Question

我有一个BorderCrossingData表。 A希望获取PassportNames至少有一个BorderCrossingDateTime-interval，该间隔超过4个月。

BorderCrossingID    PassportNumber  BorderCrossingDateTime
1                    ER-2222        2019-01-07 22:11:12.000
2                    ER-2222        2019-01-07 23:11:12.000
3                    KL-5233        2018-10-03 17:10:39.000
130                  FF-4444        2019-01-08 11:11:11.000
5                    ER-1111        NULL
6                    KL-5686        NULL
7                    ER-1111        NULL
8                    KL-5235        NULL
9                    QW-5656        NULL
10                   DF-5685        NULL
11                   KL-4558        NULL
--------
113                 LL-8989         2019-01-15 16:24:26.333
114                 ZZ-0005         2019-01-17 16:18:12.273
115                 LL-0223         2019-01-17 16:19:12.000
116                 ER-2222         2019-01-03 08:24:29.000
117                 ER-2222         2019-02-01 08:25:03.873
118                 ER-2222         2019-03-13 08:25:17.000
119                 ER-2222         2019-04-10 08:25:32.000
120                 ER-2222         2019-09-30 08:25:47.000

我已经获得BorderCrossings的BorderCrossingDateTime并将它们置于顺序中。

SELECT DISTINCT PassportNumber, BorderCrossingDateTime FROM Passports
WHERE DATEDIFF(Compare 2 upcoming DateTimes)
EXCEPT
SELECT PassportNumber, BorderCrossingDateTime FROM Passports
WHERE BorderCrossingDateTime IS NULL 
ORDER BY BorderCrossingDateTime

结果应该是这样的：

PassportName
ER-2222
TO-0140
NN-4444
TP-0140
TT-0140
WU-5645

Answer 1

下面的代码将起作用。我已经在SQL Server中进行了测试。我正在使用LEAD，LAG函数来查找上一个，下一个过境日期。

    CREATE TABLE #borderCrossing (BorderCrossingID INT, PassportNumber VARCHAR(10), BorderCrossingDateTime DATETIME)

INSERT INTO #borderCrossing
VALUES (1, 'ER-2222', '2019-01-07 22:11:12.000'), (2, 'ER-2222', '2019-03-07 22:11:12.000'), (3, 'ER-2222', '2019-08-07 22:11:12.000');

SELECT DISTINCT PassportNumber
FROM (
    SELECT BorderCrossingId, BorderCrossingDateTime AS CurrentBorderCrossingDateTime, passportNumber, lag(BorderCrossingDateTime, 1, NULL) OVER (
            PARTITION BY passportNumber ORDER BY BorderCrossingDateTime
            ) AS prevBorderCrossingDateTime, lead(BorderCrossingDateTime, 1, NULL) OVER (
            PARTITION BY passportNumber ORDER BY BorderCrossingDateTime
            ) AS nextBorderCrossingDateTime
    FROM #borderCrossing
    ) AS t
WHERE DATEDIFF(mm, prevBorderCrossingDateTime, nextBorderCrossingDateTime) > 4

在同一列中找到Datediff

1 个答案: