I am facing issue is solving on oracle logic
I have below requirement, I have one table with Some Numbers and Corresponding ID. I want to write logic where few rows are rolled same No. Please let me know what is best way to solve this. I have used left outer join on same table but its giving me cross join.
NO ID
1 A
1 B
1 C
2 A
2 B
2 C
NO ID RoID
1 A C
1 B
2 A C
2 B
You could achieve your desired output using Analytic functions:
MAX() OVER()
LAG() OVER()
For example,
SQL> WITH sample_data AS(
2 SELECT 1 NO, 'A' ID FROM dual UNION ALL
3 SELECT 1 NO, 'B' ID FROM dual UNION ALL
4 SELECT 1 NO, 'C' ID FROM dual UNION ALL
5 SELECT 2 NO, 'A' ID FROM dual UNION ALL
6 SELECT 2 NO, 'B' ID FROM dual UNION ALL
7 SELECT 2 NO, 'C' ID FROM dual
8 )
9 -- end of sample_data mocking a real table
10 SELECT no,
11 id,
12 lag(rn) over(PARTITION BY NO ORDER BY ID DESC) rn
13 FROM
14 ( SELECT t.*, MAX(ID) OVER(PARTITION BY NO ORDER BY ID DESC) rn FROM sample_data t
15 )
16 WHERE ID <> rn
17 ORDER BY no,
18 id;
NO I R
---------- - -
1 A C
1 B
2 A C
2 B
Related
Here is the table where ORGID/USERID makes unique combination:
ORGID USERID
1 1
1 2
1 3
1 4
2 1
2 5
2 6
2 7
3 9
3 10
3 11
I need to select all records (organizations and users) wherever USERID 1 is present. So USERID=1 is present in ORGID 1 and 2 so then select all users for these organizations including user 1 itself, i.e.
ORGID USERID
1 1
1 2
1 3
1 4
2 1
2 5
2 6
2 7
Is it possible to do it with one SQL query rather than SELECT *.. WHERE USERID IN (SELECT...
You could use exists:
select *
from mytable t
where exists (select 1 from mytable t1 where t1.orgid = t.orgid and t1.userid = 1)
Another option is window functions. In Postgres:
select *
from (
select t.*,
bool_or(userid = 1) over(partition by orgid) has_user_1
from mytable t
) t
where has_user_1
Or a more generic approach, that uses portable expressions:
select *
from (
select t.*,
max(case when userid = 1 then 1 else 0 end) over(partition by orgid) has_user_1
from mytable t
) t
where has_user_1 = 1
Yes, you can do it with a single select statement - no in or exists conditions, no analytic or aggregate functions in a subquery, etc. Why you want to do it that way is not clear; in any case, it is possible that the solution below is also more efficient than the alternatives. You will have to test on your real-life data to see if that is true.
The solution below has two potential disadvantages: it only works in Oracle (it uses a proprietary extension of SQL, the match_recognize clause); and it only works in Oracle 12.1 or higher.
with
my_table(orgid, userid) as (
select 1, 1 from dual union all
select 1, 2 from dual union all
select 1, 3 from dual union all
select 1, 4 from dual union all
select 2, 1 from dual union all
select 2, 5 from dual union all
select 2, 6 from dual union all
select 2, 7 from dual union all
select 3, 9 from dual union all
select 3, 10 from dual union all
select 3, 11 from dual
)
-- End of SIMULATED data (for testing), not part of the solution.
-- In real life you don't need the WITH clause; reference your actual table.
select *
from my_table
match_recognize(
partition by orgid
all rows per match
pattern (x* a x*)
define a as userid = 1
);
Output:
ORGID USERID
---------- ----------
1 1
1 2
1 3
1 4
2 1
2 5
2 7
2 6
You can use exists:
select ou.*
from orguser ou
where exists (select 1
from orguser ou ou2
where ou2.orgid = ou.orgid and ou2.userid = 1
);
Apart from Exists and windows function, you can use IN as follows:
select *
from your_table t
where t.orgid in (select t1.orgid from your_table t1 where t1.userid = 1)
I need to generate a running number / group sequence inside a select statement for a group of data.
For example
Group Name Sequence
1 a 1
1 b 2
1 c 3
2 d 1
2 e 2
2 f 3
So for each group the sequence should be a running number starting with 1 depending on the order of column"Name".
I already pleayed around with Row_Number() and Level but I couldn't get a solution.
Any idea how to do it?
Analytic functions help.
SQL> with test (cgroup, name) as
2 (select 1, 'a' from dual union all
3 select 1, 'b' from dual union all
4 select 1, 'c' from dual union all
5 select 2, 'd' from dual union all
6 select 2, 'e' from dual union all
7 select 2, 'f' from dual
8 )
9 select cgroup,
10 name,
11 row_number() over (partition by cgroup order by name) sequence
12 from test
13 order by cgroup, name;
CGROUP N SEQUENCE
---------- - ----------
1 a 1
1 b 2
1 c 3
2 d 1
2 e 2
2 f 3
6 rows selected.
SQL>
Try this
SELECT
"Group",
Name,
DENSE_RANK() OVER (PARTITION BY "Group" ORDER BY Name) AS Sequence
FROM table;
I know basics of Oracle and I am a java developer, I can do the following operation/task in java by fetching the data and iterate over it. But I would like to know that is there any way to show start and end of the sequence and the difference in between start and end using SQL(Oracle) query.
Let's say I have a table TB1 with a column seq which contains some sequential numbers
SEQ
------
1
2
3
7
8
9
14
19
20
Is there any way to display the sequence start, end and the count as follows.
Start | end | count
---------------------
1 3 3
7 9 3
14 14 1
19 20 2
Please give me some pointer if it is achievable or not.
Thanks in advance.
Yes. You could do it easily using TABIBITOSAN method.
SELECT MIN(seq)
,MAX(seq)
,count(*)
FROM (
SELECT seq
,seq - row_number() OVER (
ORDER BY seq
) grp
FROM t
)
GROUP BY grp
ORDER BY 1;
Demo
You should write a procedure, and loop in through cursor. Keep three temp variable, wherever there is break in seq(make sure you do Order by on Seq column), and insert the start,end and count into other table.
See similar example that may be helpful to you.
SQL> WITH cte_table (seq) AS (
2 SELECT 1 FROM dual UNION ALL
3 SELECT 2 FROM dual UNION ALL
4 SELECT 3 FROM dual UNION ALL
5 SELECT 7 FROM dual UNION ALL
6 SELECT 8 FROM dual UNION ALL
7 SELECT 9 FROM dual UNION ALL
8 SELECT 14 FROM dual UNION ALL
9 SELECT 19 FROM dual UNION ALL
10 SELECT 20 FROM dual),
11 table_ AS (
12 SELECT seq, seq - row_number() OVER (ORDER BY seq) grp FROM cte_table)
13 SELECT MIN(seq) "START",
14 MAX(seq) "END",
15 COUNT(*) "COUNT"
16 FROM table_
17 GROUP BY grp
18 ORDER BY 1;
Output:
START END COUNT
---------- ---------- ----------
1 3 3
7 9 3
14 14 1
19 20 2
Using your table, the query will be
WITH table_ AS (
SELECT seq, seq - row_number() OVER (ORDER BY seq) grp FROM tb1)
SELECT MIN(seq) "START",
MAX(seq) "END",
COUNT(*) "COUNT"
FROM table_
GROUP BY grp
ORDER BY 1;
I want to fetch the records in order passed for IN condition.
select * from table where id in(6,3,7,1);
is returning the rows as
id name
1 abc
3 xy
6 ab
7 ac
but I want to display the records in same orders as ids passed in condition in Oracle
id name
6 ab
3 xy
7 ac
1 abc
Please help me in fetching the records in same order as in condition ids in oracle. The values in IN condition may change dynamically.
You can do this with a case statement in the order by clause or using a join.
select *
from table
where id in(6,3,7,1)
order by (case id when 6 then 1 when 3 then 2 when 7 then 3 when 1 then 4 end);
Or:
with ids as (
select 6 as id, 1 as ordering from dual union all
select 3 as id, 2 as ordering from dual union all
select 7 as id, 3 as ordering from dual union all
select 1 as id, 4 as ordering from dual
)
select *
from table t join
ids
on t.ids = ids.id
order by ids.ordering;
Note that you don't need the in in this case, because the join does the filtering.
you can use trick
select * from table where id in(6,3,7,1) order by case when id = 6 then 1
id = 3 then 2
id = 7 then 3
id = 1 then 4
end
For example, I have table:
ID | Value
1 hi
1 yo
2 foo
2 bar
2 hehe
3 ha
6 gaga
I want my query to get ID, Value; meanwhile the returned set should be in the order of frequency count of each ID.
I tried the query below but don't know how to get the ID and Value column at the same time:
SELECT COUNT(*) FROM TABLE group by ID order by COUNT(*) desc;
The count number doesn't matter to me, I just need the data to be in such order.
Desire Result:
ID | Value
2 foo
2 bar
2 hehe
1 hi
1 yo
3 ha
6 gaga
As you can see because ID:2 appears most times(3 times), it's first on the list,
then ID:1(2 times) etc.
you can try this -
select id, value, count(*) over (partition by id) freq_count
from
(
select 2 as ID, 'foo' as value
from dual
union all
select 2, 'bar'
from dual
union all
select 2, 'hehe'
from dual
union all
select 1 , 'hi'
from dual
union all
select 1 , 'yo'
from dual
union all
select 3 , 'ha'
from dual
union all
select 6 , 'gaga'
from dual
)
order by 3 desc;
select t.id, t.value
from TABLE t
inner join
(
SELECT id, count(*) as cnt
FROM TABLE
group by ID
)
x on x.id = t.id
order by x.cnt desc
How about something like
SELECT t.ID,
t.Value,
c.Cnt
FROM TABLE t INNER JOIN
(
SELECT ID,
COUNT(*) Cnt
FROM TABLE
GROUP BY ID
) c ON t.ID = c.ID
ORDER BY c.Cnt DESC
SQL Fiddle DEMO
I see the question is already answered, but since the most obvious and most simple solution is missing, I'm posting it anyway. It doesn't use self joins nor subqueries:
SQL> create table t (id,value)
2 as
3 select 1, 'hi' from dual union all
4 select 1, 'yo' from dual union all
5 select 2, 'foo' from dual union all
6 select 2, 'bar' from dual union all
7 select 2, 'hehe' from dual union all
8 select 3, 'ha' from dual union all
9 select 6, 'gaga' from dual
10 /
Table created.
SQL> select id
2 , value
3 from t
4 order by count(*) over (partition by id) desc
5 /
ID VALU
---------- ----
2 bar
2 hehe
2 foo
1 yo
1 hi
6 gaga
3 ha
7 rows selected.