1

I have a set that looks something like this 

ID  date_IN  date_out
1    1/1/18   1/2/18
1    1/3/18   1/4/18
1    1/5/18   1/8/18
2    1/1/18   1/5/18
2    1/7/18   1/9/18

I began by 

SELECT ID, date_IN, Date_out, lead(date_out) over ( partition by (ID)
order by ID) as next_out
From table

And get something like this...

ID  date_IN  date_out  next_out
1    1/1/18   1/2/18   1/4/18
1    1/3/18   1/4/18   1/8/18
1    1/5/18   1/8/18    Null
2    1/1/18   1/5/18    1/9/18
2    1/7/18   1/9/18   Null

The problem I’m going to to have is that in my actual data many of the ID’s have A LOT of entries. The goal is to have all of the date_out’s appear on one row per ID....

ID  date_IN  date_out  next_out  next_out1  etc.   etc.
1    1/1/18   1/2/18   1/4/18     1/8/18     X      X
2    1/1/18   1/5/18   1/7/18     X         Null    Null

Is there a way to loop the lead() through the entire partition, order by ID drop everything but the first row then move on to the next ID?

4

3 回答 3

0

无需循环，但使用偏移选项。以下摘自文档。

offset
Optional. It is the physical offset from the current row in the table.
If this parameter is omitted, the default is 1.

example; lead(date_out) means next value
lead(date_out, 2) means 2nd row after current row
lead(date_out, 3) 3rd row after current row and so on.

在您的代码中；使用以下代码段；

lead(date_out) over ( partition by (ID) order by ID) as next_out,
lead(date_out, 2) over ( partition by (ID) order by ID) as next_out2,
lead(date_out, 3) over ( partition by (ID) order by ID) as next_out3
于 2018-04-17T01:24:40.470 回答
0

这是一种方法，它假设您只希望每个ID. 您可以分配一个行号，然后按以下方式聚合ID：

WITH cte AS (
    SELECT ID, date_IN, date_out,
        ROW_NUMBER() OVER (PARTITION BY ID ORDER BY date_IN) rn
    FROM yourTable
)

SELECT
    ID,
    MAX(CASE WHEN rn = 1 THEN date_IN END) AS date_IN,
    MAX(CASE WHEN rn = 1 THEN date_out END) AS date_out,
    MAX(CASE WHEN rn = 2 THEN next_IN END) AS next_in_1,
    MAX(CASE WHEN rn = 2 THEN date_out END) AS next_out_2,
    MAX(CASE WHEN rn = 3 THEN date_IN END) AS next_in_2,
    MAX(CASE WHEN rn = 3 THEN date_out END) AS next_out_2
FROM cte
GROUP BY ID
于 2018-04-17T01:27:24.953 回答
0 
   WITH TAB AS(
SELECT 1 ID,  CAST('2018/01/01' AS DATE) DATE_IN, CAST('2018/01/02' AS DATE) DATE_OUT FROM DUAL
UNION
SELECT 1, CAST('2018/01/03' AS DATE)   , CAST('2018/01/04' AS DATE) FROM DUAL
UNION
SELECT 1, CAST('2018/01/05' AS DATE)   , CAST('2018/01/08' AS DATE) FROM DUAL
UNION
SELECT 1, CAST('2018/01/09' AS DATE)   , CAST('2018/01/10' AS DATE) FROM DUAL
UNION
SELECT 1, CAST('2018/01/11' AS DATE)   , CAST('2018/01/12' AS DATE) FROM DUAL
UNION
SELECT 2, CAST('2018/01/01' AS DATE)   , CAST('2018/01/05' AS DATE) FROM DUAL
UNION
SELECT 2, CAST('2018/01/07' AS DATE)   , CAST('2018/01/09' AS DATE) FROM DUAL
) --select * from tab;
, LEAF_CALC AS(   --CONNECTING THE DATE_OUTS
SELECT 
ID
,SYS_CONNECT_BY_PATH(DATE_OUT, '$') HRCHY
, LEVEL LVL
, CONNECT_BY_ISLEAF ISLEAF
FROM TAB
CONNECT BY PRIOR DATE_OUT < DATE_IN 
START WITH ID = 1
) --SELECT * FROM LEAF_CALC;
, DATA_SORT AS( --ADDING ALL DATE_OUTS IN 1 ROW
SELECT
P.ID, P.HRCHY
FROM LEAF_CALC P,
(SELECT ID, MAX(LVL) MAXLVL FROM
LEAF_CALC
GROUP BY ID) C
WHERE P.ID = C.ID 
AND P.LVL = C.MAXLVL
)--SELECT * FROM DATA_SORT
--SEGREGATING ALL DATES USING REGEXP_SUBSTR
SELECT
ID
, REGEXP_SUBSTR(HRCHY, '[^$]+', 1, 1) DATE_IN
, REGEXP_SUBSTR(HRCHY, '[^$]+', 1, 2) NEXT_OUT
, REGEXP_SUBSTR(HRCHY, '[^$]+', 1, 3) NEXT_OUT2
, COALESCE(REGEXP_SUBSTR(HRCHY, '[^$]+', 1, 4), 'NA') NEXT_OUT3
, COALESCE(REGEXP_SUBSTR(HRCHY, '[^$]+', 1, 5), 'NA') NEXT_OUT4
FROM DATA_SORT;
于 2018-04-17T04:00:22.640 回答