oracle - 查找当前月份和上个月值的总和

Question

我有一个源表，其中包含每个月的员工帐户详细信息，日期是字符串类型(yyyyMMdd)。尝试查找每个帐户当前月份值和上个月值的总和。

Source data:

+-----------+-------------+-----------+----------+
|  date     | account     | division  |  amount  |
+-----------+-------------+-----------+----------+
| 20190331  | 123         | AB0       | 100      |
+-----------+-------------+-----------+----------+
| 20190331  | 123         | AB1       | 110      |
+-----------+-------------+-----------+----------+
| 20190331  | 123         | AB2       | 120      |
+-----------+-------------+-----------+----------+
| 20190228  | 123         | AB4       | 100      |
+-----------+-------------+-----------+----------+
| 20190228  | 123         | AB1       | 100      |
+-----------+-------------+-----------+----------+
| 20190228  | 123         | AB2       | 100      |
+-----------+-------------+-----------+----------+
| 20190131  | 123         | AB0       | 100      |
+-----------+-------------+-----------+----------+

在 impala 中运行以下查询，但这返回了当前和上个月相同的结果。

select distinct * from (
SELECT 
sum(amount) over (partition BY account, a.date) AS asset_current,
sum(amount) over (partition BY account, from_unixtime(unix_timestamp(to_date(LAST_DAY(ADD_MONTHS(to_timestamp(data_as_of_date,'yyyyMMdd'),-1))),'yyyy-MM-dd'),'yyyyMMdd')) AS asset_previous,
     account,
     date,
FROM employee_assets a
)x ;

预期输出:

+-----------+-------------+--------------------+----------------------+
|  date     | account     | current_month_sum  |  previous_month_sum  |
+-----------+-------------+--------------------+----------------------+
| 20190331  | 123         | 330                | 300                  |
+-----------+-------------+--------------------+----------------------+
| 20190228  | 123         | 300                | 100                  |
+-----------+-------------+--------------------+----------------------+
| 20190131  | 123         | 100                | 0                    |
+-----------+-------------+--------------------+----------------------+

<小时/>

我使用了以下查询，但如果上个月的数据不可用，它会返回 asset_previous 作为上个月的值。

SELECT
    x.*,
    LAG(current_month_sum, 1, 0) OVER(PARTITION BY account ORDER BY adate) previous_month_sum  
FROM (
    SELECT adate, account, SUM(amount) current_month_sum  
    FROM employee_assets
    GROUP BY adate, account
) x
ORDER BY adate DESC

例如:我们没有账户 123 的 20181231 的输入数据，因此 1 月的 asset_prev 应为 0，但查询返回 500(这是 2018 年 11 月的金额) 输入数据:

+-----------+-------------+-----------+----------+
|  date     | account     | division  |  amount  |
+-----------+-------------+-----------+----------+
| 20190331  | 123         | AB0       | 100      |
+-----------+-------------+-----------+----------+
| 20190331  | 123         | AB1       | 110      |
+-----------+-------------+-----------+----------+
| 20190331  | 123         | AB2       | 120      |
+-----------+-------------+-----------+----------+
| 20190228  | 123         | AB4       | 100      |
+-----------+-------------+-----------+----------+
| 20190228  | 123         | AB1       | 100      |
+-----------+-------------+-----------+----------+
| 20190228  | 123         | AB2       | 100      |
+-----------+-------------+-----------+----------+
| 20190131  | 123         | AB0       | 100      |
+-----------+-------------+-----------+----------+
| 20181130  | 123         | ABX       | 500      |
+-----------+-------------+-----------+----------+

查询正在返回:

+-----------+-------------+--------------------+----------------------+
|  date     | account     | current_month_sum  |  previous_month_sum  |
+-----------+-------------+--------------------+----------------------+
| 20190331  | 123         | 330                | 300                  |
+-----------+-------------+--------------------+----------------------+
| 20190228  | 123         | 300                | 100                  |
+-----------+-------------+--------------------+----------------------+
| 20190131  | 123         | 100                | 500                  |
+-----------+-------------+--------------------+----------------------+
| 20191131  | 123         | 500                | 0                    |
+-----------+-------------+--------------------+----------------------+

预期输出:

+-----------+-------------+--------------------+----------------------+
|  date     | account     | current_month_sum  |  previous_month_sum  |
+-----------+-------------+--------------------+----------------------+
| 20190331  | 123         | 330                | 300                  |
+-----------+-------------+--------------------+----------------------+
| 20190228  | 123         | 300                | 100                  |
+-----------+-------------+--------------------+----------------------+
| 20190131  | 123         | 100                | 0                    |
+-----------+-------------+--------------------+----------------------+
| 20191131  | 123         | 500                | 0                    |
+-----------+-------------+--------------------+----------------------+

Answer 1

您可以在内部查询中使用聚合，并在外部查询中使用 LAG() 来获取 account 分区中上个月的值。 LAG() 的三参数形式允许您指定默认值。

SELECT
    x.*,
    LAG(current_month_sum, 1, 0) OVER(PARTITION BY account ORDER BY adate) previous_month_sum  
FROM (
    SELECT adate, account, SUM(amount) current_month_sum  
    FROM employee_assets
    GROUP BY adate, account
) x
ORDER BY adate DESC

注意:date 对于列名来说不是一个好的选择，因为它可能与保留字冲突。我在查询中将该列重命名为 date。