Rank function in MySQL

Question

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?

Answer 1

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
...

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Here is a generic solution that sorts a table based on a column and assigns rank; rows with ties are assigned same rank (uses an extra variable for this purpose):

SET @prev_value = NULL;
SET @rank_count = 0;
SELECT id, rank_column, CASE
    WHEN @prev_value = rank_column THEN @rank_count
    WHEN @prev_value := rank_column THEN @rank_count := @rank_count + 1
END AS rank
FROM rank_table
ORDER BY rank_column

Note that there are two assignment statements in the second WHEN clause. Sample data:

CREATE TABLE rank_table(id INT, rank_column INT);
INSERT INTO rank_table (id, rank_column) VALUES
(1, 10),
(2, 20),
(3, 30),
(4, 30),
(5, 30),
(6, 40),
(7, 50),
(8, 50),
(9, 50);

Output:

+------+-------------+------+
| id   | rank_column | rank |
+------+-------------+------+
|    1 |          10 |    1 |
|    2 |          20 |    2 |
|    3 |          30 |    3 |
|    4 |          30 |    3 |
|    5 |          30 |    3 |
|    6 |          40 |    4 |
|    7 |          50 |    5 |
|    8 |          50 |    5 |
|    9 |          50 |    5 |
+------+-------------+------+

SQL Fiddle