3 回答

TA貢獻1859條經驗 獲得超6個贊
一種選擇是使用排名變量,例如:
SELECT first_name,
age,
gender,
@curRank := @curRank + 1 AS rank
FROM person p, (SELECT @curRank := 0) r
ORDER BY age;
該(SELECT @curRank := 0)部分允許變量初始化而無需單獨的SET命令。
測試用例:
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');
結果:
+------------+------+--------+------+
| 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)

TA貢獻1827條經驗 獲得超9個贊
這是一個通用的解決方案,它將分區上的密集排名分配給行。它使用用戶變量:
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;
請注意,變量賦值放在CASE表達式中。這(理論上)負責評估問題的順序。將IS NOT NULL被添加到處理數據類型轉換和短路的問題。
PS:通過刪除檢查綁定的所有條件,可以輕松地將其轉換為分區上的行號。
| 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 |

TA貢獻1772條經驗 獲得超8個贊
雖然最受歡迎的答案排名,但它沒有分區,你可以自己加入以獲得整個分區的東西:
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
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