I am struggling with some complex hierarchical data. I have successfully used a CONNECT BY query to limit the rows down to the subset that i want - and i have used SYS_CONNECT_BY_PATH to return the full tree up to the nodes of interest.
this gives me essentially some rows like this (delimited by '|'):
id path
-------------------
1, '|10|11|12|13'
2, '|10|14|15'
3, '|16|11|12|13'
4, '|16|17'
now - my challenge is to unwrap or UNPIVOT these values back into a structure like this:
id ord node
-------------
1, 1, 10
1, 2, 11
1, 3, 12
1, 4, 13
2, 1, 10
2, 2, 14
2, 3, 15
3, 1, 16
3, 2, 11
3, 3, 12
3, 4, 13
4, 1, 16
4, 2, 17
I think i am unable to use UNPIVOT directly as that is working on a fixed set of columns - which this is not.
I am playing with a PIPELINE function to unwrap this, but frankly - passing all these rows to the function is an issue since they come from another query. I am wondering if anyone has a way to UNPIVOT the values from a SYS_CONNECT_BY_PATH result set back into rows that is maybe a pure sql solution - probably with REGEX parsing...
help always appreciated - thanks
Yes, UNPIVOT operator wont do much here to help you produce the desired output.
As one of the approaches you could user regexp_count()(11g R1 version and up) regular
expression function to count all occurrences of numbers and then use regexp_substr() regular
expression function to extract the numbers as follows:
-- sample of data
SQL> with t1(id1, path1) as(
2 select 1, '|10|11|12|13' from dual union all
3 select 2, '|10|14|15' from dual union all
4 select 3, '|16|11|12|13' from dual union all
5 select 4, '|16|17' from dual
6 ),
7 occurrences(ocr) as( -- occurrences
8 select level
9 from ( select max(regexp_count(path1, '[^|]+')) as mx_ocr
10 from t1
11 ) t
12 connect by level <= t.mx_ocr
13 )
14 select id1
15 , row_number() over(partition by id1 order by id1) as ord
16 , node
17 from ( select q.id1
18 , regexp_substr(q.path1, '[^|]+', 1, o.ocr) as node
19 from t1 q
20 cross join occurrences o
21 )
22 where node is not null
23 order by id1, 2, node
24 ;
Result:
ID1 ORD NODE
---------- ---------- ------------------------------------------------
1 1 10
1 2 11
1 3 12
1 4 13
2 1 10
2 2 14
2 3 15
3 1 11
3 2 12
3 3 13
3 4 16
4 1 16
4 2 17
13 rows selected
As another approach, starting from 10g version and up, you could use model clause:
SQL> with t1(id1, path1) as(
2 select 1, '|10|11|12|13' from dual union all
3 select 2, '|10|14|15' from dual union all
4 select 3, '|16|11|12|13' from dual union all
5 select 4, '|16|17' from dual
6 )
7 select id1
8 , ord
9 , node
10 from t1
11 model
12 partition by ( rownum as id1)
13 dimension by ( 1 as ord)
14 measures( path1
15 , cast(null as varchar2(11)) as node
16 , nvl(regexp_count(path1, '[^|]+'), 0) as ocr )
17 rules(
18 node[for ord from 1 to ocr[1] increment 1] =
19 regexp_substr(path1[1], '[^|]+', 1, cv(ord))
20 )
21 order by id1, ord, node
22 ;
Result:
ID1 ORD NODE
---------- ---------- -----------
1 1 10
1 2 11
1 3 12
1 4 13
2 1 10
2 2 14
2 3 15
3 1 16
3 2 11
3 3 12
3 4 13
4 1 16
4 2 17
13 rows selected
SQLFiddle Demo
Related
SOURCE
Rowid_object
Rowid_object_matched
1
2
1
3
3
2
2
4
4
6
6
5
7
8
9
8
Target
Rowid_object
Rowid_object_matched
1
1
2
1
3
1
4
1
5
1
6
1
7
7
8
7
9
7
Here, we have Source like data and we want Target like result.
There are two groups in source which are in transitive match.
Need to identify these kind of record.
In Oracle, you can use:
SELECT rowid_object,
MIN(root) AS rowid_object_matched
FROM (
SELECT CONNECT_BY_ROOT(rowid_object) AS root,
rowid_object,
rowid_object_matched
FROM source
CONNECT BY NOCYCLE
PRIOR rowid_object IN (rowid_object, rowid_object_matched)
OR PRIOR rowid_object_matched IN (rowid_object, rowid_object_matched)
)
UNPIVOT (
rowid_object FOR key IN (rowid_object, rowid_object_matched)
)
GROUP BY rowid_object;
Which, for the sample data:
CREATE TABLE SOURCE (Rowid_object, Rowid_object_matched) AS
SELECT 1, 2 FROM DUAL UNION ALL
SELECT 1, 3 FROM DUAL UNION ALL
SELECT 3, 2 FROM DUAL UNION ALL
SELECT 2, 4 FROM DUAL UNION ALL
SELECT 4, 6 FROM DUAL UNION ALL
SELECT 6, 5 FROM DUAL UNION ALL
SELECT 7, 8 FROM DUAL UNION ALL
SELECT 9, 8 FROM DUAL;
Outputs:
ROWID_OBJECT
ROWID_OBJECT_MATCHED
1
1
2
1
3
1
4
1
6
1
5
1
7
7
8
7
9
7
fiddle
This question already has answers here:
Dynamic Pivot in Oracle's SQL
(10 answers)
Closed last year.
Here is my data
Id type count
1 jim 2
1 bim 2
1 sim 3
1 pim 1
2 jim 2
2 bim 1
Want to convert this data into below
Id jim bim sim pim
1 2 2 3 1
2 2 1 0 0
Tried this, its now working
select * FROM table
PIVOT
(
Min(Id)
FOR Id IN (select distinct(type) from table)
)
I'm trying to convert all distinct values of type row into columns and then assign respective values against every ID. Any suggestion please ?
Error
ORA-00936: missing expression
00936. 00000 - "missing expression"
*Cause:
*Action:
Error at Line: 26 Column: 16
Would this do?
SQL> with test (id, type, count) as
2 (select 1, 'jim', 2 from dual union all
3 select 1, 'bim', 2 from dual union all
4 select 1, 'sim', 3 from dual union all
5 select 1, 'pim', 1 from dual union all
6 select 2, 'jim', 2 from dual union all
7 select 2, 'bim', 1 from dual
8 )
9 select id, nvl(jim, 0) jim,
10 nvl(bim, 0) bim,
11 nvl(sim, 0) sim,
12 nvl(pim, 0) pim
13 from test
14 pivot
15 (min(count)
16 for type in ('jim' as jim, 'bim' as bim, 'sim' as sim, 'pim' as pim)
17 );
ID JIM BIM SIM PIM
---------- ---------- ---------- ---------- ----------
1 2 2 3 1
2 2 1 0 0
SQL>
I would like to know if the following is possible
For example I have a shoe factory. In this factory I have a production line, Every step in this production line is recorded into the oracle database.
if the shoe has completed a production step the result is = 1
example table
Shoe_nr production step result
1 1 1
1 2 1
1 3
2 1 1
2 2 1
2 3
3 1
3 2
3 3
Now the question, is it possible to filter out production step 3 where only the shoes have passed production step 2 which is equal to 1 in result.
I know if it can be done it's probably very easy but if you dont know i found out it's a little bit tricky.
Thanks,
Chris
Yes, you can do it with IN and a Subselect
select *
from shoes
where shoe.id in (
select shoe.id
from shoes
where production_step = 2
and result = 1
)
and production_step = 3
This might be one option; see comments within code (lines #1 - 12 represent sample data; you already have that and don't type it. Query you might be interested in begins at line #13).
SQL> with shoes (shoe_nr, production_step, result) as
2 -- sample data
3 (select 1, 1, 1 from dual union all
4 select 1, 2, 1 from dual union all
5 select 1, 3, null from dual union all
6 select 2, 1, 1 from dual union all
7 select 2, 2, 1 from dual union all
8 select 2, 3, null from dual union all
9 select 3, 1, null from dual union all
10 select 3, 2, null from dual union all
11 select 3, 3, null from dual
12 ),
13 -- which shoes' production step #3 should be skipped?
14 skip as
15 (select shoe_nr
16 from shoes
17 where production_step = 2
18 and result = 1
19 )
20 -- finally:
21 select a.shoe_nr, a.production_step, a.result
22 from shoes a
23 where (a.shoe_nr, a.production_step) not in (select b.shoe_nr, 3
24 from skip b
25 )
26 order by a.shoe_nr, a.production_step;
SHOE_NR PRODUCTION_STEP RESULT
---------- --------------- ----------
1 1 1
1 2 1
2 1 1
2 2 1
3 1
3 2
3 3
7 rows selected.
SQL>
If you just want the shoe_nr that satisfy the condition, you can use aggregation and a having clause:
select shoe_nr
from mytable
group by shoe_nr
having
max(case when production_step = 2 then result end) = 0
and max(case when production_step = 3 then 1 end) = 1
If you want the entire row corresponding to this shoe_nr at step 3, use window functions instead:
select 1
from (
select
t.*,
max(case when production_step = 2 then result end)
over(partition by shoe_nr) as has_completed_step_2
from mytable t
) t
where production_step = 3 and has_completed_step_2 = 0
Consider the following table:
ID Feature
1 1
1 2
1 3
2 3
2 4
2 6
3 5
3 10
3 12
4 12
4 18
5 10
5 30
I would like to group the individuals based on overlapping features. If two of these groups again have overlapping features, I would consider both as one group. This process should be repeated until there are no overlapping features between groups. The result of this procedure on the table above would be:
ID Feature Flag
1 1 A
1 2 A
1 3 A
2 3 A
2 4 A
2 6 A
3 5 B
3 10 B
3 12 B
4 12 B
4 18 B
5 10 B
5 30 B
So actually the problem I am trying to solve is finding connected components in a graph. Here [1,2,3] is the graph with ID 1 (see https://en.wikipedia.org/wiki/Connectivity_(graph_theory)). The problem is equivalent to this problem, however I would like to solve it with Oracle SQL.
Here is one way to do this, using a hierarchical ("connect by") query. The first step is to extract the initial relationships from the base data; the hierarchical query is built on the result from this first step. I added one more row to the inputs to illustrate a node that is a connected component by itself.
You marked the connected components as A and B - of course, that won't work if you have, say, 30,000 connected components. In my solution, I use the minimum node name as the marker for each connected component.
with
sample_data (id, feature) as (
select 1, 1 from dual union all
select 1, 2 from dual union all
select 1, 3 from dual union all
select 2, 3 from dual union all
select 2, 4 from dual union all
select 2, 6 from dual union all
select 3, 5 from dual union all
select 3, 10 from dual union all
select 3, 12 from dual union all
select 4, 12 from dual union all
select 4, 18 from dual union all
select 5, 10 from dual union all
select 5, 30 from dual union all
select 6, 40 from dual
)
-- select * from sample_data; /*
, initial_rel(id_base, id_linked) as (
select distinct s1.id, s2.id
from sample_data s1 join sample_data s2
on s1.feature = s2.feature and s1.id <= s2.id
)
-- select * from initial_rel; /*
select id_linked as id, min(connect_by_root(id_base)) as id_group
from initial_rel
start with id_base <= id_linked
connect by nocycle prior id_linked = id_base and id_base < id_linked
group by id_linked
order by id_group, id
;
Output:
ID ID_GROUP
------- ----------
1 1
2 1
3 3
4 3
5 3
6 6
Then, if you need to add the ID_GROUP as a FLAG to the base data, you can do so with a trivial join.
i have the following table with different prices in every week and need a numbering like in the last column. consecutive rows with same prices should have the same number like in weeks 11/12 or 18/19. but on the other side weeks 2 and 16 have the same prices but are not consecutive so they should get a different number.
w | price | r1 | need
===========================
1 167,93 1 1
2 180 1 2
3 164,72 1 3
4 147,42 1 4
5 133,46 1 5
6 145,43 1 6
7 147 1 7
8 147,57 1 8
9 150,95 1 9
10 158,14 1 10
11 170 1 11
12 170 2 11
13 166,59 1 12
14 161,06 1 13
15 162,88 1 14
16 180 2 15
17 183,15 1 16
18 195 1 17
19 195 2 17
i have already experimented with the analytics functions (row_number, rank, dens_rank), but didn't found a solution for this problem so far.
(oracle sql 10,11)
does anyone have a hint? thanks.
Simulating your table first:
SQL> create table mytable (w,price,r1)
2 as
3 select 1 , 167.93, 1 from dual union all
4 select 2 , 180 , 1 from dual union all
5 select 3 , 164.72, 1 from dual union all
6 select 4 , 147.42, 1 from dual union all
7 select 5 , 133.46, 1 from dual union all
8 select 6 , 145.43, 1 from dual union all
9 select 7 , 147 , 1 from dual union all
10 select 8 , 147.57, 1 from dual union all
11 select 9 , 150.95, 1 from dual union all
12 select 10, 158.14, 1 from dual union all
13 select 11, 170 , 1 from dual union all
14 select 12, 170 , 2 from dual union all
15 select 13, 166.59, 1 from dual union all
16 select 14, 161.06, 1 from dual union all
17 select 15, 162.88, 1 from dual union all
18 select 16, 180 , 2 from dual union all
19 select 17, 183.15, 1 from dual union all
20 select 18, 195 , 1 from dual union all
21 select 19, 195 , 2 from dual
22 /
Table created.
Your need column is calculated in two parts: first compute a delta column which denotes whether the previous price-column differs from the current rows price column. If you have that delta column, the second part is easy by computing the sum of those deltas.
SQL> with x as
2 ( select w
3 , price
4 , r1
5 , case lag(price,1,-1) over (order by w)
6 when price then 0
7 else 1
8 end delta
9 from mytable
10 )
11 select w
12 , price
13 , r1
14 , sum(delta) over (order by w) need
15 from x
16 /
W PRICE R1 NEED
---------- ---------- ---------- ----------
1 167.93 1 1
2 180 1 2
3 164.72 1 3
4 147.42 1 4
5 133.46 1 5
6 145.43 1 6
7 147 1 7
8 147.57 1 8
9 150.95 1 9
10 158.14 1 10
11 170 1 11
12 170 2 11
13 166.59 1 12
14 161.06 1 13
15 162.88 1 14
16 180 2 15
17 183.15 1 16
18 195 1 17
19 195 2 17
19 rows selected.
You can nest your analytic functions using inline views, so you first group the consecutive weeks with same prices and then dense_rank using those groups:
select w
, price
, r1
, dense_rank() over (
order by first_w_same_price
) drank
from (
select w
, price
, r1
, last_value(w_start_same_price) ignore nulls over (
order by w
rows between unbounded preceding and current row
) first_w_same_price
from (
select w
, price
, r1
, case lag(price) over (order by w)
when price then null
else w
end w_start_same_price
from your_table
)
)
order by w
The innermost inline view with LAG function lets the starting week of every consecutive group get it's own week number, but every consecutive week with same price gets null (weeks 12 and 19 in your data.)
The middle inline view with LAST_VALUE function then use the IGNORE NULLS feature to give the consecutive weeks the same value as the first week within each group. So week 11 and 12 both gets 11 in first_w_same_price and week 18 and 19 both gets 18 in first_w_same_price.
And finally the outer query use DENSE_RANK to give the desired result.
For each row you should count previous rows where (w-1) row price isn't the same as (w) price:
select T1.*,
(SELECT count(*)
FROM T T2
JOIN T T3 ON T2.w-1=T3.w
WHERE T2.Price<>T3.Price
AND T2.W<=T1.W)+1 rn
from t T1
SQLFiddle demo
Try this:
with tt as (
select t.*, decode(lag(price) over(order by w) - price, 0, 1, 0) diff
from t
)
select w
, price
, r1
, row_number() over (order by w) - sum(diff) over(order by w rows between UNBOUNDED PRECEDING and current row) need
from tt
SELECT w, price, r1,
ROW_NUMBER () OVER (PARTITION BY price ORDER BY price) row_column
FROM TABLE