因此,通过添加带有一些虚拟数据的 CTE,并将列名更改为安全
WITH data AS (
SELECT * FROM VALUES
(1,'a','UK','2020-12-01'),
(1,'a','UK','2021-01-01'),
(2,'a','UK','2020-12-01'),
(3,'a','UK','2021-01-01')
v( Client_Ref, zone, Market_Place, Report_Period)
)
SELECT DISTINCT d.Client_Ref,d.zone,d.Market_Place,d.Report_Period
FROM data AS d
WHERE d.Report_Period ='2020-12-01' AND d.Market_Place ='UK' AND d.Client_Ref IS NOT NULL
AND
NOT EXISTS
(
SELECT t.Client_Ref
FROM data t
WHERE t.Report_Period ='2021-01-01' AND t.Market_Place ='UK' AND t.Client_Ref IS NOT NULL AND d.Client_Ref=t.Client_Ref
);
您的 SQL 基本表单有效并返回:
CLIENT_REF ZONE MARKET_PLACE REPORT_PERIOD
2 a UK 2020-12-01
这是预期的结果。
此查询是关联子查询,Snowflake 对其支持有限。因此,虽然这可行,但当您更改查询时,它可能会遇到Unsupported subquery type cannot be evaluated 错误,请参阅SO correlated sub-query question。
通过使用LEFT JOIN 和WHERE x IS NULL 模式,可以以不相关的形式编写基本查询:
WITH data AS (
SELECT * FROM VALUES
(1,'a','UK','2020-12-01'),
(1,'a','UK','2021-01-01'),
(2,'a','UK','2020-12-01'),
(3,'a','UK','2021-01-01')
v( Client_Ref, zone, Market_Place, Report_Period)
)
SELECT DISTINCT d.Client_Ref,d.zone,d.Market_Place,d.Report_Period
FROM data AS d
LEFT JOIN data AS t
ON t.Report_Period ='2021-01-01' AND t.Market_Place ='UK' AND d.Client_Ref=t.Client_Ref
WHERE d.Report_Period ='2020-12-01' AND d.Market_Place ='UK' AND d.Client_Ref IS NOT NULL
AND t.Client_Ref IS NULL;
如果您的数据源有很多行不在目标结果范围内,可以重写它以首先进行一些过滤,如下所示:
WITH data AS (
SELECT * FROM VALUES
(1,'a','UK','2020-12-01'),
(1,'a','UK','2021-01-01'),
(2,'a','UK','2020-12-01'),
(3,'a','UK','2021-01-01')
v( Client_Ref, zone, Market_Place, Report_Period)
), wanted_data AS (
SELECT DISTINCT Client_Ref, zone, Market_Place, Report_Period
FROM data
WHERE Report_Period BETWEEN '2020-12-01' AND '2021-01-01'
AND Market_Place ='UK' AND Client_Ref IS NOT NULL
)
SELECT DISTINCT d.Client_Ref,d.zone,d.Market_Place,d.Report_Period
FROM wanted_data AS d
LEFT JOIN wanted_data AS t
ON t.Report_Period ='2021-01-01'AND d.Client_Ref=t.Client_Ref
WHERE d.Report_Period ='2020-12-01'
AND t.Client_Ref IS NULL;
但是对于我来说,如果我像您一样将列命名为 "Client Ref",我的 SQL 将无法正常工作,因此我无法回答这部分,但这就是您构建 SQL 的方式。