Oracle SQL loop LEAD() through partition - sql

Question

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?

Answer 1

Here is one approach, which assumes that you only expect to have a maximum of three date pairs per ID. You may assign a row number and then aggregate by 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

Answer 2

No need to do a loop but use the offset option. Below is lifted from the documentation.
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.
in your code; use below snippet;
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

Answer 3

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;