MySQL Select 基于优先级逻辑的唯一记录

Question

我有 table 个包含以下列的 ITEM：id、status、hash、value。如果我有以下记录：

1, "NEW", 111111, "value1"
2, "BOOKMARKED", 111111, "value2"
3, "PREPARING", 111111, "value3"
4, "NEW", 222222, "value4"
5, "BOOKMARKED", 222222, "value5"
6, "NEW", 333333, "value6"

我需要按照以下逻辑获取记录：如果 HASH 列相同，则 return 只有一条记录。 "PREPARING" 优先于 "BOOKMARKED"，"BOOKMARKED" 优先于 "NEW"。

所以查询结果会 return:

3, "PREPARING", 111111, "value3"
5, "BOOKMARKED", 222222, "value5"
6, "NEW", 333333, "value6"

谢谢。

Answer 1

一种方法是根据规则枚举值，然后取第一个值：

select t.*
from (select t.*,
             (@rn := if(@h = hash, @rn + 1,
                        if(@h := hash, 1, 1)
                       )
             ) as rn
      from t cross join
           (select @rn := 0, @h := '') params
      order by hash,
               field(status, 'PREPARING', 'BOOKMARKED', 'NEW')
     ) t
where rn = 1;

Answer 2

这是可行的，您可以更改案例结构中的优先级：

select  `ITEM`.*
from `ITEM`
inner join 
(
  SELECT `ITEM`.`hash`,
    max(
      CASE `ITEM`.`status` 
        WHEN 'PREPARING' THEN 2
        WHEN 'BOOKMARKED' THEN 1
        ELSE 0 
      END
    ) as `g`
  FROM `ITEM`
  group by `ITEM`.`hash`
) as `t` ON 
  `t`.`hash` = `ITEM`.`hash` AND
  `t`.`g` = (
    CASE `ITEM`.`status` 
        WHEN 'PREPARING' THEN 2
        WHEN 'BOOKMARKED' THEN 1
        ELSE 0 
    END
  )

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