Question

我需要有关SQL测试中问题的帮助。

假设我有两个表，即Album（album_id）和DailySells（album_id，date，albums_sold），并加入了他们

select id
from album join daily using(id)
where (date, albsold) in (select date, min(albsold)
    from daily
    group by date
    having min(albsold) = (select max(worstdaily)
                           from (select date, min(albsold) as worstdaily
                                 from daily
                                 group by date) as inner))

问题是：每天显示最畅销专辑中最畅销的专辑。

因此，我需要每天获取销量最差的专辑，然后提取其中销量最好的一张。我看到答案应该显示为1111（日期22，albums_sold 3）。

您能帮我编写一个可以解决此问题的查询吗？

谢谢！

编辑：很抱歉，我为此花了3个多小时的时间，但没有显示代码。这是我一直在尝试的方法：

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var count;

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  console.log(colors[count]);
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function stringLength(string) {
  if (string.length > 4) {
    console.log("greater than 4");
  } else if (string.length < 4) {
    console.log("less than 4");
  } else {
    console.log("is equal to 4");
  }
}

Answer 1

在子查询中，我们仅按日期记录MIN（albums_sold）值的记录，然后按albums_sold DESC排序，并使用TOP（1）取MAX：

SELECT TOP(1) mt.* 
FROM alb mt INNER JOIN
(
    SELECT date, MIN(albums_sold) MinSold
    FROM alb
    GROUP BY date
) t ON mt.date = t.date AND mt.albums_sold = t.MinSold
ORDER BY albums_sold DESC

Answer 2

有几种方法可以解决此问题。无论如何，我们必须牢记可能存在联系（一天中有多个最差的卖家，或者最差的多个卖家），并决定如何处理。我正在考虑所有相关的行，因此结果可能会跨越多行。

这是一些标准的SQL查询。您的DBMS可能支持也可能不支持。

查询1

select album_id
from mytable
where not exists
(
  select *
  from mytable other
  where other.date = mytable.date
  and other.albums_sold < mytable.albums_sold
)
order by albums_sold desc
fetch first 1 row with ties;

查询2

select album_id
from
(
  select album_id, albums_sold, rank() over (partition by date order by albums_sold) as rn
  from mytable other
) ranked
where rn = 1
order by albums_sold desc
fetch first 1 row with ties;

查询3

select album_id
from
(
  select album_id, albums_sold
  from mytable other
  order by rank() over (partition by date order by albums_sold)
  fetch first 1 row with ties
) worst
order by albums_sold desc
fetch first 1 row with ties;

您会看到可以组合使用以下方法：子查询，排名，先获取。

在最小的之中找到最大的

2 个答案: