Neea a help with grouping in sql.
I've table like
id1 id2 type
1 1 300
1 3 300
1 2 300
1 5 300
2 2 100
2 5 200
2 7 300
4 3 100
4 9 300
4 2 300
I need id1 that is mapped to one type only,
For eg, id1 '1' is mapped only to type 300, so it should only be retrieved If there is more than one type mapped to an id1 it shouldnt be retrieved. Please help.
Here is what I have attempted. But it will handle only for type 300.I need to retrieve all the id1's which are mapped to one particular type alone. So if id1 '2' is mapped for type '100' alone, it should also be retrieved.
SELECT distinct id1 from ID_TABLE where type = 300 and id1 not in
(SELECT id1 from type_table where type in (100, 200, 250))
and id1 in ( SELECT id1 FROM ID_TABLE type=300)
order by id1
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE tbl ( id1, id2, type ) AS
SELECT 1, 1, 300 FROM DUAL
UNION ALL SELECT 1, 3, 300 FROM DUAL
UNION ALL SELECT 1, 2, 300 FROM DUAL
UNION ALL SELECT 1, 5, 300 FROM DUAL
UNION ALL SELECT 2, 2, 100 FROM DUAL
UNION ALL SELECT 2, 5, 200 FROM DUAL
UNION ALL SELECT 2, 7, 300 FROM DUAL
UNION ALL SELECT 4, 3, 100 FROM DUAL
UNION ALL SELECT 4, 9, 300 FROM DUAL
UNION ALL SELECT 4, 2, 300 FROM DUAL
UNION ALL SELECT 4, 4, 200 FROM DUAL
UNION ALL SELECT 5, 2, 200 FROM DUAL
UNION ALL SELECT 5, 4, 200 FROM DUAL;
Query 1:
SELECT id1,
MIN( type )
FROM tbl
GROUP BY id1
HAVING COUNT( DISTINCT type ) = 1
Results:
| ID1 | MIN(TYPE) |
|-----|-----------|
| 1 | 300 |
| 5 | 200 |
After the group by below you may still have multiple id1 so having will limit to only those situations when there is a single id1 for type. Using MIN as an arbitrary aggregator here as there is only one value anyway:
SELECT MIN(id1), type
FROM type_table
GROUP BY type
HAVING COUNT(id1)=1;
Related
I have a RoadInsp table in Oracle 18c. I've put the data in a CTE for purpose of this question:
with roadinsp (objectid, asset_id, date_) as (
select 1, 1, to_date('2016-04-01','YYYY-MM-DD') from dual union all
select 2, 1, to_date('2019-03-01','YYYY-MM-DD') from dual union all
select 3, 1, to_date('2022-01-01','YYYY-MM-DD') from dual union all
select 4, 2, to_date('2016-04-01','YYYY-MM-DD') from dual union all
select 5, 2, to_date('2021-01-01','YYYY-MM-DD') from dual union all
select 6, 3, to_date('2022-03-01','YYYY-MM-DD') from dual union all
select 7, 3, to_date('2016-04-01','YYYY-MM-DD') from dual union all
select 8, 3, to_date('2018-03-01','YYYY-MM-DD') from dual union all
select 9, 3, to_date('2013-03-01','YYYY-MM-DD') from dual union all
select 10, 3, to_date('2010-06-01','YYYY-MM-DD') from dual
)
select * from roadinsp
OBJECTID ASSET_ID DATE_
---------- ---------- ----------
1 1 2016-04-01
2 1 2019-03-01
3 1 2022-01-01 --select this row
4 2 2016-04-01
5 2 2021-01-01 --select this row
6 3 2022-03-01 --select this row
7 3 2016-04-01
8 3 2018-03-01
9 3 2013-03-01
10 3 2010-06-01
I'm using GIS software that only lets me use SQL in a WHERE clause/SQL expression, not a full SELECT query.
I want to select the greatest n per group using the WHERE clause. In other words, for each ASSET_ID, I want to select the row that has the latest date.
As an experiment, I want to make the selection specifically using the EXISTS operator.
The reason being: While this post technically pertains to Oracle (since that's what S.O. community members would have access to), in practice, I want to use the logic in a proprietary database called a file geodatabase. The file geodatabase has very limited SQL support; a small subset of SQL-92 syntax. But it does seem to support EXISTS and subqueries, although not correlated subqueries, joins, or any modern SQL syntax. Very frustrating.
SQL reference for query expressions used in ArcGIS
Subquery support in file geodatabases is limited to the following:
Scalar subqueries with comparison operators. A scalar subquery returns a single value, for example:
GDP2006 > (SELECT MAX(GDP2005) FROM countries)
For file geodatabases, the set functions AVG, COUNT, MIN, MAX, and
SUM can only be used in scalar subqueries.
EXISTS predicate, for example:
EXISTS (SELECT * FROM indep_countries WHERE COUNTRY_NAME = 'Mexico')
Question:
Using the EXISTS operator, is there a way to select the greatest n per group? (keeping in mind the limitations mentioned above)
Edit:
If an asset has multiple rows with the same top date, then only one of those rows should be selected.
rank analytic function does the job, if it is available to you (Oracle 18c certainly does support it).
Sample data:
SQL> with roadinsp (objectid, asset_id, date_) as (
2 select 1, 1, to_date('2016-04-01','YYYY-MM-DD') from dual union all
3 select 2, 1, to_date('2019-03-01','YYYY-MM-DD') from dual union all
4 select 3, 1, to_date('2022-01-01','YYYY-MM-DD') from dual union all
5 select 4, 2, to_date('2016-04-01','YYYY-MM-DD') from dual union all
6 select 5, 2, to_date('2021-01-01','YYYY-MM-DD') from dual union all
7 select 6, 3, to_date('2022-03-01','YYYY-MM-DD') from dual union all
8 select 7, 3, to_date('2016-04-01','YYYY-MM-DD') from dual union all
9 select 8, 3, to_date('2018-03-01','YYYY-MM-DD') from dual union all
10 select 9, 3, to_date('2013-03-01','YYYY-MM-DD') from dual union all
11 select 10, 3, to_date('2010-06-01','YYYY-MM-DD') from dual
12 ),
Query begins here: first rank rows per asset_id by date in descending order:
13 temp as
14 (select r.*,
15 rank() over (partition by asset_id order by date_ desc) rnk
16 from roadinsp r
17 )
Finally, fetch rows that rank as the highest:
18 select *
19 from temp
20 where rnk = 1;
OBJECTID ASSET_ID DATE_ RNK
---------- ---------- ---------- ----------
3 1 2022-01-01 1
5 2 2021-01-01 1
6 3 2022-03-01 1
SQL>
If you can't use that, how about a subquery?
<snip>
13 select r.objectid, r.asset_id, r.date_
14 from roadinsp r
15 where (r.asset_id, r.date_) in (select t.asset_id, t.max_date
16 from (select a.asset_id, max(a.date_) max_date
17 from roadinsp a
18 group by a.asset_id
19 ) t
20 );
OBJECTID ASSET_ID DATE_
---------- ---------- ----------
6 3 2022-03-01
5 2 2021-01-01
3 1 2022-01-01
SQL>
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 small issue with my report and I need to know if its even possible to do it?
Im using Oracle12c and the tool OBIEE, im trying to create a custom column with numbers values (1 and 2) that are matching my results from my "Percent" column in a way I described below.
Here is my results in table:
I will give u an example of how it should work:
Emilian is an owner of few customers, the customers have their annual revenue listed and the column next to it its the Percent value of the total customer revenue for Emilian. Now, in my custom column I need to show "1" for customers that contribute more than (or exact) 80% of his total and "2" for the rest. So in Emilian Case, his first two customers will be "1" since 78% + 14% is already above 80% and the rest will be "2". For other Owners that only have one customer, all of them logically would be matched with "1" since their contribution is 100%
Hope I made this clear, will be veery grateful for the help with coding it :)
Alex
There's probably a much more efficient way to do this. I built up what you need to get at with a series of sub-selects. This still doesn't handle the equal percents, but you said that isn't an expected problem. I'd still watch out for it.
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE t1 ( ownerId int, customerId int, revenue int ) ;
INSERT INTO t1 ( ownerid, customerid, revenue )
SELECT 1, 1, 99 FROM dual UNION ALL
SELECT 1, 2, 200 FROM dual UNION ALL
SELECT 1, 3, 300 FROM dual UNION ALL
SELECT 1, 4, 400 FROM dual UNION ALL
SELECT 2, 5, 100 FROM dual UNION ALL
SELECT 2, 6, 100 FROM dual UNION ALL
SELECT 2, 7, 200 FROM dual UNION ALL
SELECT 2, 8, 600 FROM dual UNION ALL
SELECT 3, 9, 100 FROM dual UNION ALL
SELECT 3, 10, 900 FROM dual UNION ALL
SELECT 4, 11, 1000 FROM dual UNION ALL
SELECT 5, 12, 1000 FROM dual UNION ALL
SELECT 6, 13, 200 FROM dual UNION ALL
SELECT 6, 14, 200 FROM dual UNION ALL
SELECT 6, 15, 200 FROM dual UNION ALL
SELECT 6, 16, 200 FROM dual UNION ALL
SELECT 6, 17, 200 FROM dual UNION ALL
SELECT 42, 736784, 1480000 FROM dual UNION ALL
SELECT 42, 736580, 280160 FROM dual UNION ALL
SELECT 42, 1040137, 112486 FROM dual UNION ALL
SELECT 42, 738685, 22903 FROM dual UNION ALL
SELECT 42, 736781, 56 FROM dual
;
Query 1:
SELECT s3.ownerID, s3.customerID, s3.revenue, s3.OwnerRevenue
, CAST(s3.customerRevPct AS decimal(5,2)) AS customerRevPct
, CASE WHEN s3.PctRT < 80 OR s3.custCount = 1 THEN 1 ELSE 2 END AS customCol
/* Do the running pcts add up to 80+? 1 customer = 100% == 1. What if all are pcts are equal? */
FROM (
SELECT s2.*
, 100-SUM(nvl(s2.customerRevPct,0)) OVER (PARTITION BY s2.ownerID ORDER BY s2.customerRevPct, s2.customerID RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS pctRT
, COUNT(*) OVER (PARTITION BY s2.ownerID ORDER BY (s2.ownerID) ) AS custCount /* Is there only 1 customer? */
FROM (
SELECT s1.*
, ( ( ( s1.revenue * 1.0 ) / s1.ownerRevenue ) * 100 ) AS customerRevPct
FROM (
SELECT t1.ownerID, t1.customerID, t1.revenue
, SUM(t1.revenue) OVER ( PARTITION BY t1.ownerID ) AS ownerRevenue
FROM t1
) s1
) s2
) s3
WHERE ownerID = 42 /* REMOVE THIS LINE - TESTING ONLY */
ORDER BY s3.ownerID, s3.customerRevPct DESC
Results:
| OWNERID | CUSTOMERID | REVENUE | OWNERREVENUE | CUSTOMERREVPCT | CUSTOMCOL |
|---------|------------|---------|--------------|----------------|-----------|
| 42 | 736784 | 1480000 | 1895605 | 78.08 | 1 |
| 42 | 736580 | 280160 | 1895605 | 14.78 | 1 |
| 42 | 1040137 | 112486 | 1895605 | 5.93 | 2 |
| 42 | 738685 | 22903 | 1895605 | 1.21 | 2 |
| 42 | 736781 | 56 | 1895605 | 0 | 2 |
EDIT: I changed the Fiddle to illustrate your data example.
create table custrev(owner varchar2(100), cust_id number, revenue number);
insert into custrev values('Emilian',1,1480000);
insert into custrev values('Emilian',2,280160);
insert into custrev values('Emilian',3,112486);
insert into custrev values('Emilian',4,22903);
insert into custrev values('Emilian',5,56);
insert into custrev values('Andy',6,1378);
insert into custrev values('Sandy',7,560000);
commit;
Below is the SQL for your requirement.
select owner, cust_id, revenue, pct,
case when pct = 100 then 1
when flg is null or flg < 80 then 1
else 2 end flag_col
from (select owner, cust_id, revenue, pct,--cumulative_sum,
lag(cumulative_sum) over(partition by owner
order by revenue desc) flg
from (select owner, cust_id, revenue, pct,
sum(pct) over(partition by owner
order by revenue desc
rows between unbounded preceding
and current row) cumulative_sum
from (select owner, cust_id, revenue,
round(ratio_to_report(revenue) over(partition by owner)*100,2) pct
from custrev)
)
)
order by owner, revenue desc;
Output:
OWNER CUST_ID REVENUE PCT FLAG_COL
Andy 6 1378 100 1
Emilian 1 1480000 78.08 1
Emilian 2 280160 14.78 1
Emilian 3 112486 5.93 2
Emilian 4 22903 1.21 2
Emilian 5 56 0 2
Sandy 7 560000 100 1
Alex,
OBIEE is based on models. Not on SQL.
So sorry to say this but the SQL code will help you exactly zero...
I have a table which have 4 dimensions for a foreignid.
I want to find unique combination based on 2 dimensions.
TABLE1
-----------------------------
ID NAME VALUE TABLE2ID
-----------------------------
1 TYPE 10 1
2 DIR IN 1
3 STATE MA 1
4 COUNT 100 1
5 TYPE 10 2
6 DIR IN 2
7 STATE SA 2
8 COUNT 200 2
9 TYPE 20 3
10 DIR OUT 3
11 STATE MA 3
12 COUNT 300 3
-----------------------------
Here, I want the TABLE2IDs based on the combination of TYPE and DIR rows which is unique.
So, here if you aggregate the row values based on TYPE and DIR you will get
-----------------------------
TYPE DIR TABLE2ID
-----------------------------
10 IN 1
10 IN 2
20 OUT 3
-----------------------------
Note:
The above question is answered
Additional Question related to this.
I have another table which have the count for table2 id based on hour.
I want to group all the count for all hours in a day for unique combination in table1(Don't worry about table 2 structure).
TABLE3
-----------------------------
ID TIME COUNT TABLE2ID
-----------------------------
1 2016101601 10 1
2 2016101602 20 1
3 2016101603 30 1
4 2016101604 40 1
5 2016101601 10 2
6 2016101602 20 2
7 2016101603 30 2
8 2016101604 40 2
9 2016101601 10 3
10 2016101602 20 3
11 2016101603 30 3
12 2016101604 40 3
-----------------------------
Here, I want the output be grouped based on unique value of table 1 according to type and name(regardless of table2id)
----------------------------------
TYPE DIR DATE COUNT
----------------------------------
10 IN 20161016 200
20 OUT 20161016 100
---------------------------------
Use a PIVOT:
Oracle Setup:
CREATE TABLE table1 ( id, name, value, table2id ) AS
SELECT 1, 'TYPE', '10', 1 FROM DUAL UNION ALL
SELECT 2, 'DIR', 'IN', 1 FROM DUAL UNION ALL
SELECT 3, 'STATE', 'MA', 1 FROM DUAL UNION ALL
SELECT 4, 'COUNT', '100', 1 FROM DUAL UNION ALL
SELECT 5, 'TYPE', '10', 2 FROM DUAL UNION ALL
SELECT 6, 'DIR', 'IN', 2 FROM DUAL UNION ALL
SELECT 7, 'STATE', 'SA', 2 FROM DUAL UNION ALL
SELECT 8, 'COUNT', '200', 2 FROM DUAL UNION ALL
SELECT 9, 'TYPE', '20', 3 FROM DUAL UNION ALL
SELECT 10, 'DIR', 'OUT', 3 FROM DUAL UNION ALL
SELECT 11, 'STATE', 'MA', 3 FROM DUAL UNION ALL
SELECT 12, 'COUNT', '300', 3 FROM DUAL;
Query:
SELECT *
FROM ( SELECT name, value, table2id FROM table1 )
PIVOT ( MAX(value) FOR name IN ( 'TYPE' AS type, 'DIR' AS DIR ) );
Output:
TABLE2ID TYPE DIR
-------- ---- ---
1 10 IN
2 10 IN
3 20 OUT
Or as an alternative:
SELECT table2id,
MAX( CASE WHEN name = 'TYPE' THEN value END ) AS type,
MAX( CASE WHEN name = 'DIR' THEN value END ) AS dir
FROM table1
GROUP BY table2id;
You could join two subqueries, one that selects the types and one that selects the dirs for the same id:
SELECT type, dir, a.table2id
FROM (SELECT value AS type, table2id
FROM table1
WHERE name = 'TYPE') a
JOIN (SELECT value AS dir, table2id
FROM table1
WHERE name = 'DIR') b ON a.table2id = b.table2id
Lets say I have a table like this.
ID Parent Value
1 NULL 1000
2 1 1000
3 2 1000
4 2 1000
5 2 1000
6 2 1000
7 2 1000
8 1 1000
9 8 1000
10 8 1000
11 8 1000
I want to add every child value of a given id recursively. The correct output would be.
ID Parent Value
1 NULL 11000
2 1 6000
3 2 1000
4 2 1000
5 2 1000
6 2 1000
7 2 1000
8 1 4000
9 8 1000
10 8 1000
11 8 1000
There is only one "top" parent and it has Parent value of "Null". I'm very new to SQL so any kind of help would be appreciated. I'm using Oracle 11 if that helps.
Yes, you can do it using the CONNECT_BY_ROOT operator.
Basically, "START WITH" every row, sum up the children for each root, and then group by root. Like this:
with test_data (id, parent, value) as (
SELECT 1, NULL, 1000 FROM DUAL UNION ALL
SELECT 2, 1, 1000 FROM DUAL UNION ALL
SELECT 3, 2, 1000 FROM DUAL UNION ALL
SELECT 4, 2, 1000 FROM DUAL UNION ALL
SELECT 5, 2, 1000 FROM DUAL UNION ALL
SELECT 6, 2, 1000 FROM DUAL UNION ALL
SELECT 7, 2, 1000 FROM DUAL UNION ALL
SELECT 8, 1, 1000 FROM DUAL UNION ALL
SELECT 9, 8, 1000 FROM DUAL UNION ALL
SELECT 10, 8, 1000 FROM DUAL UNION ALL
SELECT 11, 8, 1000 FROM DUAL)
SELECT root_id id, root_parent parent, sum(value) value
FROM (
SELECT connect_by_root(id) root_id, connect_by_root(parent) root_parent, value
FROM test_data td
connect by parent = prior id
-- notice there is no "start with" clause
)
group by root_id, root_parent
order by root_id