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Rank function in MySQL

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sql

mysql

rank

People also ask

Does MySQL have rank function?

The ranking functions in MySql are used to rank each row of a partition. The ranking functions are also part of MySQL windows functions list. These functions are always used with OVER() clause. The ranking functions always assign rank on basis of ORDER BY clause.

What is rank function in SQL?

RANK() Function in SQL Server The RANK() function is a window function could be used in SQL Server to calculate a rank for each row within a partition of a result set. The same rank is assigned to the rows in a partition which have the same values. The rank of the first row is 1.

What is rank function?

The RANK function is an OLAP ranking function that calculates a ranking value for each row in an OLAP window. The return value is an ordinal number, which is based on the required ORDER BY expression in the OVER clause.

What is DENSE_RANK function in MySQL?

The DENSE_RANK() is a window function that assigns a rank to each row within a partition or result set with no gaps in ranking values. The rank of a row is increased by one from the number of distinct rank values which come before the row.


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)

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


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

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 |
+----+-------+------+---------------+---------------+-----------------+

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 |
+------------+------+--------+------+--------------------------+-----------------+