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）。尝试查找每个帐户的当月值和上月值之和。

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      |
+-----------+-------------+-----------+----------+

在黑斑羚的查询下面，但对于当前和上个月，我返回的结果相同。

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                    |
+-----------+-------------+--------------------+----------------------+

我使用了以下查询，但是如果上个月的数据不可用，它将返回前一个月的asset_previous。

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：我们没有帐户123的20181231输入数据，因此月份1的asset_prev应该为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()以在account分区中获取上个月的值。 LAG()的三个参数形式使您可以指定默认值。

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adate。

查找当前月和上个月值的总和

1 个答案: