I want to write the code by Having a node at level N (for example 3) of the tree, find all members up to level M (for example 6).How can I write this code? what is the best idea for this problem?
I am new in oracle , please explain me and help me for resolve this problem .
CREATE TABLE tree (
node_id INT,
parent_id INT
)
INSERT INTO
tree
SELECT 1, NULL FROM DUAL
UNION ALL SELECT 2, 1 FROM DUAL
UNION ALL SELECT 3, 1 FROM DUAL
UNION ALL SELECT 4, 2 FROM DUAL
UNION ALL SELECT 5, 2 FROM DUAL
UNION ALL SELECT 6, 3 FROM DUAL
UNION ALL SELECT 7, 3 FROM DUAL
UNION ALL SELECT 8, 4 FROM DUAL
UNION ALL SELECT 9, 4 FROM DUAL
UNION ALL SELECT 10, 5 FROM DUAL
UNION ALL SELECT 11, 5 FROM DUAL
UNION ALL SELECT 12, 6 FROM DUAL
UNION ALL SELECT 13, 6 FROM DUAL
UNION ALL SELECT 14, 7 FROM DUAL
UNION ALL SELECT 15, 7 FROM DUAL
15 rows affected
SELECT
node_id, parent_id, level
FROM
tree
WHERE
level < 3
CONNECT BY
PRIOR node_id = parent_id
START WITH
parent_id = 3
ORDER BY
level, node_id
NODE_ID
PARENT_ID
LEVEL
6
3
1
7
3
1
12
6
2
13
6
2
14
7
2
15
7
2
fiddle
Related
I am looking for ideas on how to group numbers into low and high ranges in Oracle SQL. I looking to to avoid cursors...any ideas welcome
Example input
ID
LOW
HIGH
A
0
2
A
2
3
A
3
5
A
9
11
A
11
13
A
13
15
B
0
1
B
1
4
B
7
9
B
11
12
B
12
17
B
17
18
Which would result in the following grouping into ranges
ID
LOW
HIGH
A
0
5
A
9
15
B
0
4
B
7
9
B
11
18
This is a Gaps & Islands problem. You can use the traditional solution.
For example:
select max(id) as id, min(low) as low, max(high) as high
from (
select x.*, sum(i) over(order by id, low) as g
from (
select t.*,
case when low = lag(high) over(partition by id order by low)
and id = lag(id) over(partition by id order by low)
then 0 else 1 end as i
from t
) x
) y
group by g
Result:
ID LOW HIGH
--- ---- ----
A 0 5
A 9 15
B 0 4
B 7 9
B 11 18
See running example at db<>fiddle.
From Oracle 12, you should use MATCH_RECOGNIZE for row-by-row pattern matching:
SELECT *
FROM table_name
MATCH_RECOGNIZE(
PARTITION BY id
ORDER BY low, high
MEASURES
FIRST(low) AS low,
MAX(high) AS high
PATTERN (overlapping* last_row)
DEFINE
overlapping AS NEXT(low) <= MAX(high)
)
Which, for the sample data:
CREATE TABLE table_name (id, low, high) AS
SELECT 'A', 0, 2 FROM DUAL UNION ALL
SELECT 'A', 2, 3 FROM DUAL UNION ALL
SELECT 'A', 3, 5 FROM DUAL UNION ALL
SELECT 'A', 9, 11 FROM DUAL UNION ALL
SELECT 'A', 11, 13 FROM DUAL UNION ALL
SELECT 'A', 13, 15 FROM DUAL UNION ALL
SELECT 'B', 0, 1 FROM DUAL UNION ALL
SELECT 'B', 1, 4 FROM DUAL UNION ALL
SELECT 'B', 7, 9 FROM DUAL UNION ALL
SELECT 'B', 11, 12 FROM DUAL UNION ALL
SELECT 'B', 12, 17 FROM DUAL UNION ALL
SELECT 'B', 17, 18 FROM DUAL UNION ALL
SELECT 'C', 0, 10 FROM DUAL UNION ALL
SELECT 'C', 1, 3 FROM DUAL UNION ALL
SELECT 'C', 5, 8 FROM DUAL UNION ALL
SELECT 'C', 9, 15 FROM DUAL UNION ALL
SELECT 'C', 10, 14 FROM DUAL UNION ALL
SELECT 'C', 11, 13 FROM DUAL;
Outputs:
ID
LOW
HIGH
A
0
5
A
9
15
B
0
4
B
7
9
B
11
18
C
0
15
fiddle
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
I know a little bit of sql, only the basic, now I need to create a analytic query but can't do this yet.
I have 2 tables on my db oracle, client and exams:
I am tried a lot of ways to get the mean of exams by client, but no success yet.4
The result expected is:
exams = 13
clients = 6
13/6= 2.166666666...7
How can I do that?
If you have clients who have not taken any exams then you want:
SELECT AVG(COUNT(e.nu_ordem)) AS avg_exames_by_client
FROM cliente c
LEFT OUTER JOIN exames e
ON (c.id = e.id_cliente)
GROUP BY c.id;
or:
SELECT (SELECT COUNT(*) FROM exames) / (SELECT COUNT(*) FROM cliente)
AS avg_exames_by_client
FROM DUAL;
Which, for the sample data:
CREATE TABLE cliente (id PRIMARY KEY) AS
SELECT 1 FROM DUAL UNION ALL
SELECT 2 FROM DUAL UNION ALL
SELECT 3 FROM DUAL UNION ALL
SELECT 4 FROM DUAL UNION ALL
SELECT 5 FROM DUAL UNION ALL
SELECT 6 FROM DUAL;
CREATE TABLE exames (nu_ordem PRIMARY KEY, id_cliente) AS
SELECT 1, 1 FROM DUAL UNION ALL
SELECT 2, 5 FROM DUAL UNION ALL
SELECT 3, 5 FROM DUAL UNION ALL
SELECT 4, 2 FROM DUAL UNION ALL
SELECT 5, 6 FROM DUAL UNION ALL
SELECT 6, 1 FROM DUAL UNION ALL
SELECT 7, 1 FROM DUAL UNION ALL
SELECT 8, 4 FROM DUAL UNION ALL
SELECT 9, 5 FROM DUAL UNION ALL
SELECT 10, 3 FROM DUAL UNION ALL
SELECT 11, 6 FROM DUAL UNION ALL
SELECT 12, 2 FROM DUAL UNION ALL
SELECT 13, 1 FROM DUAL;
Both output:
AVG_EXAMES_BY_CLIENT
2.166666666666666667
If you then add a couple of clients but no more exams:
INSERT INTO cliente (id)
SELECT 7 FROM DUAL UNION ALL
SELECT 8 FROM DUAL
Then the average is:
AVG_EXAMES_BY_CLIENT
1.625
db<>fiddle here
You can try below formula to get the result -
SELECT COUNT(*)/COUNT(DISTINCT id_cliente)
FROM exams;
I am trying to figure out the root parent in a table with hierarchical data. The following example works as expected but I need to do something extra. I want to avoid the query to ignore null id1 and show the (root parent - 1) if the root parent is null.
with table_a ( id1, child_id ) as (
select null, 1 from dual union all
select 1, 2 from dual union all
select 2, 3 from dual union all
select 3, NULL from dual union all
select 4, NULL from dual union all
select 5, 6 from dual union all
select 6, 7 from dual union all
select 7, 8 from dual union all
select 8, NULL from dual
)
select connect_by_root id1 as id, id1 as root_parent_id
from table_a
where connect_by_isleaf = 1
connect by child_id = prior id1
order by id 1
This brings up the following data
4 4
6 5
7 5
8 5
5 5
3 null
null null
2 null
1 null
what I want is
3 1
1 1
2 1
4 4
7 5
8 5
5 5
6 5
is it possible?
Thanks for the help
Using a recursive CTE you can do:
with table_a ( id1, child_id ) as (
select null, 1 from dual union all
select 1, 2 from dual union all
select 2, 3 from dual union all
select 3, NULL from dual union all
select 4, NULL from dual union all
select 5, 6 from dual union all
select 6, 7 from dual union all
select 7, 8 from dual union all
select 8, NULL from dual
),
n (s, e) as (
select id1 as s, child_id as e from table_a where id1 not in
(select child_id from table_a
where id1 is not null and child_id is not null)
union all
select n.s, a.child_id
from n
join table_a a on a.id1 = n.e
)
select
coalesce(e, s) as c, s
from n
order by s
Result:
C S
- -
3 1
1 1
2 1
4 4
5 5
7 5
8 5
6 5
As a side note, "Recursive CTEs" are more flexible than the old-school CONNECT BY.
This looks like it works but it may be incorrect, as I do not quite understand the logic behind choosing 1 for this, looks arbitrary to me, not much like real data will be.
As Hogan has asked already, it would be helpful if you could perhaps provide an explanation or an expanded data set to test this hierarchy.
with table_a ( id1, child_id ) as (
select null, 1 from dual union all
select 1, 2 from dual union all
select 2, 3 from dual union all
select 3, NULL from dual union all
select 4, NULL from dual union all
select 5, 6 from dual union all
select 6, 7 from dual union all
select 7, 8 from dual union all
select 8, NULL from dual
)
select connect_by_root id1 as id, id1 as root_parent_id
from table_a
where connect_by_isleaf = 1 and connect_by_root id1 is not null
connect by nocycle child_id = prior nvl(id1, 1)
order by 2, 1;
Sample execution:
FSITJA#dbd01 2019-07-19 13:51:13> with table_a ( id1, child_id ) as (
2 select null, 1 from dual union all
3 select 1, 2 from dual union all
4 select 2, 3 from dual union all
5 select 3, NULL from dual union all
6 select 4, NULL from dual union all
7 select 5, 6 from dual union all
8 select 6, 7 from dual union all
9 select 7, 8 from dual union all
10 select 8, NULL from dual
11 )
12 select connect_by_root id1 as id, id1 as root_parent_id
13 from table_a
14 where connect_by_isleaf = 1 and connect_by_root id1 is not null
15 connect by nocycle child_id = prior nvl(id1, 1)
16 order by 2, 1;
ID ROOT_PARENT_ID
---------- --------------
1 1
2 1
3 1
4 4
5 5
6 5
7 5
8 5
8 rows selected.
I have a parent child relationship stored in a table
If I query with 1, I want to retrieve all the LAST LEVEL Children of 1.
I should get back 5, 6,8.
Ignore any cyclical data that comes along.
Please see the attached image
Use a hierarchical query and restrict the output to only the non-cyclic leaf rows using the pseudocolumns CONNECT_BY_ISLEAF and CONNECT_BY_ISCYCLE:
Oracle Setup:
CREATE TABLE your_table ( parent_column, child_column ) AS
SELECT 1, 2 FROM DUAL UNION ALL
SELECT 2, 3 FROM DUAL UNION ALL
SELECT 3, 4 FROM DUAL UNION ALL
SELECT 4, 5 FROM DUAL UNION ALL
SELECT 3, 6 FROM DUAL UNION ALL
SELECT 3, 7 FROM DUAL UNION ALL
SELECT 7, 8 FROM DUAL;
Query:
SELECT child_column
FROM your_table
WHERE CONNECT_BY_ISLEAF = 1
AND CONNECT_BY_ISCYCLE = 0
START WITH parent_column = 1
CONNECT BY NOCYCLE PRIOR child_column = parent_column
Output:
CHILD_COLUMN
------------
5
6
8