用於確定連續訪問天數的不同時段的SQL？

Question

Jeff最近問了this question並得到了一些很好的答案。

傑夫的問題圍繞着找到已連續幾天登錄到系統的用戶。使用數據庫表結構如下：

 
Id  UserId CreationDate 
------ ------ ------------ 
750997  12 2009-07-07 18:42:20.723 
750998  15 2009-07-07 18:42:20.927 
751000  19 2009-07-07 18:42:22.283 

讀第一爲清楚起見，然後...

我被確定有多少不同（N） - 天爲一個週期的問題很感興趣用戶。

是否可以創建一個可以返回用戶列表的快速SQL查詢以及它們具有的不同（n）天期數？

編輯：根據下面的評論如果某人有連續2天，那麼差距，然後連續4天，然後一個差距，然後連續8天。這將是3「不同的4天時期」。 8天的時間應該算作兩個背靠背的4天時間。

Answer 1

這與我的測試數據非常吻合。

DECLARE @days int 
SET @days = 30 

SELECT DISTINCT l.UserId, (datediff(d,l.CreationDate, -- Get first date in contiguous range 
(
    SELECT min(a.CreationDate) as CreationDate 
    FROM UserHistory a 
     LEFT OUTER JOIN UserHistory b 
      ON a.CreationDate = dateadd(day, -1, b.CreationDate) AND 
      a.UserId = b.UserId 
    WHERE b.CreationDate IS NULL AND 
     a.CreationDate >= l.CreationDate AND 
     a.UserId = l.UserId 
))+1)/@days as cnt 
INTO #cnttmp 
FROM UserHistory l 
    LEFT OUTER JOIN UserHistory r 
     ON r.CreationDate = dateadd(day, -1, l.CreationDate) AND 
     r.UserId = l.UserId 
WHERE r.CreationDate IS NULL 
ORDER BY l.UserId 

SELECT UserId, sum(cnt) 
FROM #cnttmp 
GROUP BY UserId 
HAVING sum(cnt) > 0 

Answer 2

我的答案似乎還沒出現......

我再試一次......

羅布·法利的回答原來的問題有，包括連續數的便利好處天。

with numberedrows as 
(
     select row_number() over (partition by UserID order by CreationDate) - cast(CreationDate-0.5 as int) as TheOffset, CreationDate, UserID 
     from tablename 
) 
select min(CreationDate), max(CreationDate), count(*) as NumConsecutiveDays, UserID 
from numberedrows 
group by UserID, TheOffset 

使用整數除法，簡單地將天的連續數給出「天期間不同（n）的」由整個連續週期覆蓋的數量...
- 2/4 = 0
- 第4/4 = 1
- 8/4 = 2
- 9/4 = 2
- 等，等

因此，這裏是我對羅布的回答您的需求...
（我真的很喜歡Rob's answer，去閱讀的解釋，它的靈感思維）

with 
    numberedrows (
     UserID, 
     TheOffset 
    ) 
as 
(
    select 
     UserID, 
     row_number() over (partition by UserID order by CreationDate) 
      - DATEDIFF(DAY, 0, CreationDate) as TheOffset 
    from 
     tablename 
), 
    ConsecutiveCounts(
     UserID, 
     ConsecutiveDays 
    ) 
as 
(
    select 
     UserID, 
     count(*) as ConsecutiveDays 
    from 
     numberedrows 
    group by 
     UserID, 
     TheOffset 
) 
select 
    UserID, 
    SUM(ConsecutiveDays/@period_length) AS distinct_n_day_periods 
from 
    ConsecutiveCounts 
group by 
    UserID 

唯一真正的區別是，我把羅布的結果，然後通過另一GROUP BY運行它...

Answer 3

所以 - ！我要去從我最後一個問題的查詢開始，其中列出了連續幾天的每次運行。然後，我將按用戶ID和NumConsecutiveDays將其分組，以計算這些用戶有多少天的運行時間。

with numberedrows as 
(
     select row_number() over (partition by UserID order by CreationDate) - cast(CreationDate-0.5 as int) as TheOffset, CreationDate, UserID 
     from tablename 
) 
, 
runsOfDay as 
(
select min(CreationDate), max(CreationDate), count(*) as NumConsecutiveDays, UserID 
from numberedrows 
group by UserID, TheOffset 
) 
select UserID, NumConsecutiveDays, count(*) as NumOfRuns 
from runsOfDays 
group by UserID, NumConsecutiveDays 
; 

和當然，如果你想這個過濾器只考慮一定長度的運行，然後把「裏NumConsecutiveDays> = @days」在過去的查詢。

現在，如果您想要將連續16天計爲三次爲期5天的運行，那麼每次運行都會計爲NumConsecutiveDays/@runlength（這將對每個整數進行舍入）。因此，現在不是隻計算每個數的數量，而是使用SUM。你可以使用上面的查詢並使用SUM（NumOfRuns * NumConsecutiveDays/@runlength），但是如果你理解了邏輯，那麼下面的查詢就容易一些。

with numberedrows as 
(
     select row_number() over (partition by UserID order by CreationDate) - cast(CreationDate-0.5 as int) as TheOffset, CreationDate, UserID 
     from tablename 
) 
, 
runsOfDay as 
(
select min(CreationDate), max(CreationDate), count(*) as NumConsecutiveDays, UserID 
from numberedrows 
group by UserID, TheOffset 
) 
select UserID, sum(NumConsecutiveDays/@runlength) as NumOfRuns 
from runsOfDays 
where NumConsecutiveDays >= @runlength 
group by UserID 
; 

希望這有助於

羅布