0 on
  , RANK() OVER (PARTITION BY customerID ORDER BY dateFrom) AS rank, 

  -- Erandac's method in combination of Daniel's and Salman's
  -- count all items in sequence, maximum reaches row count.
  , IF(customerID=@_lastRank, @_curRank:=@_curRank, @_curRank:=@_sequence+1) AS sequenceRank
  , @_sequence:=@_sequence+1 as sequenceOverAll

  -- Dense partition ranking, works also with MySQL 5.7
  -- remember to set offset values in from clause
  , IF(customerID=@_lastRank, @_nxtRank:=@_nxtRank, @_nxtRank:=@_nxtRank+1 ) AS partitionRank
  , IF(customerID=@_lastRank, @_overPart:=@_overPart+1, @_overPart:=1 ) AS partitionSequence

  , @_lastRank:=customerID
FROM myCustomers, 
  (SELECT @_curRank:=0, @_sequence:=0, @_lastRank:=0, @_nxtRank:=0, @_overPart:=0 ) r
ORDER BY customerID, myDate

The partition ranking in the 3rd variant in this code snippet will return continous ranking numbers. this will lead to a data structur similar to the rank() over partition by result. As an example, see below. In particular, the partitionSequence will always start with 1 for each new partitionRank, using this method:

customerID    myDate   sequenceRank (Erandac)
                          |    sequenceOverAll
                          |     |   partitionRank
                          |     |     | partitionSequence
                          |     |     |    | lastRank
... lines ommitted for clarity
40    09.11.2016 11:19    1     44    1   44    40
40    09.12.2016 12:08    1     45    1   45    40
40    09.12.2016 12:08    1     46    1   46    40
40    09.12.2016 12:11    1     47    1   47    40
40    09.12.2016 12:12    1     48    1   48    40
40    13.10.2017 16:31    1     49    1   49    40
40    15.10.2017 11:00    1     50    1   50    40
76    01.07.2015 00:24    51    51    2    1    76
77    04.08.2014 13:35    52    52    3    1    77
79    15.04.2015 20:25    53    53    4    1    79
79    24.04.2018 11:44    53    54    4    2    79
79    08.10.2018 17:37    53    55    4    3    79
117   09.07.2014 18:21    56    56    5    1   117
119   26.06.2014 13:55    57    57    6    1   119
119   02.03.2015 10:23    57    58    6    2   119
119   12.10.2015 10:16    57    59    6    3   119
119   08.04.2016 09:32    57    60    6    4   119
119   05.10.2016 12:41    57    61    6    5   119
119   05.10.2016 12:42    57    62    6    6   119
...

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萌无敌 2024-09-18 05:54:26

select id,first_name,gender,age,
rank() over(partition by gender order by age) rank_g
from person

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');
INSERT INTO person VALUES (9,'AKSH',32,'M');

select id,first_name,gender,age,
rank() over(partition by gender order by age) rank_g
from person

CREATE TABLE person (id int, first_name varchar(20), age int, gender char(1));

INSERT INTO person VALUES (1, 'Bob', 25, 'M');
INSERT INTO person VALUES (2, 'Jane', 20, 'F');
INSERT INTO person VALUES (3, 'Jack', 30, 'M');
INSERT INTO person VALUES (4, 'Bill', 32, 'M');
INSERT INTO person VALUES (5, 'Nick', 22, 'M');
INSERT INTO person VALUES (6, 'Kathy', 18, 'F');
INSERT INTO person VALUES (7, 'Steve', 36, 'M');
INSERT INTO person VALUES (8, 'Anne', 25, 'F');
INSERT INTO person VALUES (9,'AKSH',32,'M');

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伤感在游骋 2024-09-18 05:54:26

SELECT FirstName,Age,Gender, RANK() OVER (partition by Gender order by Age desc) AS 'Partition by Gender' FROM Person

您可以根据您想要的顺序（升序还是降序）使用 asc/desc

SELECT FirstName,Age,Gender, RANK() OVER (partition by Gender order by Age desc) AS 'Partition by Gender' FROM Person

you can use asc/desc depending on which order you want if whether ascending or descending

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隱形的亼 2024-09-18 05:54:26

对丹尼尔的答案进行调整，
一些 MySQL 版本在 IF 语句中抛出错误

SELECT    first_name,
          age,
          gender,
          (IF(age=@_last_age,@curRank:=@curRank,@curRank:=@_sequence)) AS `rank`,
          @_sequence:=@_sequence+1,@_last_age:=age
FROM      person p, (SELECT @curRank := 1, @_sequence:=1, @_last_age:=0) r
ORDER BY  age;

所以我必须用大括号将 IF 条件测试包装在 Ranking 中

这 anly 计算排名但不计算 Dense_rank

In a tweak to Daniel's answer,
Some MySQL Versions throw errors at the IF statement

SELECT    first_name,
          age,
          gender,
          (IF(age=@_last_age,@curRank:=@curRank,@curRank:=@_sequence)) AS `rank`,
          @_sequence:=@_sequence+1,@_last_age:=age
FROM      person p, (SELECT @curRank := 1, @_sequence:=1, @_last_age:=0) r
ORDER BY  age;

So I had to wrap the IF condition testing in Ranking with braces

This anly calculates ranking but not Dense_rank

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淡水深流 2024-09-18 05:54:26

set @insert_rank = -2;

SELECT c_updated_time,
@insert_rank := IF(true = true, @insert_rank + 1, 1) 
from my_table limit 5;

该行将打印上面的选择查询；

select @v_query1;

show variables like '%read_only%';
SHOW GRANTS FOR 'USERNAME'@'localhost';

1660616f-f908-4d7e-ab60-c00b77adaf96

select count(*) from INFORMATION_SCHEMA.COLUMNS;
SHOW DATABASES WHERE `Database` IN ('information_schema', 'mysql', 'performance_schema');
use INFORMATION_SCHEMA;
show tables;
desc INFORMATION_SCHEMA.COLUMNS;
SELECT TABLE_SCHEMA, TABLE_NAME, COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME LIKE '%columnname%' and TABLE_SCHEMA like '%Database%' ORDER BY TABLE_NAME ,COLUMN_NAME;
SELECT TABLE_SCHEMA, TABLE_NAME, COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME = 'TableName' and TABLE_SCHEMA <> 'test' ORDER BY TABLE_NAME ,COLUMN_NAME;

对于日期列

SELECT CURRENT_DATE() AS current_date;

对于时间戳列

select now();

set @insert_rank = -2;

SELECT c_updated_time,
@insert_rank := IF(true = true, @insert_rank + 1, 1) 
from my_table limit 5;

This line will print the select above query;

select @v_query1;

show variables like '%read_only%';
SHOW GRANTS FOR 'USERNAME'@'localhost';

1660616f-f908-4d7e-ab60-c00b77adaf96

select count(*) from INFORMATION_SCHEMA.COLUMNS;
SHOW DATABASES WHERE `Database` IN ('information_schema', 'mysql', 'performance_schema');
use INFORMATION_SCHEMA;
show tables;
desc INFORMATION_SCHEMA.COLUMNS;
SELECT TABLE_SCHEMA, TABLE_NAME, COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME LIKE '%columnname%' and TABLE_SCHEMA like '%Database%' ORDER BY TABLE_NAME ,COLUMN_NAME;
SELECT TABLE_SCHEMA, TABLE_NAME, COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME = 'TableName' and TABLE_SCHEMA <> 'test' ORDER BY TABLE_NAME ,COLUMN_NAME;

For Date column