I'm writing simple SQL Statement where I need to found my boss's boss. For example if there is three ranks in the column like (lil boss,boss and big boss) and I input lil boss id it need to return the the big boss id. I think that this query must be write white case or decode statements but i have struggles with that. In this code everything would run fine except if I input lil boss ID.Can I use nested decode or case statements? I think that UNION statement would be solution also .
If my table has three column:
id
status
boss_id
1
big boss
null
2
big boss
null
3
lil boss
4
4
boss
2
If I input 3 it have to return 2
select decode(status,'big boss', id,
'boss',boss_id,
'lil boss',boss_id)
from bosses
where id=123113;
I need to found my boss's boss
Use a hierarchical query:
SELECT CONNECT_BY_ROOT id AS root_id,
CONNECT_BY_ROOT status AS root_status,
id,
status
FROM bosses
WHERE LEVEL = 3
OR (LEVEL < 3 AND CONNECT_BY_ISLEAF = 1)
START WITH id = 2
CONNECT BY PRIOR boss_id = id;
Which, for the sample data:
CREATE TABLE bosses (id, status, boss_id) AS
SELECT 1, 'big boss', null FROM DUAL UNION ALL
SELECT 2, 'big boss', null FROM DUAL UNION ALL
SELECT 3, 'lil boss', 4 FROM DUAL UNION ALL
SELECT 4, 'boss', 2 FROM DUAL;
Outputs:
ROOT_ID
ROOT_STATUS
ID
STATUS
3
lil boss
2
big boss
and if you START WITH id = 2:
SELECT CONNECT_BY_ROOT id AS root_id,
CONNECT_BY_ROOT status AS root_status,
id,
status
FROM bosses
WHERE LEVEL = 3
OR (LEVEL < 3 AND CONNECT_BY_ISLEAF = 1)
START WITH id = 2
CONNECT BY PRIOR boss_id = id;
Then it outputs:
ROOT_ID
ROOT_STATUS
ID
STATUS
2
big boss
2
big boss
fiddle
I was able to come to a solution to this problem. For this purpose I used hierarchical queries and the solution of #MT0. Here is the code that not need to use level. I think it is the best and simpliest solution to the problem. Thanks for helping me!
select id as bossid,
status as boss_status
from bosses b
where b.status='big boss'
start with id ='4'
connect by prior boss_id = id;
Starting from the generic node (or leaf) how to get the root node?
SELECT
ID,MSG_ID,PARENT_ID
FROM
TABLE_X
CONNECT BY PRIOR PARENT_ID = ID;
ID MSG_ID PARENT_ID
4 3 NULL
5 93bea0f71b07-4037-9009-f148fa39bb62 4
4 3 NULL
6 6f5f5d4ab1ec-4f00-8448-7a6dfa6461b2 4
4 3 NULL
7 3 NULL
8 7e0fae569637-4d29-9075-c273eb39ae8e 7
7 3 NULL
9 8a3e7485b3e8-45b1-a31d-c52fd32111c0 7
7 3 NULL
10 fcc622d5af92-4e61-8d7c-add3da359a8b 7
7 3 NULL
How to get the root msg_id?
You were on the right path, but you are missing two essential ingredients.
First, to indicate the starting point, you need to use the start with clause. Obviously, it will be something like start with id = <input value>. You didn't tell us how you will provide the input value. The most common approach is to use bind variables (best for many reasons); I named the bind variable input_id in the query below.
Second, you only want the "last" row (since you are navigating the tree in the opposite direction: towards the root, not from the root; so the root is now the "leaf" as you navigate this way). For that you can use the connect_by_isleaf pseudocolumn in the where clause.
So, the query should look like this: (note that I am only selecting the root message id, as that is all you requested; if you need more columns, include them in select)
select msg_id
from table_x
where connect_by_isleaf = 1 -- keep just the root row (leaf in this traversal)
start with id = :input_id -- to give the starting node
connect by prior parent_id = id
;
You can use a recursive query:
with cte (id, msg_id, parent_id) as (
select id, msg_id, parent_id, from mytable where id = ?
union all
select t.id, t.msg_id, t.parent_id
from mytable t
inner join cte c on c.parent_id = t.id
)
select * from cte where parent_id is null
I would like to get help for a hierarchical query (Oracle 11gR2). I have a hard time with those kind of queries...
In fact, it's a 2 in 1 question (2 different approches needed).
I’m looking for a way to get the distance from all individials records to the root (not the opposite). My data are in a tree like structure:
CREATE TABLE MY_TREE
(ID_NAME VARCHAR2(1) PRIMARY KEY,
PARENT_ID VARCHAR2(1),
PARENT_DISTANCE NUMBER(2)
);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('A',NULL,NULL);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('B','A',1);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('C','B',3);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('D','B',5);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('E','C',7);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('F','D',11);
INSERT INTO MY_TREE (ID_NAME,PARENT_ID,PARENT_DISTANCE) VALUES('G','D',13);
Hierarchically, my data look like this (but I have multiple independents roots and many more levels):
In the first approch, I'm looking for a query that will give me this result:
LEVEL ROOT NODE ID_NAME ROOT_DISTANCE
----- ---- ---- ------- -------------
1 A null A null
2 A null B 1
3 A B C 4
4 A B E 11
3 A B D 6
4 A D F 17
4 A D G 19
In this result,
the "NODE" column mean the ID_NAME of the closest split element
the "ROOT_DISTANCE" column mean the distance from an element to the root (ex: the ROOT_DISTANCE for ID_NAME=G is the distance from G to A: G(13)+D(5)+B(1)=19)
In this approch, I will always specify a maximum of 2 roots.
The second approch must be a PL/SQL script that will do the same calculation (ROOT_DISTANCE), but in a iterative way, and that will write the result in a new table. I want to run this script one time, so all roots (~1000) will be processed.
Here's the way I see the script:
For all roots, we need to find associated leafs and then calculate the distance from the leaf to the root (for all element between the leaf and the root) and put this into a table.
This script is needed for "performance perspectives", so if an element have been already calculated (ex: a split node that was calculated by another leaf), we need to stop the calculation and pass to the next leaf because we already know the result from there to the root. For example, if the system calculates E-C-B-A, and then F-D-B-A, the B-A section should not be calcultated again because it was done in the first pass.
You can awnser one or both of those questions, but i will need the awnser to those two questions.
Thank you!
Try this one:
WITH brumba(le_vel,root,node,id_name,root_distance) AS (
SELECT 1 as le_vel, id_name as root, null as node, id_name, to_number(null) as root_distance
FROM MY_TREE WHERE parent_id IS NULL
UNION ALL
SELECT b.le_vel + 1, b.root,
CASE WHEN 1 < (
SELECT count(*) FROM MY_TREE t1 WHERE t1.parent_id = t.parent_id
)
THEN t.parent_id ELSE b.node
END,
t.id_name, coalesce(b.root_distance,0)+t.parent_distance
FROM MY_TREE t
JOIN brumba b ON b.id_name = t.parent_id
)
SELECT * FROM brumba
Demo: https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=d5c231055e989c3cbcd763f4b3d3033f
There is no need for "the second approch" using PL/SQL - the above SQL will calculate results for all root nodes (which have null in parent_id column) at once.
Just add a prefix either INSERT INTO tablename(col1,col2, ... colN) ... or CREATE TABLE name AS ... to the above query.
The above demo contains an example for the latter option CREATE TABLE xxx AS query
Here's one option which shows how to get the first part of your question:
SQL> with temp as
2 (select level lvl,
3 ltrim(sys_connect_by_path(id_name, ','), ',') path
4 from my_tree
5 start with parent_id is null
6 connect by prior id_name = parent_id
7 ),
8 inter as
9 (select t.lvl,
10 t.path,
11 regexp_substr(t.path, '[^,]+', 1, column_value) col
12 from temp t,
13 table(cast(multiset(select level from dual
14 connect by level <= regexp_count(path, ',') + 1
15 ) as sys.odcinumberlist ))
16 )
17 select i.lvl,
18 i.path,
19 sum(m.parent_distance) dis
20 from inter i join my_tree m on m.id_name = i.col
21 group by i.lvl, i.path
22 order by i.path;
LVL PATH DIS
---- ---------- ----------
1 A
2 A,B 1
3 A,B,C 4
4 A,B,C,E 11
3 A,B,D 6
4 A,B,D,F 17
4 A,B,D,G 19
7 rows selected.
SQL>
Here is how you can solve this with a hierarchical (connect by) query.
In most hierarchical problems, hierarchical queries will be faster than recursive queries (recursive with clause). However, your question is not purely hierarchical - you need to compute the distance to root, and unlike recursive with, you can't get that in a single pass with hierarchical queries. So it will be interesting to hear - from you! - whether there are any significant performance differences between the approaches. For what it's worth, on the very small data sample you provided, the Optimizer estimated a cost of 5 using connect by, vs. 48 for the recursive solution; whether that means anything in real life you will find out and hopefully you will let us know, too.
In the hierarchical query, I flag out the parents that have two or more children (I use an analytic function for this, to avoid a join). Then I build the hierarchy, and in the last step I aggregate to get the required bits. As in the recursive solution, this should give you everything you need - for all roots and all nodes - in a single SQL query; there is no need for PL/SQL code.
with
branching (id_name, parent_id, parent_distance, b) as (
select id_name, parent_id, parent_distance,
case when count(*) over (partition by parent_id) > 1 then 'y' end
from my_tree
)
, hierarchy (lvl, leaf, id_name, parent_id, parent_distance, b) as (
select level, connect_by_root id_name, id_name, parent_id, parent_distance, b
from branching
connect by id_name = prior parent_id
)
select max(lvl) as lvl,
min(id_name) keep (dense_rank last order by lvl) as root,
leaf as id_name,
min(decode(b, 'y', parent_id))
keep (dense_rank first order by decode(b, 'y', lvl)) as node,
sum(parent_distance) as root_distance
from hierarchy
group by leaf;
LVL ROOT ID_NAME NODE ROOT_DISTANCE
--- ------- ------- ------- -------------
1 A A
2 A B 1
3 A C B 4
3 A D B 6
4 A E B 11
4 A F D 17
4 A G D 19
I have a table that shows the relationships between people a bit like this.
id linked_id
1 2
1 3
3 4
4 1
There is no apparent order to the table or the relationships.
I'm trying to find a way to list all the ids that have any kind of link to a given id. So for examples from the table above:
id = 1 would return 1, 2, 3 and 4.
id = 2 would also return 1, 2, 3 and 4
It's an oracle database, and the query would have to be in plain SQL. Thanks for your help, this has been driving me nuts.
You could use something like this:
SELECT linked_id
FROM DATA
START WITH ID = :1
CONNECT BY NOCYCLE PRIOR ID = linked_id
OR ID = PRIOR linked_id
UNION
SELECT ID
FROM DATA
START WITH linked_id = :1
CONNECT BY NOCYCLE PRIOR ID = linked_id
OR ID = PRIOR linked_id
UNION
SELECT :1 FROM dual
I am using SQL Server 2008. I have a table like this:
UnitId ParentId UnitName
---------------------------
1 0 FirstUnit
2 1 SecondUnit One
3 1 SecondUnit Two
4 3 B
5 2 C
6 4 D
7 6 E
8 5 F
I want to get second parent of the record. For example:
If I choose unit id that equal to 8, It will bring unit id is equal to 2 to me. It needs to be SecondUnit One. or If I choose unit id that equal to 7, It will bring unit id is equal to 3 to me. It needs to be SecondUnit Two.
How can I write a SQL query this way?
It took me a while, but here it is :)
with tmp as (
select unitId, parentId, unitName, 0 as iteration
from t
where unitId = 7
union all
select parent.unitId, parent.parentId, parent.unitName, child.iteration + 1
from tmp child
join t parent on child.parentId = parent.unitId
where parent.parentId != 0
)
select top 1 unitId, parentId, unitName from tmp
order by iteration desc
Here is also a fiddle to play with.
SELECT t.*, tParent1.UnitId [FirstParent], tParent2.UnitId [SecondParent]
FROM Table t
LEFT JOIN Table tParent1 ON t.ParentId = tParent1.UnitId
LEFT JOIN Table tParent2 ON tParent1.ParentId = tParent2.UnitId
WHERE t.UnitId = <Unit ID search here>
AND NOT tParent2.UnitId IS NULL
Edit: And leave out second part of the WHERE clause if you want results returned even if they don't have a second parent.