Oracle:我需要从表的每k行中选择n行 [英] Oracle: I need to select n rows from every k rows of a table

问题描述

例如: 我的表有10000行.首先，我将其分为5组2000(k)行.然后，从每组2000行中，我将仅选择前100(n)行. 通过这种方法，我尝试使用特定模式扫描表的某些行.

For example: My table has 10000 rows. First I will divide it in 5 sets of 2000(k) rows. Then from each set of 2000 rows I will select only top 100(n) rows. With this approach I am trying to scan some rows of table with a specific pattern.

推荐答案

假设您使用某种逻辑对它们1 - 10000进行排序，并且只想输出1-100,2001-2100,4001-4100,etc行，则可以使用ROWNUM伪列:

Assuming you are ordering them 1 - 10000 using some logic and want to output only rows 1-100,2001-2100,4001-4100,etc then you can use the ROWNUM pseudocolumn:

SELECT *
FROM   (
  SELECT t.*,
         ROWNUM AS rn            -- Secondly, assign a row number to the ordered rows
  FROM   (
    SELECT *
    FROM   your_table
    ORDER BY your_condition      -- First, order the data
  ) t
)
WHERE MOD( rn - 1, 2000 ) < 100; -- Finally, filter the top 100 per 2000.

或者您可以使用ROW_NUMBER()分析功能:

Or you could use the ROW_NUMBER() analytic function:

SELECT *
FROM   (
  SELECT t.*,
         ROW_NUMBER() OVER ( ORDER BY your_condition ) AS rn
  FROM   your_table
)
WHERE  MOD( rn - 1, 2000 ) < 100;

是否有可能以指数方式增加样本数据集.像1k，2k，4k，8k ....然后从中获取一些行.

Is it possible to increase the set of sample data exponentially. Like 1k, 2k, 4k,8k....and then fetch some rows from these.

将WHERE子句替换为:

WHERE rn - POWER(
             2,
             TRUNC( CAST( LOG( 2, CEIL( rn / 1000 ) ) AS NUMBER(20,4) ) )
           ) * 1000 + 1000 <= 100

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