Question

见表：

表名：A

Call Session ID           Sequence No       Source Application
STUD-213                     1              Application A
STUD-213                     2              Application B
STUD-213                     3              Application A
STUD-213                     4              Application C
STUD-321                     1              Application A
STUD-321                     2              Application C
STUD-321                     3              Application A
STUD-321                     4              Application B

这是一个示例数据。问题是，我需要计算这些项目之间的项目数。例如，有多少 应用程序A 转到 应用程序B ？我想要的输出应该是：

Source Application      End Application         Count
Application A           Application B            7
Application B           Application C            10
Application A           Application C            6

如果我在SQL中尝试这个，我认为这会很棘手。表A的记录将是巨大的，每月约1000万，所以我也考虑了性能。

您对如何处理此问题有什么建议吗？

编辑：更改了帖子标题。希望它有助于将来偶然发现这个问题的人。

Answer 1

CountApp是AppA最终以AppB结束的次数，
VersionCount是从AppA到AppB所需的步骤。

希望其中一个是你所追求的。

declare @CS table (CallSessionID varchar(8), SequenceNo int, SourceApplication varchar(15))
insert into @CS
values
    ('STUD-213',1,'Application A'),
    ('STUD-213',2,'Application B'),
    ('STUD-213',3,'Application A'),
    ('STUD-213',4,'Application C'),
    ('STUD-321',1,'Application A'),
    ('STUD-321',2,'Application C'),
    ('STUD-321',3,'Application A'),
    ('STUD-321',4,'Application B')

select 
    mini.SourceApplication as StartApp, 
    maxi.SourceApplication as EndApp, 
    Count(*) as CountApp,
    mm.VersionCount
from
    (select CallSessionID, min(SequenceNo) MinSeq, Max(SequenceNo) MaxSeq, count(SequenceNo) VersionCount
    from @CS
    group by CallSessionID) mm
    inner join @CS mini
        on mini.CallSessionID = mm.CallSessionID
        and mini.SequenceNo = mm.MinSeq
    inner join @CS maxi
        on maxi.CallSessionID = mm.CallSessionID
        and maxi.SequenceNo = mm.MaxSeq
group by 
    mini.SourceApplication, 
    maxi.SourceApplication

编辑版本2
此版本查看序列中从一个应用程序到下一个应用程序的转换，然后在每次从AppA转换到AppB时进行计数。

select  
    c1.SourceApplication as StartApp, 
    c2.SourceApplication as EndApp, 
    Count(*) as CountApp
from
    @CS c1
    inner join @CS c2
        on c1.CallSessionID = c2.CallSessionID
        and c1.SequenceNo = (c2.SequenceNo - 1)
group by
    c1.SourceApplication, 
    c2.SourceApplication

Answer 2

结束解决我的问题。我使用循环来获取每个流程。仍然不确定这一次在生产中的表现。我尝试在2.05秒内查询250万条记录的表格。

我将把这个留在这里，以帮助将来徘徊于此问题的任何人：

DECLARE @MAXSEQNO INT
            , @MINSEQNO INT
            , @INCREMENT INT;

DECLARE @TABLE TABLE ([SourceApplication] VARCHAR(50)
                                        , [EndApplication] VARCHAR(50)
                                        , [ApplicationCount] INT);

SET @MAXSEQNO = (SELECT MAX(SequenceNo) FROM A)
SET @MINSEQNO = (SELECT MIN(SequenceNo) FROM A)

WHILE(@MINSEQNO <> @MAXSEQNO)
BEGIN

    SET @INCREMENT = (SELECT @MINSEQNO + 1)

    INSERT INTO @TABLE
                ([Sourceapplication], [EndApplication], [ApplicationCount])
    SELECT A.SourceApplication AS [StartApplication]
                , B.SourceApplication AS [EndApplication]
                , COUNT(*) AS [ApplicationCount]
    FROM A AS Main
    INNER JOIN A B ON (Main.CallSessionID = B.CallSessionID)
    WHERE A.ApplicationSequenceNo = @MINSEQNO
                AND B.ApplicationSequenceNo = @INCREMENT
    GROUP BY A.SourceApplication
                , B.SourceApplication

    SET @MINSEQNO = (SELECT @MINSEQNO + 1)

END

SELECT [SourceApplication]
            , [EndApplication]
            , SUM(ApplicationCount) AS [ApplicationCount]
FROM @TABLE
GROUP BY [SourceApplication]
                , [EndApplication]

如何查询连续数字序列的记录

2 个答案: