Mysql按if条件排序

编程入门行业动态更新时间:2024-10-27 06:31:04

Mysql按if条件排序 - 错误的顺序(Mysql order by if condition - wrong order)

我有一张简单的桌子：

CREATE TABLE `dev_a4a`.`test` ( `id` INT UNSIGNED NOT NULL AUTO_INCREMENT , `type` VARCHAR(45) NOT NULL , `priority` INT NOT NULL , PRIMARY KEY (`id`) );

以下行：

id|type |priority 1 |long |8 2 |long |3 3 |short|9 4 |short|1

我想获得按条件排序的行：

SELECT (RAND() * priority) as prio, type, priority FROM test ORDER BY (IF(type = 'short', '2', prio)) DESC, id DESC

结果我获得了不按if条件排序的行。每次看起来都是随机的。这是可能的结果之一：

prio | type | priority '0.05013570194145264', 'long', '8' '2.9015473750434326', 'long', '3' '0.320064320527077', 'short', '1' '7.598900996706356', 'short', '9'

我究竟做错了什么？

预期结果：

prio | type | priority '2.9015473750434326', 'long', '3' <- order by prio '7.598900996706356', 'short', '9' <- order by common value 2 '0.320064320527077', 'short', '1' <- order by common value 2 '0.05013570194145264', 'long', '8' <- order by prio

I got a simple table:

CREATE TABLE `dev_a4a`.`test` ( `id` INT UNSIGNED NOT NULL AUTO_INCREMENT , `type` VARCHAR(45) NOT NULL , `priority` INT NOT NULL , PRIMARY KEY (`id`) );

with following rows:

id|type |priority 1 |long |8 2 |long |3 3 |short|9 4 |short|1

I want to get rows ordered by condition:

SELECT (RAND() * priority) as prio, type, priority FROM test ORDER BY (IF(type = 'short', '2', prio)) DESC, id DESC

As a result I got rows not ordered by if condition. Every time it looks like it's random. Here's one of possible results:

prio | type | priority '0.05013570194145264', 'long', '8' '2.9015473750434326', 'long', '3' '0.320064320527077', 'short', '1' '7.598900996706356', 'short', '9'

What am I doing wrong?

Expected result: