Oracle SQL 计算每个月每周的天数答案

【问题标题】：Oracle SQL count days in week for every monthOracle SQL 计算每个月每周的天数
【发布时间】：2020-06-19 23:17:44
【问题描述】：

我有这两张表（时间和销售额）：

次

TIME_ID   |   DAY_NAME  |  DAY_NUMBER_IN_WEEK    |  CALENDAR_MONTH_NAME |  CALENDAR_MONTH_ID
1998-01-10|   Monday    |  1                     |  January             |  1684
1998-01-10|   Tuesday   |  2                     |  January             |  1684
1998-01-10|   Wednesday |  3                     |  January             |  1684
...
1998-01-11|   Monday    |  1                     |  February            |  1685
1998-01-11|   Tuesday   |  2                     |  January             |  1685
1998-01-11|   Wednesday |  3                     |  January             |  1685

销售

PROD_ID   |  TIME_ID     |  AMOUNT_SOLD
13        |  1998-01-10  |  1232          
13        |  1998-01-11  |  1233 
14        |  1998-01-11  |  1233

我需要为一周中的每一天（星期一、星期二、星期三...）和每个月的每一天的每个 PROD_ID 的 AMOUNT_SOLD 的总和创建列。

SELECT SUM(times.day_number_in_week), times.calendar_month_name, times.day_name, times.calendar_year
FROM sales
INNER JOIN times  ON times.time_id = sales.time_id
GROUP BY times.calendar_month_number, times.calendar_month_name, times.day_name, times.calendar_year

输出：

5988    March   Wednesday   1998
9408    April   Thursday    1998
7532    June    Sunday  1998
9220    July    Thursday    1998
7490    July    Sunday  1998
12540   August  Saturday    1998

但是所有年份的所有星期三的总和，我需要一个月的所有日期（星期三，星期一...）的 1 个月的总和。

你能帮帮我吗？

【问题讨论】：

标签： sql oracle join group-by sum

【解决方案1】：

你可以做条件聚合：

SELECT 
    t.calendar_year
    t.calendar_month_name, 
    SUM(case when t.day_number_in_week = 1 then s.amount_sold else 0 end) amount_sold_mon,
    SUM(case when t.day_number_in_week = 2 then s.amount_sold else 0 end) amount_sold_tue,
    SUM(case when t.day_number_in_week = 3 then s.amount_sold else 0 end) amount_sold_wed,
    ...
FROM sales s
INNER JOIN times t ON t.time_id = sales.time_id
GROUP BY t.calendar_year, t.calendar_month_number, t.calendar_month_name

【讨论】：

SELECT times.calendar_year_id, times.calendar_month_id, SUM(times.day_number_in_week = 1 然后 sales.amount_sold 否则 0 end) amount_sold_mon, SUM(times.day_number_in_week = 2 然后是 sales 的情况.amount_sold else 0 end) amount_sold_tue, SUM(case when times.day_number_in_week = 3 then sales.amount_sold else 0 end) amount_sold_wed FROM sales INNER JOIN times ON times.time_id = sales.time_id GROUP BY times.calendar_year_id, times.calendar_month_id, times .day_number_in_week 工作，但我需要为特定的 prod_id 计算它
@H.Ivanov：只需添加一个where 子句，过滤您想要结果的prod_id。
工作完美，但我需要从 SUM 函数中删除所有 0 我该怎么做？我尝试使用 HAVING 但不工作...你能帮帮我吗？