Oracle - Convert value from rows into ranges - sql

Are there any techniques that would allow a row set like this
WITH
base AS
(
SELECT 1 N FROM DUAL UNION ALL
SELECT 2 N FROM DUAL UNION ALL
SELECT 3 N FROM DUAL UNION ALL
SELECT 6 N FROM DUAL UNION ALL
SELECT 7 N FROM DUAL UNION ALL
SELECT 17 N FROM DUAL UNION ALL
SELECT 18 N FROM DUAL UNION ALL
SELECT 19 N FROM DUAL UNION ALL
SELECT 21 N FROM DUAL
)
SELECT a.N
FROM base a
to yield results
1 3
6 7
17 19
21 21
It is in effect a rows to ranges operation.
I'm playing in Oracle Land, and would appreciate any suggestions.

I feel like this can probably be improved on, but it works:
WITH base AS (
SELECT 1 N FROM DUAL UNION ALL
SELECT 2 N FROM DUAL UNION ALL
SELECT 3 N FROM DUAL UNION ALL
SELECT 6 N FROM DUAL UNION ALL
SELECT 7 N FROM DUAL UNION ALL
SELECT 17 N FROM DUAL UNION ALL
SELECT 18 N FROM DUAL UNION ALL
SELECT 19 N FROM DUAL UNION ALL
SELECT 21 N FROM DUAL
)
, lagged AS
(
SELECT n, LAG(n) OVER (ORDER BY n) lag_n FROM base
)
, groups AS
(
SELECT n, row_number() OVER (ORDER BY n) groupnum
FROM lagged
WHERE lag_n IS NULL OR lag_n < n-1
)
, grouped AS
(
SELECT n, (SELECT MAX(groupnum) FROM groups
WHERE groups.n <= base.n
) groupnum
FROM base
)
SELECT groupnum, MIN(n), MAX(n)
FROM grouped
GROUP BY groupnum
ORDER BY groupnum

Another way:
WITH base AS
(
SELECT 1 N FROM DUAL UNION ALL
SELECT 2 N FROM DUAL UNION ALL
SELECT 3 N FROM DUAL UNION ALL
SELECT 6 N FROM DUAL UNION ALL
SELECT 7 N FROM DUAL UNION ALL
SELECT 17 N FROM DUAL UNION ALL
SELECT 18 N FROM DUAL UNION ALL
SELECT 19 N FROM DUAL UNION ALL
SELECT 21 N FROM DUAL
)
select min(n), max(n) from
(
select n, connect_by_root n root from base
connect by prior n = n-1
start with n not in (select n from base b
where exists (select 1 from base b1 where b1.n = b.n-1)
)
)
group by root
order by root

Yet another way:
with base as (
select 1 n from dual union all
select 2 n from dual union all
select 3 n from dual union all
select 6 n from dual union all
select 7 n from dual union all
select 17 n from dual union all
select 18 n from dual union all
select 19 n from dual union all
select 21 n from dual)
select a,b
from (select a
,case when b is not null and a is not null
then b
else lead(n) over (order by n)
end b
from (select n
,a
,b
from (select n
,case n-1 when lag (n) over (order by n) then null else n end a
,case n+1 when lead (n) over (order by n) then null else n end b
from base)
where a is not null
or b is not null))
where a is not null
order by a

Related

Get parent id from level with Oracle SQL

I have a hierarchical structure defined by level and order of elements. Is it possible to create "parent_id" column with Oracle SQL without using procedures?
I need to generate red values:
test data:
with t as
(
select 1 id, 'element1' name, 1 level_ from dual union all
select 2 id, 'element2' name, 2 level_ from dual union all
select 3 id, 'element3' name, 3 level_ from dual union all
select 4 id, 'element4' name, 3 level_ from dual union all
select 5 id, 'element5' name, 3 level_ from dual union all
select 6 id, 'element6' name, 3 level_ from dual union all
select 7 id, 'element7' name, 2 level_ from dual union all
select 8 id, 'element8' name, 3 level_ from dual union all
select 9 id, 'element9' name, 4 level_ from dual union all
select 10 id, 'element10' name, 4 level_ from dual union all
select 11 id, 'element11' name, 1 level_ from dual union all
select 12 id, 'element12' name, 2 level_ from dual union all
select 13 id, 'element13' name, 3 level_ from dual union all
select 14 id, 'element14' name, 4 level_ from dual union all
select 15 id, 'element15' name, 4 level_ from dual union all
select 16 id, 'element16' name, 3 level_ from dual union all
select 17 id, 'element17' name, 4 level_ from dual union all
select 18 id, 'element18' name, 4 level_ from dual union all
select 19 id, 'element19' name, 1 level_ from dual
)
select * from t
From Oracle 12, you can use MATCH_RECOGNIZE:
select *
from t
MATCH_RECOGNIZE (
ORDER BY id DESC
MEASURES
child.id AS id,
child.name AS name,
child.lvl AS lvl,
parent.id AS parent_id
ONE ROW PER MATCH
AFTER MATCH SKIP TO NEXT ROW
PATTERN (child ancestors*? (parent | $))
DEFINE
parent AS lvl = child.lvl - 1
)
ORDER BY id
Or, again from Oracle 12, a LATERAL join:
select *
from t c
LEFT OUTER JOIN LATERAL(
SELECT p.id AS parent_id
FROM t p
WHERE c.id > p.id
AND c.lvl = p.lvl + 1
ORDER BY id DESC
FETCH FIRST ROW ONLY
)
ON (1 = 1)
ORDER BY id
Or, in earlier versions:
SELECT id, name, lvl, parent_id
FROM (
SELECT c.*,
p.id AS parent_id,
ROW_NUMBER() OVER (PARTITION BY c.id ORDER BY p.id DESC) AS rn
FROM t c
LEFT OUTER JOIN t p
ON (c.id > p.id AND c.lvl = p.lvl + 1)
)
WHERE rn = 1
ORDER BY id
Which, for the sample data:
CREATE TABLE t (id, name, lvl ) as
select 1, 'element1', 1 from dual union all
select 2, 'element2', 2 from dual union all
select 3, 'element3', 3 from dual union all
select 4, 'element4', 3 from dual union all
select 5, 'element5', 3 from dual union all
select 6, 'element6', 3 from dual union all
select 7, 'element7', 2 from dual union all
select 8, 'element8', 3 from dual union all
select 9, 'element9', 4 from dual union all
select 10, 'element10', 4 from dual union all
select 11, 'element11', 1 from dual union all
select 12, 'element12', 2 from dual union all
select 13, 'element13', 3 from dual union all
select 14, 'element14', 4 from dual union all
select 15, 'element15', 4 from dual union all
select 16, 'element16', 3 from dual union all
select 17, 'element17', 4 from dual union all
select 18, 'element18', 4 from dual union all
select 19, 'element19', 1 from dual;
All output:
ID
NAME
LVL
PARENT_ID
1
element1
1
null
2
element2
2
1
3
element3
3
2
4
element4
3
2
5
element5
3
2
6
element6
3
2
7
element7
2
1
8
element8
3
7
9
element9
4
8
10
element10
4
8
11
element11
1
null
12
element12
2
11
13
element13
3
12
14
element14
4
13
15
element15
4
13
16
element16
3
12
17
element17
4
16
18
element18
4
16
19
element19
1
null
db<>fiddle here

Impala list all dates between 2 dates

Can HUE Impala create a column which shows all dates between a specified start and end dates?
I want to list a column with date values.
You can use this sql.
select a.Date_Range
from (
select date1 - INTERVAL (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.a) ) DAY as Date_Range
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as d
) a
where a.Date_Range <= date2
Explanation -
You first create a range of numbers. And then add it to the date1 to get a range. Then you can pick your date range less than date2.

