SQL组范围值

Question

我已经看了这里的其他几个问题/答案，但我无法将它们应用于我的问题。我正试图根据一个关键列来识别多个连续中断。我发现的大多数例子都不处理同一个键列的序列中的多个中断。

Sample data:
Location     Number
------------------------
300          15
300          16
300          17
300          18
300          21
300          22
300          23
400          10
400          11
400          14
400          16
Here is the result I am looking for:
Location     StartNumber    StartNumber
------------------------------------------
300                   15             18
300                   21             23
400                   10             11
400                   14             14
400                   16             16

Answer 1

这是一个相对可移植的SQL解决方案，因为您没有指定DB

Create Table  SampleData  (Location int, Number Int)
INSERT INTO SampleData VALUES (300, 15)
INSERT INTO SampleData VALUES (300, 16)
INSERT INTO SampleData VALUES (300, 17)
INSERT INTO SampleData VALUES (300, 18)
INSERT INTO SampleData VALUES (300, 21)
INSERT INTO SampleData VALUES (300, 22)
INSERT INTO SampleData VALUES (300, 23)
INSERT INTO SampleData VALUES (400, 10)
INSERT INTO SampleData VALUES (400, 11)
INSERT INTO SampleData VALUES (400, 14)
INSERT INTO SampleData VALUES (400, 16)

SELECT 
        t1.Location,
        t1.Number      AS startofgroup, 
       MIN(t2.Number) AS endofgroup 
FROM   (SELECT Number , Location
        FROM   SampleData tbl1 
        WHERE  NOT EXISTS(SELECT * 
                          FROM   SampleData tbl2 
                          WHERE  tbl1.Number - tbl2.Number = 1
                                 and tbl1.Location = tbl2.Location)) t1 
       INNER JOIN (SELECT Number , Location
                   FROM   SampleData tbl1 
                   WHERE  NOT EXISTS(SELECT * 
                                     FROM   SampleData tbl2 
                                     WHERE  tbl2.Number - tbl1.Number = 1
                                     and tbl1.Location = tbl2.Location)) t2 
         ON t1.Number <= t2.Number 
            and t1.Location = t2.Location
GROUP  BY 
    t1.Location,
    t1.Number 
ORDER BY 
   Location,
   startofgroup

输出

Location    startofgroup endofgroup
----------- ------------ -----------
300         15           18
300         21           23
400         10           11
400         14           14
400         16           16

它是清单2的修改版本。用于识别岛屿的基于集合的解决方案。亚历山大·科扎克的《从岛屿和序列数的差距》

如果您正在寻找SQL Server 2005及更高版本的更多选项，您应该搜索短语"Itzik Ben Gan gaps and islands"

Answer 2

如果您使用的是支持lag（）函数的RDBMS，那么它应该会告诉您中断的位置。然后，您应该能够使用它，以及一些case语句和小心使用min（）和max（）函数，来获得您想要的查询。

select location, lag_number as startnumber, number as endnumber
from(select location, number, lag_number
from(
    select location, number
    , lag(number) over (partition by location order by number) as lag_number
    from table
    )a
    where number is not null and lag_number is not null
)b
where number-lag_number>1 order by 1,2,3;