Question

我想填写需要开始和结束日期信息的日历对象。我有一列包含一系列日期。有些日期是连续的（有一天差异），有些则没有。

InfoDate  

2013-12-04  consecutive date [StartDate]
2013-12-05  consecutive date
2013-12-06  consecutive date [EndDate]

2013-12-09                   [startDate]
2013-12-10                   [EndDate]

2014-01-01                   [startDate]
2014-01-02 
2014-01-03                   [EndDate]

2014-01-06                   [startDate]
2014-01-07                   [EndDate]

2014-01-29                   [startDate]
2014-01-30 
2014-01-31                   [EndDate]

2014-02-03                   [startDate]
2014-02-04                   [EndDate]

我想选择每个连续日期范围的开始和结束日期（块中的第一个和最后一个）。

StartDate     EndDate

2013-12-04    2013-12-06
2013-12-09    2013-12-10
2014-01-01    2014-01-03
2014-01-06    2014-01-07
2014-01-29    2014-01-31
2014-02-03    2014-02-04

我想仅使用SQL来解决问题。

Answer 1

不需要连接或递归CTE。标准的gap-and-island解决方案是分组（值减去row_number），因为它在连续序列中是不变的。开始日期和结束日期只是组的MIN（）和MAX（）。

WITH t AS (
  SELECT InfoDate d,ROW_NUMBER() OVER(ORDER BY InfoDate) i
  FROM @d
  GROUP BY InfoDate
)
SELECT MIN(d),MAX(d)
FROM t
GROUP BY DATEDIFF(day,i,d)

Answer 2

你去..

;WITH CTEDATES
AS
(
    SELECT ROW_NUMBER() OVER (ORDER BY Infodate asc ) AS ROWNUMBER,infodate FROM YourTableName  

),
 CTEDATES1
AS
(
   SELECT ROWNUMBER, infodate, 1 as groupid FROM CTEDATES WHERE ROWNUMBER=1
   UNION ALL
   SELECT a.ROWNUMBER, a.infodate,case datediff(d, b.infodate,a.infodate) when 1 then b.groupid else b.groupid+1 end as gap FROM CTEDATES A INNER JOIN CTEDATES1 B ON A.ROWNUMBER-1 = B.ROWNUMBER
)

select min(mydate) as startdate, max(infodate) as enddate from CTEDATES1 group by groupid

如果这回答了你的问题，请不要忘记将其标记为答案。

Answer 3

我已将这些值插入名为#consec的表中，然后执行以下操作：

select t1.*
,t2.infodate as binfod
into #temp1
from #consec t1
left join #consec t2 on dateadd(DAY,1,t1.infodate)=t2.infodate

select t1.*
,t2.infodate as binfod
into #temp2
from #consec t1
left join #consec t2 on dateadd(DAY,1,t2.infodate)=t1.infodate
;with cte as(
select infodate,  ROW_NUMBER() over(order by infodate asc) as seq from #temp1
where binfod is null
),
cte2 as(
select infodate, ROW_NUMBER() over(order by infodate asc) as seq from #temp2
where binfod is null
)

select t2.infodate as [start_date]
,t1.infodate as [end_date] from cte t1
left join cte2 t2 on t1.seq=t2.seq

只要您的日期不重叠，那就应该为您完成工作。

Answer 4

这是我的测试数据样本：

--required output
-- 01 - 03
-- 08 - 09
-- 12 - 14

DECLARE @maxRN int;
WITH #tmp AS (
                SELECT CAST('2013-01-01' AS date) DT
    UNION ALL   SELECT CAST('2013-01-02' AS date)
    UNION ALL   SELECT CAST('2013-01-03' AS date)
    UNION ALL   SELECT CAST('2013-01-05' AS date)
    UNION ALL   SELECT CAST('2013-01-08' AS date)
    UNION ALL   SELECT CAST('2013-01-09' AS date)
    UNION ALL   SELECT CAST('2013-01-12' AS date)
    UNION ALL   SELECT CAST('2013-01-13' AS date)
    UNION ALL   SELECT CAST('2013-01-14' AS date)
),
#numbered AS (
    SELECT 0 RN, CAST('1900-01-01' AS date) DT
    UNION ALL
    SELECT ROW_NUMBER() OVER (ORDER BY DT) RN, DT
    FROM #tmp
)

