如何进行连接以获取前几天的值？答案

【问题标题】：How to do a join to get previous days values?如何进行连接以获取前几天的值？
【发布时间】：2021-03-19 14:55:13
【问题描述】：

我想加入如下表格：

Report_Period	Entity	Tag	Users Count	Report_Period_M-1	Report_Period_D-1	...
2021-03-31	entity 1	X	471	2017-05-31	2021-03-18	...
2020-12-31	entity 2	A	135	2020-11-30	2021-03-18	...
2020-11-30	entity 3	X	402	2020-10-31	2021-03-18	...

包含第 1 天结果的视图如下：

Report_Period	Entity	Tag	Users Count	Report_Period_D-1
2021-03-31	entity 1	X	445	2021-03-18
2021-03-31	entity 2	A	135	2021-03-18
2021-03-31	entity 3	X	402	2021-03-18

我的目标是至少返回这样的结果：

Report_Period	Entity	Tag	Users Count	Users Count D-1	Report_Period_M-1	...
2021-03-31	entity 1	X	471	445	2021-02-28	...
2020-12-31	entity 2	A	135	NULL	2020-11-30	...
2020-11-30	entity 3	X	402	NULL	2020-10-31	...

如果我在 Report_Period、Entity & Tag 上使用经典联接，我将只返回包含报告期 2021-03-31 的结果行。

是否可以替换当前期间的用户计数值？如果我想用当月的用户数 D-1 替换用户数值？

欢迎您的帮助，感谢您的宝贵时间！

【问题讨论】：

标签： sql join snowflake-cloud-data-platform

【解决方案1】：

根据您的示例，您似乎想要一个 LEFT JOIN

WITH table_a(report_period, entity, tag, users_count, report_period_m1, report_period_d1) AS (
    SELECT * FROM VALUES
    ('2021-03-31', 'entity 1','X',471,'2017-05-31','2021-03-18'),
    ('2020-12-31', 'entity 2','A',135,'2020-11-30','2021-03-18'),
    ('2020-11-30', 'entity 3','X',402,'2020-10-31','2021-03-18')
), table_b(report_period, entity, tag, users_count, report_period_d1) AS (
    SELECT * FROM VALUES
    ('2021-03-31','entity 1','X',445, '2021-03-18'),
    ('2021-03-31','entity 2','A',135, '2021-03-18'),
    ('2021-03-31','entity 3','X',402, '2021-03-18')
)
SELECT a.*
    ,b.*
FROM table_a AS a
LEFT JOIN table_b AS b
    ON a.report_period = b.report_period AND a.entity = b.entity
ORDER BY 1 desc, 2;

给予：

REPORT_PERIOD   ENTITY      TAG   USERS_COUNT   REPORT_PERIOD_M1    REPORT_PERIOD_D1    REPORT_PERIOD   ENTITY      TAG   USERS_COUNT   REPORT_PERIOD_D1
2021-03-31      entity 1    X     471           2017-05-31          2021-03-18          2021-03-31      entity 1    X     445           2021-03-18
2020-12-31      entity 2    A     135           2020-11-30          2021-03-18          null            null        null  null           null
2020-11-30      entity 3    X     402           2020-10-31          2021-03-18          null            null        null  null           null

但是对于“替换值”的声明，我可以假设您的意思是，当b 不匹配时，您想将b.users_count 与a.users_count 合并，因此代码变为..

WITH table_a(report_period, entity, tag, users_count, report_period_m1, report_period_d1) AS (
    SELECT * FROM VALUES
    ('2021-03-31', 'entity 1','X',471,'2017-05-31','2021-03-18'),
    ('2020-12-31', 'entity 2','A',135,'2020-11-30','2021-03-18'),
    ('2020-11-30', 'entity 3','X',402,'2020-10-31','2021-03-18')
), table_b(report_period, entity, tag, users_count, report_period_d1) AS (
    SELECT * FROM VALUES

    ('2021-03-31','entity 1','X',445, '2021-03-18'),
    ('2021-03-31','entity 2','A',135, '2021-03-18'),
    ('2021-03-31','entity 3','X',402, '2021-03-18')
)
SELECT a.*
    ,COALESCE(b.USERS_COUNT, a.USERS_COUNT) as b_USERS_COUNT
FROM table_a AS a
LEFT JOIN table_b AS b
    ON a.report_period = b.report_period AND a.entity = b.entity
ORDER BY 1,2;

【讨论】：