在 BigQuery 中按 DATETIME 排名
Rank by DATETIME in BigQuery
我有以下 table 架构:
Field name Type Mode Policy tags Description
order_id STRING NULLABLE
customer_id INTEGER NULLABLE
order_datetime TIMESTAMP NULLABLE %Y-%m-%d %H:%M:%E*S
item_id STRING NULLABLE
order_quantity INTEGER NULLABLE
客户可以在一个日期下多个订单。
我正在尝试编写一个查询,为每个客户在他们下订单的每个日期排名 order_id:
SELECT customer_id,
order_datetime as order_date,
order_id,
RANK() OVER(PARTITION BY customer_id, CAST(order_datetime AS DATETIME) ORDER BY 2 DESC) as rank
FROM `SQL_sets.orders`
其中returns以下:
customer_id order_date order_id rank
21456 2019-01-12 9:28:35 A-005 1
21456 2019-01-12 12:28:35 A-005 1
21456 2019-01-12 19:28:35 A-005 1
31874 2020-11-15 0:00:00 A-009 1
32483 2020-11-21 0:00:00 A-001 1
32483 2020-11-22 0:00:00 A-001 1
42491 2019-01-16 2:52:07 A-006 1
42491 2019-01-20 2:52:07 A-0101 1
42491 2019-12-01 9:52:07 A-007 1
42491 2019-12-01 15:12:07 A-008 1
42491 2020-11-19 0:00:00 A-006 1
55400 2019-01-11 12:52:07 A-0088 1
但我expect/wanted得到:
customer_id order_date order_id rank
21456 2019-01-12 9:28:35 A-005 1
21456 2019-01-12 12:28:35 A-005 2
21456 2019-01-12 19:28:35 A-005 3
31874 2020-11-15 0:00:00 A-009 1
32483 2020-11-21 0:00:00 A-001 1
32483 2020-11-22 0:00:00 A-001 1
42491 2019-01-16 2:52:07 A-006 1
42491 2019-01-20 2:52:07 A-0101 1
42491 2019-12-01 9:52:07 A-007 1
42491 2019-12-01 15:12:07 A-008 2
42491 2020-11-19 0:00:00 A-006 1
55400 2019-01-11 12:52:07 A-0088 1
我做错了什么?
你没有说,但我假设你正在使用 SQL 服务器。这可能更多关于你想要什么。在您的示例中,您的分区是按客户和日期时间划分的,因此除非客户同时订购两件东西,否则排名将始终为 1。如果您只对同一日期发生的订单感兴趣,则需要将其转换为那种...
SELECT customer_id,
order_datetime as order_date,
order_id,
RANK() OVER(PARTITION BY customer_id ORDER BY CAST(order_datetime AS DATETIME) DESC) rank
FROM `SQL_sets.orders`
我有以下 table 架构:
Field name Type Mode Policy tags Description
order_id STRING NULLABLE
customer_id INTEGER NULLABLE
order_datetime TIMESTAMP NULLABLE %Y-%m-%d %H:%M:%E*S
item_id STRING NULLABLE
order_quantity INTEGER NULLABLE
客户可以在一个日期下多个订单。 我正在尝试编写一个查询,为每个客户在他们下订单的每个日期排名 order_id:
SELECT customer_id,
order_datetime as order_date,
order_id,
RANK() OVER(PARTITION BY customer_id, CAST(order_datetime AS DATETIME) ORDER BY 2 DESC) as rank
FROM `SQL_sets.orders`
其中returns以下:
customer_id order_date order_id rank
21456 2019-01-12 9:28:35 A-005 1
21456 2019-01-12 12:28:35 A-005 1
21456 2019-01-12 19:28:35 A-005 1
31874 2020-11-15 0:00:00 A-009 1
32483 2020-11-21 0:00:00 A-001 1
32483 2020-11-22 0:00:00 A-001 1
42491 2019-01-16 2:52:07 A-006 1
42491 2019-01-20 2:52:07 A-0101 1
42491 2019-12-01 9:52:07 A-007 1
42491 2019-12-01 15:12:07 A-008 1
42491 2020-11-19 0:00:00 A-006 1
55400 2019-01-11 12:52:07 A-0088 1
但我expect/wanted得到:
customer_id order_date order_id rank
21456 2019-01-12 9:28:35 A-005 1
21456 2019-01-12 12:28:35 A-005 2
21456 2019-01-12 19:28:35 A-005 3
31874 2020-11-15 0:00:00 A-009 1
32483 2020-11-21 0:00:00 A-001 1
32483 2020-11-22 0:00:00 A-001 1
42491 2019-01-16 2:52:07 A-006 1
42491 2019-01-20 2:52:07 A-0101 1
42491 2019-12-01 9:52:07 A-007 1
42491 2019-12-01 15:12:07 A-008 2
42491 2020-11-19 0:00:00 A-006 1
55400 2019-01-11 12:52:07 A-0088 1
我做错了什么?
你没有说,但我假设你正在使用 SQL 服务器。这可能更多关于你想要什么。在您的示例中,您的分区是按客户和日期时间划分的,因此除非客户同时订购两件东西,否则排名将始终为 1。如果您只对同一日期发生的订单感兴趣,则需要将其转换为那种...
SELECT customer_id,
order_datetime as order_date,
order_id,
RANK() OVER(PARTITION BY customer_id ORDER BY CAST(order_datetime AS DATETIME) DESC) rank
FROM `SQL_sets.orders`