SQL练习：SQL61_较难

题目： 对于employees表中，输出first_name排名(按first_name升序排序)为奇数的first_name

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');

输出格式:

因为Georgi按first_name排名为3，Anneke按first_name排名为1，所以会输出这2个，且输出时不需排序。

first_name
Georgi
Anneke

题解：

# 方式1：WHERE+子查询+COUNT
SELECT A.first_name
FROM employees A
WHERE (
    SELECT COUNT(*) # COUNT为奇数，则此first_name排序为奇
    FROM employees B
    WHERE A.first_name >= B.first_name) & 1 # 计算所在字段B.first_name的排序（>=的字段数）

# 方式2：FROM+双重子查询+COUNT
SELECT e1.first_name
FROM (SELECT e2.first_name,(SELECT COUNT(*)
                            FROM employees AS e3
                            WHERE e3.first_name <= e2.first_name) AS rowid
      FROM employees AS e2) AS e1
WHERE e1.rowid & 1

# 方式3：FROM+子查询+ROW_NUMBER
SELECT A.first_name
FROM employees A,(SELECT first_name, ROW_NUMBER() OVER(ORDER BY first_name) AS r_num
      FROM employees) AS B
WHERE B.r_num & 1 AND A.first_name = B.first_name

## 求奇数：按位与【& 1】，或求余【% 2 = 1】。不过当然位运算比较快啦。
## 求排序：聚合函数COUNT，或窗口函数ROW_NUMBER

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• 题目来源于：牛客网