Question

当前设计

Table: 1_notes
------------------------------------------
| id  | text           | created_at          |
------------------------------------------
| 1_1 | u1 first note  | 2018-01-01 10:00:00 |
| 1_2 | u1 second note | 2018-01-03 10:00:00 |


Table: 1_note_timeline
---------------------------------------------------------------------
| note_id  | note_created_at     | likes_count | created_at          |
---------------------------------------------------------------------
| 1_1      | 2018-01-01 10:00:00 | 10          | 2018-01-01 10:00:00 |
| 1_1      | 2018-01-01 10:00:00 | 20          | 2018-01-02 10:00:00 |
| 1_2      | 2018-01-03 10:00:00 | 10          | 2018-01-03 10:00:00 |
| 1_1      | 2018-01-01 10:00:00 | 15          | 2018-01-03 10:00:00 |


Table: 2_notes
--------------------------------------------
| id  | text           | created_at            |
--------------------------------------------
| 2_1 | u2 first note  | 2018-01-01 10:00:00 |
| 2_2 | u2 second note | 2018-01-03 10:00:00 |


Table: 2_note_timeline
---------------------------------------------------------------------
| note_id | note_created_at     | likes_count | created_at          |
---------------------------------------------------------------------
| 2_1     | 2018-01-01 10:00:00 | 10          | 2018-01-01 10:00:00 |
| 2_1     | 2018-01-01 10:00:00 | 20          | 2018-01-02 10:00:00 |
| 2_2     | 2018-01-03 10:00:00 | 10          | 2018-01-03 10:00:00 |
| 2_1     | 2018-01-01 10:00:00 | 15          | 2018-01-03 10:00:00 |

对于每个用户，其笔记数据有2个表。

{{userId}} _ notes表包含ID为userId的用户的注释
{{userId}} _ note_timeline表每天跟踪记录的likes_count数据

所需结果应具有：

最喜欢的两个用户的前2个笔记
点数必须是最新的（并且不能为MAX_VALUE，因为likes_count的注释会随着时间而减少）

最终输出

Output
---------------------------------------------------------------
| note_id | note_created_at     | likes_count | text          |
---------------------------------------------------------------
| 1_1     | 2018-01-01 10:00:00 | 15          | u1 first note |
| 2_1     | 2018-01-01 10:00:00 | 15          | u2 first note |

Answer 1

以下是用于BigQuery标准SQL

#standardSQL
WITH `project.dataset.user_notes` AS (
  SELECT * FROM `project.dataset.user1_notes` UNION ALL
  SELECT * FROM `project.dataset.user2_notes`
), `project.dataset.user_note_timeline` AS (
  SELECT * FROM `project.dataset.user1_note_timeline` UNION ALL
  SELECT * FROM `project.dataset.user2_note_timeline`
)
SELECT note_id, note_created_at, likes_count, text 
FROM (
  SELECT note_id, ARRAY_AGG(STRUCT(note_created_at, likes_count, created_at) ORDER BY created_at DESC LIMIT 1)[OFFSET(0)].*
  FROM `project.dataset.user_note_timeline`
  GROUP BY note_id
  ORDER BY likes_count DESC, note_created_at
  LIMIT 2
) t
JOIN `project.dataset.user_notes` n
ON note_id  = id

Answer 2

希望这可以解决您的问题

SELECT TOP 1 t1.note_id, t1.note_created_at, t1.likes_count, t2.[text]
  FROM 1_note_timeline t1 INNER JOIN 1_notes t2 ON t1.note_id = t2.Id
 ORDER BY t1.created_at DESC

UNION

SELECT TOP 1 t1.note_id, t1.note_created_at, t1.likes_count, t2.[text]
  FROM 2_note_timeline t1 INNER JOIN 2_notes t2 ON t1.note_id = t2.Id
 ORDER BY t1.created_at DESC

最好为表引入索引，以提高性能。

在BigQuery中的联接表上，使用Group By子句获取“最新行”的最有效查询是什么？

2 个答案:

希望这可以解决您的问题