SQL Server:检查上一条记录的日期

我有一个SQL Server表，每当有人签出和签入卡片时，它都会在数据库中存储事务。我试图提出一个视图，以显示所有的卡交易，其中一个人没有签入一张卡在7天内的结帐。我的桌子如下所示：

+-------------+-------------+------------------+
| Card_Number |   Status    | Transaction_Date |
+-------------+-------------+------------------+
|      123456 | Checked In  | 7/8/2018         |
|      123456 | Checked Out | 7/1/2018         |
|      234567 | Checked In  | 7/8/2018         |
|      234567 | Checked Out | 7/1/2018         |
|      456789 | Checked In  | 7/5/2018         |
|      456789 | Checked Out | 7/1/2018         |
+-------------+-------------+------------------+

在本例中，我希望记录的卡号为123456和234567，但由于某些原因，我无法找到一个好方法来实现这一点。任何帮助都将不胜感激。先谢谢你

首先想到的是不存在：

select t.*
from t
where status = 'Checked Out' and
      not exists (select 1
                  from t t2
                  where t2.card_number = t.card_number and
                        t2.status = 'Checked In' and
                        t2.transaction_date >= t.transaction_date and
                        t2.transaction_date < dateadd(day, 7, t.transaction_date)
                 );

---

您是否研究过SQL Server的超前/滞后功能？这些允许轻松访问上一行或下一行

例如：

DECLARE @data TABLE ( [card_number] VARCHAR(15), [status] VARCHAR(15), [transaction_date] DATETIME );

INSERT INTO @data (
            [card_number], [status], [transaction_date]
        ) VALUES
        ( '123456', 'Checked In', '07/08/2018' )
        , ( '123456', 'Checked Out', '07/01/2018' )
        , ( '234567', 'Checked In', '07/08/2018' )
        , ( '234567', 'Checked Out', '07/01/2018' )
        , ( '456789', 'Checked In', '07/05/2018' )
        , ( '456789', 'Checked Out', '07/01/2018' );

        SELECT
            [card_number]
            , [status]
            , CONVERT( VARCHAR(10), [transaction_date], 101 ) AS [txn_date]
            , CONVERT( VARCHAR(10), LAG( [transaction_date], 1 ) OVER ( ORDER BY [card_number], [transaction_date] ), 101 ) AS [prev_txn_date]
            , CONVERT( VARCHAR(10), LEAD( [transaction_date], 1 ) OVER ( ORDER BY [card_number], [transaction_date] ), 101 ) AS [next_txn_date]
        FROM @data 
        WHERE
            [card_number] = '123456'
        ORDER BY 
            [card_number], [transaction_date];

返回：

+-------------+-------------+------------+---------------+---------------+
| card_number |   status    |  txn_date  | prev_txn_date | next_txn_date |
+-------------+-------------+------------+---------------+---------------+
|      123456 | Checked Out | 07/01/2018 | NULL          | 07/08/2018    |
|      123456 | Checked In  | 07/08/2018 | 07/01/2018    | NULL          |
+-------------+-------------+------------+---------------+---------------+

---

不确定我是否做对了，但这就是我所尝试的。希望有帮助

select card_number, transaction_date, sta
from
(
select t1.card_number, t1.transaction_date, t1.sta, datediff(DAY, t1.transaction_date, t2.transaction_date) as diff
from trans t1
join trans t2
on t1.card_number = t2.card_number
) result
where diff=7 and sta = 'Checked Out'

---

嗨，戈登，你能再解释一下关于“不存在”的用法吗？我试过你的例子，但当我使用“不存在”时，我没有得到任何回报。如果我使用“Exists”，我会返回行。@Jestes。你的问题是，我正试图提出一个视图，以显示所有卡交易，其中一个人没有在结账后7天内签入卡。这表明不存在。这接近于我正在寻找的，但我不想有一个使用卡号的where语句。可以在where语句中使用lag或lead，以便在计算中比较日期吗？更不用说我能够使其工作并满足我的需要。谢谢你的帮助。回答你的问题，是的。你已经决定了，不客气。

select card_number, transaction_date, sta
from
(
select t1.card_number, t1.transaction_date, t1.sta, datediff(DAY, t1.transaction_date, t2.transaction_date) as diff
from trans t1
join trans t2
on t1.card_number = t2.card_number
) result
where diff=7 and sta = 'Checked Out'