To find the starting and ending points in a sequence

I've a table (T1) with one column (C1) with the below values
1
2
3
5
6
8
9
10
I want the output to print the continuous sequences with start and ending points like below.
1-3
5-6
8-10
Could you please help?
Any Database is fine.
Oracle: sample data first, while code you really need begins at line #11.
SQL> with t1 (c1) as
2 (select 1 from dual union all
3 select 2 from dual union all
4 select 3 from dual union all
5 select 5 from dual union all
6 select 6 from dual union all
7 select 8 from dual union all
8 select 9 from dual union all
9 select 10 from dual
10 )
11 select min(c1), max(c1)
12 from (select c1, c1 - row_number() over (order by c1) rn
13 from t1
14 )
15 group by rn
16 order by rn;
MIN(C1) MAX(C1)
---------- ----------
1 3
5 6
8 10
SQL>
You can use the following query. I have tested it with SQL Server, but I think it will work without modifications in Oracle:
create table t1(c1 int);
insert into t1
select *
from (values(1),(2),(3),(5),(6),(8),(9),(10))t(x);
select case when count(*) >1 then
concat(min(c1),'-',max(c1))
else concat(max(c1),'')
end as concat_cs
from (
select c1
,ROW_NUMBER() over(order by c1 asc) as rnk
,c1 - ROW_NUMBER() over(order by c1 asc) as grp
from t1
)x
group by x.grp
Output
concat_cs
1-3
5-6
8-10
with stab as (
select 1 as val from dual union all
select 2 as val from dual union all
select 3 as val from dual union all
select 5 as val from dual union all
select 6 as val from dual union all
select 8 as val from dual union all
select 9 as val from dual union all
select 10 as val from dual union all
select 13 as val from dual union all
select 15 as val from dual union all
select 16 as val from dual union all
select 17 as val from dual union all
select 18 as val from dual union all
select 19 as val from dual union all
select 23 as val from dual
),sq2 as(
select
row_number() over(order by 1) as rownumber,val
from stab
)
select
a.val,b.val
from sq2 A
join sq2 b on b.rownumber = a.rownumber+2
where mod(A.rownumber,3)=1
Output:
1 3
5 8
9 13
15 17
18 23

Count in each row the number of second column

Here is answer to request
The question is how to count by each selected_date e.x:
2012-02-10: 1
2012-02-15: 0
2012-02-14: 3
2012-02-11: 0
How to make this request
Here is the request to get above answer
select selected_date, date1 from
(select selected_date from
(select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) selected_date from
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
where selected_date between '2012-02-10' and '2012-02-15' ) vv left join clicker on clicker.date1=vv.selected_date
This might work:
SELECT selected_date, SUM(CASE WHEN date1 IS NULL THEN 0 ELSE 1 END) FROM table
GROUP BY selected_date
So , basically this?
SELECT t.selected_date, COUNT(t.date1)
FROM ( Your Query Here )
GROUP BY t.selected_date
COUNT() ignores NULL values by default, so it will count only matches .

How can I find unoccupied id numbers in a table?

In my table I want to see a list of unoccupied id numbers in a certain range.
For example there are 10 records in my table with id's: "2,3,4,5,10,12,16,18,21,22" and say that I want to see available ones between 1 and 25. So I want to see a list like:
1,6,7,89,11,13,14,15,17,19,20,23,24,25
How should I write my sql query?
Select the numbers form 1 to 25 and show only those that are not in your table
select n from
( select rownum n from dual connect by level <= 25)
where n not in (select id from table);
Let's say you a #numbers table with three numbers -
CREATE TABLE #numbers (num INT)
INSERT INTO #numbers (num)
SELECT 1
UNION
SELECT 3
UNION
SELECT 6
Now, you can use CTE to generate numbers recursively from 1-25 and deselect those which are in your #numbers table in the WHERE clause -
;WITH n(n) AS
(
SELECT 1
UNION ALL
SELECT n+1 FROM n WHERE n < 25
)
SELECT n FROM n
WHERE n NOT IN (select num from #numbers)
ORDER BY n
OPTION (MAXRECURSION 25);
You can try using the "NOT IN" clause:
select
u1.user_id + 1 as start
from users as u1
left outer join users as u2 on u1.user_id + 1 = u2.id
where
u2.id is null
see also SQL query to find Missing sequence numbers
You need LISTAGG to get the output in a single row.
SQL> WITH DATA1 AS(
2 SELECT LEVEL rn FROM dual CONNECT BY LEVEL <=25
3 ),
4 data2 AS(
5 SELECT 2 num FROM dual UNION ALL
6 SELECT 3 FROM dual UNION ALL
7 SELECT 4 from dual union all
8 SELECT 5 FROM dual UNION ALL
9 SELECT 10 FROM dual UNION ALL
10 SELECT 12 from dual union all
11 SELECT 16 from dual union all
12 SELECT 18 FROM dual UNION ALL
13 SELECT 21 FROM dual UNION ALL
14 SELECT 22 FROM dual)
15 SELECT listagg(rn, ',')
16 WITHIN GROUP (ORDER BY rn) num_list FROM data1
17 WHERE rn NOT IN(SELECT num FROM data2)
18 /
NUM_LIST
----------------------------------------------------
1,6,7,8,9,11,13,14,15,17,19,20,23,24,25
SQL>