Oracle 查找当前月份和上个月值的总和

我有一个包含每个月员工帐户详细信息的源表，日期为字符串类型（yyyyMMdd）。尝试查找每个帐户的当前月值和上个月值之和

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中运行下面的查询，但这将返回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 ;

预期产出：

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

---

我已经使用了下面的查询，但是如果上个月的数据不可用，它将返回上个月的asset_

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月份的asset_prev应为0，但查询返回500（即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                    |
+-----------+-------------+--------------------+----------------------+

预期产出：

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

---

您可以在内部查询中使用聚合，在外部查询中使用

LAG（）

在

account

分区中获取上个月的值。

LAG（）

的三个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

不是列名的好选择，因为它可能与保留字冲突。我在查询中将该列重命名为

adate

。

非常感谢您的解决方案。它返回了实际数据的正确结果。欢迎@asarf！很高兴为您提供帮助。如果我们在所有月份都有可用的帐户，查询就可以了。我有一个场景，我在20190531没有账户“123”，但它在20190630和20190430。因此，在“20190630”月份，其返回的“20190430”月份数据为上月值。在这种情况下，它应该返回“0”，因为在5月份，帐户123不存在。请帮助更新此查询。请帮助更新此查询。从unix时间戳（unix时间戳）到日期（最后一天）（添加月数（到时间戳（数据作为日期的日期，'yyyyMMdd'），-1）），'yyyyy-MM-dd'），'yyyyyyymmdd'）的

的原因是什么，为什么要乱用日期和unix时间戳？
数据作为日期的数据类型是什么？