每列值限制结果

Question

我有一个非常大的数据库，看起来像这样（3米行）

+-----------+----------------+-------+
|  Category |   SubCategory  | Count |
+-----------+----------------+-------+
|  cat1     |   green        | 200   |
|  cat1     |   blue         | 215   |
|  cat1     |   red          | 20    |
|  cat2     |   something    | 540   |
|  cat2     |   chicken      | 100   |
|  cat2     |   dog          | 33    |
|  cat3     |   ball         | 404   |
|  cat3     |   bed          | 777   |
|  cat4     |   shoes        | 72    |
|  cat4     |   green        | 14    |
|  cat4     |   red          | 2     |
|  cat4     |   panda        | 4     |
+-----------+----------------+-------+

如何按Count排序然后获取cat1的前100个结果，然后获取cat2的前100个结果，依此类推。

Answer 1

您可以通过内部联接和分组来执行此操作。

注意：我将count更改为count1。在查询中不使用db保留关键字始终是一种好习惯。

的 SQLFiddle Demo

SELECT category,
       subcategory,
       count1
FROM
  (SELECT t1.category,
          max(t1.subcategory) AS subcategory,
          max(t1.count1) AS count1,
          count(t1.count1) AS rnk
   FROM tbl t1
   INNER JOIN tbl t2 ON t1.category=t2.category
   AND t1.count1<=t2.count1
   GROUP BY t1.category,
            t1.count1) cnt
WHERE rnk <=100

内部查询的工作方式与rank over partition by类似，并rank按count1 desc分组category。它的输出类似于

+----------+-------------+--------+-----+
| category | subcategory | count1 | rnk |
+----------+-------------+--------+-----+
| cat1     | red         |     20 |   3 |
| cat1     | green       |    200 |   2 |
| cat1     | blue        |    215 |   1 |
| cat2     | dog         |     33 |   3 |
| cat2     | chicken     |    100 |   2 |
| cat2     | something   |    540 |   1 |

然后你可以通过外部查询获得前100行。

Answer 2

您需要为每个组建立row_number，然后相应地进行过滤。使用mysql，您需要使用user-defined variables：

select category, subcategory, `count`
from (
    select *, 
         ( 
            case category 
            when @curCategory 
            then @curRow := @curRow + 1 
            else @curRow := 1 AND @curCategory := category end
          ) + 1 AS rank
    from yourtable cross join (select @curRow:=0, @curCategory:='') c
    order by category, `count` desc
) t
where rank <= 100

• Condensed Fiddle Demo