sql - Oracle 数据密集化填补空白

Question

试图填补表格上的空白以用于报告目的，该表格仅在表格上的值更改时才更新。

代码：

WITH 
week_list AS --List of weeks
 (
  SELECT (  (trunc(to_date('06152012','mmddyyyy'), 'IW')+7) - (LEVEL * 7) )+6 as week 
      from dual
      connect by level <= 7
  order by ( (trunc(to_date('06152012','mmddyyyy'), 'IW')+7) - (LEVEL * 7) )+6
 )  
SELECT 
  product, 
  week,
  undist_amt_eod as quantity,
  LAST_VALUE(undist_amt_eod IGNORE NULLS) OVER (PARTITION BY product ORDER BY week) repeated_quantity
FROM 
 (
    SELECT 
      product, 
      week_list.week, 
      inv_table.undist_amt_eod
    FROM 
      inv_table PARTITION BY (product)
        RIGHT OUTER JOIN week_list ON (week_list.week = inv_table.history_date)
    where 
      inv_table.tab = '*' --Includes all types of this product
  )
ORDER BY product, week;

周列表示例输出：

表格内容：请注意，表格每天可以有多个选项卡。* 是当天所有选项卡的总和，所以我只对 * 值感兴趣。

我基于此处找到的 oracle 示例中的代码。不确定为什么数据输出仍然不密集。

Answer 1

问题是您的 WHERE 子句：

where inv_table.tab = '*' --Includes all types of this product

因为您正在执行右外连接，所以 inv_table.tab 将在没有匹配项时为 NULL。将其更改为以下内容之一：

where inv_table.tab = '*' or inv_table.history_date is null --Includes all types of this product

或者，如果 inv_table.tab 永远不能为 NULL，则：

where coalesce(inv_table.tab, '*') = '*'