当 BigQuery 中的值是字符串或日期时，Case 如何与 else 一起使用？

Posted 2023-03-25

【中文标题】当 BigQuery 中的值是字符串或日期时，Case 如何与 else 一起使用？

【英文标题】：How use Case with than else when values is string or date in BigQuery?

【发布时间】：2019-09-26 11:43:03

【问题描述】：

我有一个这样的 BigQuery 表：

+------+------------+------------------+
| Name |    date    | order_id | value |
+------+------------+----------+-------+
| JONES| 2019-01-03 | 11       |    10 |
| JONES| 2019-01-05 | 12       |    5  |
| JONES| 2019-06-03 | 13       |    15 |
| JONES| 2019-07-03 | 14       |    20 |
| John | 2019-07-23 | 15       |    10 |
+------+------------+----------+-------+

并且需要添加具有特定order_id值的列（意味着order_id我们不知道，只能计算它），例如当order_count为3时。对于一阶解决方案将是这样并找到最后一个order_id 也不是什么大问题：

WITH data AS (
  SELECT "JONES" name, DATE("2019-01-03") date, "11" order_id, 10 value
  UNION ALL
  SELECT "JONES", DATE("2019-01-05"), "12", 5
  UNION ALL
  SELECT "JONES", DATE("2019-06-03"), "13", 15
  UNION ALL
  SELECT "JONES", DATE("2019-07-03"), "14", 20
  UNION ALL
  SELECT "John", DATE("2019-07-23"), "15", 10
)
SELECT name,
ARRAY_AGG(STRUCT(DATE as f_date) ORDER BY order_id LIMIT 1)[OFFSET(0)].*,
  max(order_id) AS l_id, 
  min(order_id) as f_id, 
  SUM(value) AS total
FROM DATA
GROUP BY name

输出：

+------+------+------------+-----+------+
| name | f_id | f_date     |l_id | total|
+------+------+------------+-----+------+
| JONES| 11   | 2019-01-03 | 14  | 50   | 
| John | 15   | 2019-07-23 | 15  | 10   | 
+------+------+------------+-----+------+

是否有相同的方法来获取最后一个订单的值和当 count order_id 为 3 时的值？

输出：

+------+----------+----+----+-----+-----+----------+-----------+
| name |f_date    |f_id|l_id|total|3_id |  3_date  |  l_date   |
+------+----------+----+----+-----+-----+----------+-----------+
| JONES|2019-01-03| 11 | 14 |  50 |13   |2019-06-03| 2019-07-03|
| John |2019-07-23| 15 | 15 |  10 |NULL |NULL      | 2019-07-23|
+------+----------+----+----+-----+-----+----------+-----------+

我想Case应该有一些东西，但这里不适用于非数值的聚合：

(CASE WHEN order_id = l_id THEN date ELSE Null END) AS l_date

【参考方案1】：

以下是 BigQuery 标准 SQL

WITH data AS (
  SELECT "JONES" name, DATE("2019-01-03") `date`, "11" order_id, 10 value UNION ALL
  SELECT "JONES", DATE("2019-01-05"), "12", 5 UNION ALL
  SELECT "JONES", DATE("2019-06-03"), "13", 15 UNION ALL
  SELECT "JONES", DATE("2019-07-03"), "14", 20 UNION ALL
  SELECT "John", DATE("2019-07-23"), "15", 10
)
SELECT name,
  ARRAY_AGG(STRUCT(DATE AS f_date) ORDER BY order_id LIMIT 1)[OFFSET(0)].*,
  MAX(order_id) AS l_id, 
  MIN(order_id) AS f_id, 
  SUM(value) AS total,
  ARRAY_AGG(STRUCT(order_id AS `_3_order`, `date` AS `_3_date`) ORDER BY order_id LIMIT 3)[SAFE_OFFSET(2)].*,
  ARRAY_AGG(STRUCT(`date` AS l_date) ORDER BY order_id DESC LIMIT 1)[OFFSET(0)].*
FROM DATA t
GROUP BY name   

结果

Row name    f_date      l_id    f_id    total   _3_order    _3_date     l_date   
1   JONES   2019-01-03  14      11      50      13          2019-06-03  2019-07-03   
2   John    2019-07-23  15      15      10      null        null        2019-07-23  

【讨论】：

@ Mikhail - 还有一个问题，如果我需要小于 l_id 的先前 order_id，我可以用 ARRAY_AGG 做同样的事情吗？

@Anna - 当然，请先自己尝试。如果仍然有挑战 - 请告诉我，我也会帮助解决这个问题

谢谢，所以对于 prev 将是 ARRAY_AGG(STRUCT(date AS l_date) ORDER BY order_id DESC LIMIT 2)[SAFE_OFFSET(1)].*，如果超出范围，Safe_OFFSET 也会返回 NULL