BigQuery 中的动态标头答案

【问题标题】：Dynamic Header in BigQueryBigQuery 中的动态标头
【发布时间】：2021-09-04 18:26:31
【问题描述】：

我正在创建一个仪表板，我想在其中显示每个类别每周关闭的项目。这是我的表格示例。

Category	Opened Date	Closed Date
Sales	06/01/2021	06/02/2021
Product	06/02/2021	06/07/2021
Feedback	06/07/2021	06/14/2021
Sales	05/18/2021	05/23/2021
Product	06/01/2021	06/01/2021
Feedback	06/01/2021	06/05/2021
Sales	05/21/2021	05/24/2021
Product	05/21/2021	05/26/2021
Product	06/01/2021	06/02/2021
Feedback	05/31/2021	06/13/2021
Sales	06/02/2021	06/06/2021
Product	06/04/2021	06/07/2021

这是我想要达到的结果。

每周关闭类别

Category	05/23/2021	05/30/2021	06/06/2021	06/13/2021
Sales	2	1	0	1
Product	1	2	2	0
Feedback	0	0	1	2

我尝试使用 SET 以便动态创建列，但是当我使用 @ 符号时 bigQuery 显示错误。当我使用 column_alias 时，它也会按字面显示列名

SET @column_week1 = DATE_SUBB(CURRENT_DATE, INTERVAL 7 DAYS)

还有其他选择吗？

【问题讨论】：

听起来您实际上想要创建表格的视图，您可以在其中按类别和完成周聚合 表格源，并使用完成一周作为支点。

标签： google-bigquery

【解决方案1】：

试试execute_immediate 和pivot 运算符：

DECLARE weeks_list STRING;

create temp table input_table as
select 'Sales' as category, '06/02/2021' as closed_date union all
select 'Product', '06/07/2021' union all
select 'Feedback', '06/14/2021' union all
select 'Sales', '05/23/2021' union all
select 'Product', '06/01/2021' union all
select 'Feedback', '06/05/2021' union all
select 'Sales', '05/24/2021' union all
select 'Product', '05/26/2021' union all
select 'Product', '06/02/2021' union all
select 'Feedback', '06/13/2021' union all
select 'Sales', '06/06/2021' union all
select 'Product', '06/07/2021';

create temp table normalized_table as select category, FORMAT_DATE('_%m_%d_%Y', DATE_TRUNC(PARSE_DATE('%m/%d/%Y', closed_date), WEEK)) as closed_week from input_table;
SET weeks_list = (select string_agg(distinct '"' || closed_week || '"') from normalized_table);
execute immediate 'SELECT * FROM (SELECT category, closed_week FROM normalized_table) PIVOT(count(*) FOR closed_week IN (' || weeks_list || '))';

附言

您可以跳过创建临时normalized_table 和weeks_list 变量：

execute immediate (
  select '''
    SELECT * FROM (
      SELECT 
        category,
        FORMAT_DATE("_%m_%d_%Y", DATE_TRUNC(PARSE_DATE("%m/%d/%Y", closed_date), WEEK)) as closed_week
      FROM input_table)
      PIVOT(count(*) FOR closed_week IN (''' || string_agg(distinct '"' || closed_week || '"') || '''))
  '''
  from (
    select
      category,
      FORMAT_DATE("_%m_%d_%Y", DATE_TRUNC(PARSE_DATE("%m/%d/%Y", closed_date), WEEK)) as closed_week
    from input_table
  )
);

【讨论】：

【解决方案2】：

你可以一次完成所有相同的操作

execute immediate (             
select '''select * from (
  select replace('_' || (last_day(parse_date('%m/%d/%Y', closed), week) - 6), '-', '_') week, 
    category
  from `project.dataset.table`
  )
  pivot(count(*) for week in ("''' ||  string_agg(week, '", "')  || '''"))
'''
from (select distinct replace('_' || (last_day(parse_date('%m/%d/%Y', closed), week) - 6), '-', '_') week from `project.dataset.table` order by week) 
)

有输出

还可以查看https://stackoverflow.com/a/67479622/5221944 - 可以帮助您理解上述代码

【讨论】：