SELECT * INTO #tmpTable FROM #numbered;
SELECT @maxRN = MAX(RN) FROM #tmpTable;

INSERT INTO #tmpTable
SELECT @maxRN + 1, CAST('2100-01-01' AS date);

WITH #paired AS (
    SELECT 
    ROW_NUMBER() OVER(ORDER BY TStart.DT) RN, TStart.DT DTS, TEnd.DT DTE
    FROM #tmpTable TStart
    INNER JOIN #tmpTable TEnd 
    ON TStart.RN = TEnd.RN - 1
    AND DATEDIFF(dd,TStart.DT,TEnd.DT) > 1  
)

SELECT TS.DTE, TE.DTs 
FROM #paired TS
INNER JOIN #paired TE ON TS.RN = TE.RN -1
AND TS.DTE <> TE.DTs -- you could remove this filter if you want to have start and end on the same date

DROP TABLE #tmpTable

将#tmp数据替换为您的实际表格。

Answer 5

您可以这样做，这里是sqlfiddle

select
  min(ndate) as start_date,
  max(ndate) as end_date
from
(select
  ndate,
  dateadd(day, -row_number() over (order by ndate), ndate) as rnk
 from dates
 ) t
 group by
   rnk

Answer 6

另一个可以在此处使用的简单解决方案是-

with tmp as 
(
select
datefield
, dateadd('day',-row_number() over(order by date asc),datefield) as date_group 
from table
)
select
min(datefield) as start_date
, max(datefield) as end_date 
from tmp
group by date_group

Answer 7


create table mysql.date_test ( name VARCHAR(10) , d date);

insert into mysql.date_test values
('Bill','2021-04-01'),
('Bill','2021-04-02'), 
('Bill','2021-04-03'), 
('Bill','2021-04-05'), 
('Bill','2021-04-06'), 
('Bill','2021-04-07'),
('Bill','2021-04-09'),
('Bill','2021-04-10');


Subtract row_number from the date. You will get the same date if dates are conscutive.

mysql> select name,d,row_number() over (partition by name order by d) as rn,ADDDATE(d,INTERVAL - row_number() over (partition by name order by d) DAY) as grp
    -> from   mysql.date_test;

+------+------------+----+------------+
| name | d          | rn | grp        |
+------+------------+----+------------+
| Bill | 2021-04-01 |  1 | 2021-03-31 |
| Bill | 2021-04-02 |  2 | 2021-03-31 |
| Bill | 2021-04-03 |  3 | 2021-03-31 |
| Bill | 2021-04-05 |  4 | 2021-04-01 |
| Bill | 2021-04-06 |  5 | 2021-04-01 |
| Bill | 2021-04-07 |  6 | 2021-04-01 |
| Bill | 2021-04-09 |  7 | 2021-04-02 |
+------+------------+----+------------+
7 rows in set (0.00 sec)

Then group on the static date

mysql> select name,min(d) as start_date,max(d) as end_date
    -> from
    -> (
    -> select name,d,row_number() over (partition by name order by d),ADDDATE(d,INTERVAL - row_number() over (partition by name order by d) DAY) as grp
    -> from   mysql.date_test
    -> )a
    -> group by name,grp;

+------+------------+------------+
| name | start_date | end_date   |
+------+------------+------------+
| Bill | 2021-04-01 | 2021-04-03 |
| Bill | 2021-04-05 | 2021-04-07 |
| Bill | 2021-04-09 | 2021-04-10 |
+------+------------+------------+

Answer 8

SELECT InfoDate ,
    CASE
      WHEN TRUNC(InfoDate - 1) = TRUNC(lag(InfoDate,1,InfoDate) over (order by InfoDate))
      THEN NULL
      ELSE InfoDate
    END STARTDATE,
    CASE
      WHEN TRUNC(InfoDate + 1) = TRUNC(lead(InfoDate,1,InfoDate) over (order by InfoDate))
      THEN NULL
      ELSE InfoDate
    END ENDDATE
  FROM TABLE;

使用SQL检测连续的日期范围

8 个答案: