Question

我的表ITEM包含以下列：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"

我需要获得具有以下逻辑的记录：如果HASH列相同则只返回一条记录。 “PREPARING”优先于“BOOKMARKED”，“BOOKMARKED”优先于“NEW”。

因此查询结果将返回：

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

这是有效的，您可以在案例结构中更改优先级：

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
  )

MySQL根据优先级逻辑选择唯一记录

2 个答案: