Approach I: Using flow control statement
For students with odd id, the new id is (id+1) after switch unless it is the last seat. And for students with even id, the new id is (id-1). In order to know how many seats in total, we can use a subquery:
SELECT COUNT(*) AS counts FROM seat
Then, we can use the
CASE statement and
MOD() function to alter the seat id of each student.
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;
Approach II: Using bit manipulation and
Bit manipulation expression
(id+1)^1-1 can calculate the new id after switch.
SELECT id, (id+1)^1-1, student FROM seat;
| id | (id+1)^1-1 | student | |----|------------|---------| | 1 | 2 | Abbot | | 2 | 1 | Doris | | 3 | 4 | Emerson | | 4 | 3 | Green | | 5 | 6 | Jeames |
Then, we can make a temp table and join seat with this table like below.
SELECT * FROM seat s1 LEFT JOIN seat s2 ON (s1.id+1)^1-1 = s2.id ORDER BY s1.id;
| id | student | id | student | |----|---------|----|---------| | 1 | Abbot | 2 | Doris | | 2 | Doris | 1 | Abbot | | 3 | Emerson | 4 | Green | | 4 | Green | 3 | Emerson | | 5 | Jeames | | |
Note:The first two columns are from s1 and the last two are from s2.
At last, we can output s1.id and s2.student. However, the s2.student is NULL for seat id '5' but s1.student is right. Thus, we we can use function
COALESCE() to generate the correct output for the last record.
SELECT s1.id, COALESCE(s2.student, s1.student) AS student FROM seat s1 LEFT JOIN seat s2 ON ((s1.id + 1) ^ 1) - 1 = s2.id ORDER BY s1.id;
Note: This solution comes from @FANGXIAOFANG.