Question

我在big-query（datawarehouse）中有一个表：

我希望得到以下结果：

以下是计算方法的解释：

2017-10-01 = 100美元是显而易见的，因为数据只有一个
2017-10-02 = $ 400是第一行和第三行的总和。为什么？因为第二行和第三行具有相同的发票。所以我们只使用最新的更新。
2017-10-04 = 800美元是第1,3行和第4行的总和。为什么？这是因为我们每天只收一张发票。第1行（T001），第3行（T002），第4行（T003）
2017-10-05 = 100美元是第1,5行和第6行的总和。为什么？这是因为我们每天只收一张发票。第1行（T001），第5行（T002），第6行（T003）

老实说，我完全失去了怎么做。我已多次尝试分组等等。但它们都没有按预期工作。这是我今天迄今为止的最新努力：

SELECT 
  amount,
  updatedDateOnly,
  invNo
FROM 
(
  SELECT 
    invNo,
    UpdatedDate,
    amount,
    DATE(updatedDate) as updatedDateOnly,
    row_number() OVER (PARTITION BY  invNo ORDER BY UpdatedDate DESC) AS rownum
  FROM [project:dataset.test] 
)
WHERE
  rownum = 1

仅返回上次日期。现在，我不知道如何查询每日。

感谢任何专家并愿意帮助查询的人。谢谢。

更新： json中的数据，以防您想在bigquery或其他SQL服务器中尝试：

{"UpdatedDate":"2017-10-01 01:00:00","InvNo":"T001","amount":100}
{"UpdatedDate":"2017-10-02 01:00:00","InvNo":"T002","amount":200}
{"UpdatedDate":"2017-10-02 02:00:00","InvNo":"T002","amount":300}
{"UpdatedDate":"2017-10-04 01:00:00","InvNo":"T003","amount":400}
{"UpdatedDate":"2017-10-05 01:00:00","InvNo":"T002","amount":500}
{"UpdatedDate":"2017-10-05 02:00:00","InvNo":"T003","amount":500}

Answer 1

以下是BigQuery Standard SQL

#standardSQL
WITH dates AS (
  SELECT DISTINCT DATE(UpdatedDate) UpdatedDay
  FROM `project.dataset.test`
),
qualified AS (
  SELECT DATE(UpdatedDate) UpdatedDay, InvNo, ARRAY_AGG(amount ORDER BY UpdatedDate DESC LIMIT 1)[SAFE_OFFSET(0)] amount
  FROM `project.dataset.test`
  GROUP BY UpdatedDay, InvNo
)
SELECT UpdatedDay, SUM(amount) amount
FROM (
  SELECT d.UpdatedDay UpdatedDay, InvNo, ARRAY_AGG(amount ORDER BY q.UpdatedDay DESC LIMIT 1)[SAFE_OFFSET(0)] amount
  FROM dates d
  JOIN qualified q
  ON q.UpdatedDay <= d.UpdatedDay
  GROUP BY UpdatedDay, InvNo
)
GROUP BY UpdatedDay
-- ORDER BY UpdatedDay

您可以使用以下来自问题的虚拟数据进行测试/播放

#standardSQL
WITH `project.dataset.test` AS (
  SELECT TIMESTAMP '2017-10-01 01:00:00' UpdatedDate, 'T001' InvNo, 100 amount UNION ALL
  SELECT TIMESTAMP '2017-10-02 01:00:00', 'T002', 200 UNION ALL
  SELECT TIMESTAMP '2017-10-02 02:00:00', 'T002', 300 UNION ALL
  SELECT TIMESTAMP '2017-10-04 01:00:00', 'T003', 400 UNION ALL
  SELECT TIMESTAMP '2017-10-05 01:00:00', 'T002', 500 UNION ALL
  SELECT TIMESTAMP '2017-10-05 02:00:00', 'T003', 500 
),
dates AS (
  SELECT DISTINCT DATE(UpdatedDate) UpdatedDay
  FROM `project.dataset.test`
),
qualified AS (
  SELECT DATE(UpdatedDate) UpdatedDay, InvNo, ARRAY_AGG(amount ORDER BY UpdatedDate DESC LIMIT 1)[SAFE_OFFSET(0)] amount
  FROM `project.dataset.test`
  GROUP BY UpdatedDay, InvNo
)
SELECT UpdatedDay, SUM(amount) amount
FROM (
  SELECT d.UpdatedDay UpdatedDay, InvNo, ARRAY_AGG(amount ORDER BY q.UpdatedDay DESC LIMIT 1)[SAFE_OFFSET(0)] amount
  FROM dates d
  JOIN qualified q
  ON q.UpdatedDay <= d.UpdatedDay
  GROUP BY UpdatedDay, InvNo
)
GROUP BY UpdatedDay
ORDER BY UpdatedDay

结果符合预期

UpdatedDay  amount   
2017-10-01   100     
2017-10-02   400     
2017-10-04   800     
2017-10-05  1100

Answer 2

在每个日期，您需要每张发票的最新金额。这很复杂。一种解决方案是从日期和记录的交叉连接开始。然后窗口函数可用于获取最新的金额：

select dte,
       sum(case when seqnum = 1 then amount else 0 end) as amount
from (select d.dte, t.*,
             row_number() over (partition by t.invNo order by t.UpdatedDate desc) as seqnum
      from (select distinct date(UpdatedDate) as dte
            from `project.dataset.test` t
           ) d join
           `project.dataset.test` t
           on date(t.UpdatedDate) <= d.dte
     ) td
group by dte;

每日大量汇总

2 个答案: