如何从SQL表中获取同一日期匹配的记录

Question

如何查找同一个账号ID（不管属于什么状态）在同一日期有多个状态的所有记录，时间不重要，但日期很重要

预期结果应如下表所述，我应该选择同一帐户 ID 在同一天具有多个状态的数据

账户ID	状态	上次更新时间戳
12345	DISB	2023-09-07 06:30:30.000
12345	设置	2023-09-07 06:34:30.000
32341	雷布	2023-09-07 14:30:30.000
32341	D	2023-09-07 15:31:30.000
52355	设置	2023-09-09 14:30:30.000
52355	D	2023-09-09 15:31:30.000

样本表数据：

Create table PaymentRecord(
  accountid BIGINT,
  Status varchar(10),
  lastupdatedTimestamp DATETIME
)

INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(12345, 'DISB','2023-09-07 16:30:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(12345, 'SET', '2023-09-07 16:34:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(12346, 'D',   '2023-09-07 11:30:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(22341, 'CLR', '2023-09-08 13:30:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(32341, 'REB', '2023-09-08 14:30:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(32341, 'D',   '2023-09-08 15:31:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(42325, 'CLR', '2023-09-09 11:30:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(52355, 'SET', '2023-09-09 14:30:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(52355, 'D',   '2023-09-09 15:31:30.000')
INSERT INTO PaymentRecord (accountid,Status,lastupdatedTimestamp) VALUES(52355, 'SK',  '2023-09-10 16:31:30.000')

我尝试了以下方法，但没有达到预期的结果。

select distinct 
  AccountId
, Status
, convert(date, LastUpdatedTimestamp)
from PaymentRecord
where LastUpdatedTimestamp between '2023-09-07' and '2023-09-11'
Order by
  AccountId
, convert(date, LastUpdatedTimestamp)
, Status desc

Answer 1

我花了一段时间才明白你的问题

但是用你的

WHERE

我无法重现你想要的结果

这个概念是esy进行一个子选择，您可以在某个日期获取多个状态的帐户ID和日期，并将其加入到主查询中

select p1.AccountId
  ,convert(date,p1.LastUpdatedTimestamp),Status 
from PaymentRecord p1
  JOIN
  
  (SELECT AccountId,convert(date,LastUpdatedTimestamp) as LastUpdatedTimestamp
  FROM  PaymentRecord
 where LastUpdatedTimestamp between '2023-09-08' and '2023-09-11'
  GROUP BY AccountId,convert(date,LastUpdatedTimestamp)
HAVING COUNt(DISTINCT Status) > 1) p2 
  ON p1.AccountId = p2.AccountId AND convert(date,p1.LastUpdatedTimestamp) = p2.LastUpdatedTimestamp
  Order by p1.AccountId, p1.LastUpdatedTimestamp,Status desc

账户ID	（无栏名）	状态
32341	2023-09-08	雷布
32341	2023-09-08	D
52355	2023-09-09	设置
52355	2023-09-09	D

小提琴