我有一个表，其中两列[id，value]都是数字。
在这个例子中：
[ id, value ]
[ 1, 6 ]
[ 2, 4 ]
[ 3, 10 ]
[ 4, 2 ]
[ 5, 7 ]
[ 6, 3 ]
对于给定的ID，我想要检索前3个id（具有最高值的id），它们的最高位置，如果给定的id不在前3个，也得到它的位置，id和值：
示例1：ask_id = 5返回：
[ position, id, value ]
[ 1, 3, 10 ]
[ 2, 5, 7 ]
[ 3, 1, 6 ]
示例2：ask_id = 4.返回：
[ position, id, value ]
[ 1, 3, 10 ]
[ 2, 5, 7 ]
[ 3, 1, 6 ]
[ 6, 4, 2 ]
所以重点是：
答案 0 :(得分：2)
select t2.pos, t1.id, t1.value
from test as t1
inner join
(select id, value, @pos:=if(@pos is null, 0, @pos)+1 as pos
from test order by value desc) as t2
on t1.id=t2.id
where t2.pos<=3 or t2.id={$ask_id}
order by t2.pos;
答案 1 :(得分：1)
在MySQL中测试过 检索前3个id（具有最高值的id），其位置按升序排列。
set @num = 0;
SELECT @num := @num + 1 as position_sequence,id,value FROM tablename
ORDER BY value desc
limit 3;
答案 2 :(得分：1)
基本上，这个想法是这样的：
按value对行进行排名。
检索至少满足下列条件之一的行：
position BETWEEN 1 AND 3
id = @given_id
这些帖子举例说明了如何在MySQL中替换排名函数（至少是其中最基本的函数ROW_NUMBER()）：
这种方法应该谨慎使用，正如this article所解释的那样。
尽管如此，上述步骤的一种可能实现可能如下所示：
SET @pos = 0;
SELECT
position,
id,
value
FROM (
SELECT
id,
value,
@pos := @pos + 1 AS position
FROM atable
ORDER BY value DESC
) s
WHERE position BETWEEN 1 AND 3
OR id = @given_id
ORDER BY position
答案 3 :(得分：0)
我还没有（还）测试MySQL中选定答案的前三位有关系的有趣案例，但是我已经在Informix中对这些案例进行了测试，它产生了我认为应该的答案生产。
假设该表名为leader_board：
CREATE TABLE leader_board(id INTEGER NOT NULL PRIMARY KEY, value INTEGER NOT NULL);
INSERT INTO leader_board(id, value) VALUES(1, 6);
INSERT INTO leader_board(id, value) VALUES(2, 4);
INSERT INTO leader_board(id, value) VALUES(3, 10);
INSERT INTO leader_board(id, value) VALUES(4, 2);
INSERT INTO leader_board(id, value) VALUES(5, 7);
INSERT INTO leader_board(id, value) VALUES(6, 3);
此查询适用于显示的数据，假设特殊ID为4：
SELECT b.position - c.tied + 1 AS standing, a.id, a.value
FROM leader_board AS a
JOIN (SELECT COUNT(*) AS position, d.id
FROM leader_board AS d
JOIN leader_board AS e ON (d.value <= e.value)
GROUP BY d.id
) AS b
ON a.id = b.id
JOIN (SELECT COUNT(*) AS tied, f.id
FROM leader_board AS f
JOIN leader_board AS g ON (f.value = g.value)
GROUP BY f.id
) AS c
ON a.id = c.id
WHERE (a.id = 4 OR (b.position - c.tied + 1) <= 3) -- Special ID = 4; Top N = 3
ORDER BY position, a.id;
原始数据输出：
standing id value
1 3 10
2 5 7
3 1 6
6 4 2
这两个子查询密切相关，但它们产生不同的答案。有一段时间，我使用了两个临时表来保存这些结果。特别是，第一个子查询（AS b）产生一个位置，但是当存在联系时，该位置是最低位置而不是最高位置。那就是：
ID Value
1 10
2 7
3 7
4 7
输出将是：
Position ID
1 1
4 2
4 3
4 4
但是，我们希望将它们视为：
Position ID
1 1
2 2
2 3
2 4
因此，更正的位置是原始位置减去绑定值的数量（ID∈{2,3,4}为3，ID为1为1）加1.第二个子查询返回绑定值的数量对于每个ID。可能有一种更简洁的方法来进行计算，但我不确定它目前是什么。
但是，代码应该证明它处理以下情况：
为了每次都保存重写查询，我将其转换为Informix样式的存储过程，该过程同时接收应显示的特殊ID和前N（默认为3）值，并将它们作为过程的参数。 （是的，RETURNING子句中的符号很奇怪。）
CREATE PROCEDURE leader_board_standings(extra_id INTEGER, top_n INTEGER DEFAULT 3)
RETURNING INTEGER AS standing, INTEGER AS id, INTEGER AS value;
DEFINE standing, id, value INTEGER;
FOREACH SELECT b.position - c.tied + 1 AS standing, a.id, a.value
INTO standing, id, value
FROM leader_board AS a
JOIN (SELECT COUNT(*) AS position, d.id
FROM leader_board AS d
JOIN leader_board AS e ON (d.value <= e.value)
GROUP BY d.id
) AS b
ON a.id = b.id
JOIN (SELECT COUNT(*) AS tied, f.id
FROM leader_board AS f
JOIN leader_board AS g ON (f.value = g.value)
GROUP BY f.id
) AS c
ON a.id = c.id
WHERE (a.id = extra_id OR (b.position - c.tied + 1) <= top_n)
ORDER BY position, a.id
RETURN standing, id, value WITH RESUME;
END FOREACH;
END PROCEDURE;
可以调用它来产生与之前相同的结果：
EXECUTE PROCEDURE leader_board_standings(4);
为了说明上面列出的各种情况，请添加和删除额外的行：
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
6 4 2
INSERT INTO leader_board(id, value) VALUES(10, 10);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
1 10 10
3 5 7
7 4 2
INSERT INTO leader_board(id, value) VALUES(11, 10);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
1 10 10
1 11 10
8 4 2
INSERT INTO leader_board(id, value) VALUES(12, 10);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
1 10 10
1 11 10
1 12 10
9 4 2
DELETE FROM leader_board WHERE id IN (10, 11, 12);
EXECUTE PROCEDURE leader_board_standings(6, 4); -- Special ID 6; Top 4
1 3 10
2 5 7
3 1 6
4 2 4
5 6 3
INSERT INTO leader_board(id, value) VALUES(7, 7);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
2 7 7
7 4 2
INSERT INTO leader_board(id, value) VALUES(13, 7);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
2 7 7
2 13 7
8 4 2
INSERT INTO leader_board(id, value) VALUES(14, 7);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
2 7 7
2 13 7
2 14 7
9 4 2
DELETE FROM leader_board WHERE id IN(7, 13, 14);
INSERT INTO leader_board(id, value) VALUES(8, 6);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
3 8 6
7 4 2
INSERT INTO leader_board(id, value) VALUES(9, 6);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
3 8 6
3 9 6
8 4 2
INSERT INTO leader_board(id, value) VALUES(15, 6);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
3 8 6
3 9 6
3 15 6
9 4 2
EXECUTE PROCEDURE leader_board_standings(3); -- Special ID 3 appears in top 3
1 3 10
2 5 7
3 1 6
这一切对我来说都是正确的。