Question

假设我想从名为messages的表中检索100条记录，我想通过以下方式获取它们：

1st message
100th message
2nd message
99th message
3rd message
98th message
(...)

有没有办法有效地？什么是适当的查询？或者我应该查询选择前50个，查询选择最后50个然后合并结果？

Answer 1

如果您的ID是一系列数字，请尝试：

第一

SET @half = (SELECT MAX(id) FROM messages)/2;

然后

SELECT * FROM `messages` ORDER BY (IF(id<@half,@half*2-id,id-1)) DESC,id ASC;

Answer 2

重点是创建两个虚拟列“serie_order”（变体）和“serie”（常量），您将在数据的两个部分上使用（您必须将数据分成两部分）。

SELECT * FROM (
  SELECT 1 as serie, message_id AS serie_order , *  FROM
   (SELECT message_id FROM messages ) as part_up
UNION 
  SELECT 2 as serie, 101-message_id  as serie_order, * FROM
   (SELECT message_id FROM messages) as part_down
) AS world

ORDER BY serie_order ASC, serie ASC
LIMIT 100

Answer 3

set @rank:=0; 

select id from 
(select id, @rank:=(coalesce(@rank, 0)+1) as new_order 
 from a_table 
 order by some_column limit 100) as ordering
order by if (new_order<=50, new_order-1, abs(100-new_order)) asc;

是否可以在SQL中的SELECT中合并？

3 个答案: