mysql - SQL 交换行值

Question

这是来自 leetcode 的交换席位问题.

老师想给相邻的学生换座位。 输入:

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Abbot   |
|    2    | Doris   |
|    3    | Emerson |
|    4    | Green   |
|    5    | Jeames  |
+---------+---------+

输出:

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Doris   |
|    2    | Abbot   |
|    3    | Green   |
|    4    | Emerson |
|    5    | Jeames  |
+---------+---------+

给出的解决方案:

SELECT
    (CASE
        WHEN MOD(id, 2) != 0 AND counts != id THEN id + 1
        WHEN MOD(id, 2) != 0 AND counts = id THEN id
        ELSE id - 1
    END) AS id,
    student
FROM
    seat,
    (SELECT
        COUNT(*) AS counts
    FROM
        seat) AS seat_counts
 ORDER BY id ASC;

我的解决方案:

select 
(Case 
    when mod(id,2) !=0 and id != count(*) then id + 1
    when mod(id,2) !=0 and id = count(*) then id
    else id - 1
 end) as 'id',
 student
from seat
order by id;

给定的解决方案效果很好，但我的只能得到一行输出。

我的输出:

+---------+---------+
|    id   | student |
+---------+---------+
|    2    | Abbot   |
+---------+---------+

谁能解释一下我的解决方案和给定的解决方案之间有什么区别以及为什么我的解决方案是错误的？ 谢谢。

Answer 1

当直接在查询中使用 count() 时，您将始终只有一行。 另一种解决方案是通过子查询使用 count() 作为整个查询的值。

尝试

SELECT id, count(*) FROM seat

和

SELECT id, (SELECT COUNT(*) FROM seat) FROM seat

并观察差异。