如何对mysql数据进行反向排序

Question

An odd title I know, but I'm stumped. Say if I grab 20 rows like this:

SELECT * FROM `articles` WHERE date >= UNIX_TIMESTAMP(DATE(NOW() - INTERVAL 30 DAY)) ORDER BY `views` DESC LIMIT 20

I then want to show them in reverse, from lowest to highest "views", how can I do this? I cannot simply order by "ASC" as that gives me the wrong result set. I want the highest views and to then order those from lowest to highest.

Answer 1

taking Zerkms's idea:

Select * FROM (
    SELECT * FROM `articles` 
    WHERE date >= UNIX_TIMESTAMP(DATE(NOW() - INTERVAL 30 DAY))
    ORDER BY `views` DESC 
    LIMIT 20 
) as reverse_article 
ORDER BY views ASC

will let you double sort

1. The highest will get picked (inner select)

2. The lowest will get picked off first.

HOWEVER

you can also reverse sort via PHP by first getting your fetchALL and running the count backward from count() (which is your max size array) down to 0 (which is your highest array view)

take your pick

Answer 2

You use a subquery for this. The subquery chooses the right 20 rows, the outer query then sorts them according to your final criteria:

SELECT a.*
FROM (SELECT a.*
      FROM articles a
      WHERE date >= UNIX_TIMESTAMP(DATE(NOW() - INTERVAL 30 DAY))
      ORDER BY views DESC
      LIMIT 20
     ) a
ORDER BY views ASC;

Answer 3

这应该工作

SELECT * FROM (SELECT * FROM `articles` WHERE date >= UNIX_TIMESTAMP(DATE(NOW() - INTERVAL 30 DAY)) ORDER BY `views` DESC LIMIT 20) t ORDER BY `views` ASC

Answer 4

MySQL查询：

(SELECT * FROM `articles` WHERE date >= UNIX_TIMESTAMP(DATE(NOW() - INTERVAL 30 DAY)) ORDER BY `views` DESC LIMIT 20) ORDER by `views` ASC;