I am currently having troubles with filtering my SQL records. I need something like what it results in the following concept: Table is
A B
1 1
1 3
2 1
2 2
2 3
2 4
3 1
3 2
I want to select value of A , where B=1 and B=2 And B=3 when same A .... result is
A
2
Please help
You can use aggregation:
select a
from mytable
where b in (1, 2, 3)
group by a
having count(*) = 3
This assumes no duplicates in the table - else, you need to change the having clause to:
having count(distinct b) = 3
Related
I have a table that looks something like this:
id name status
2 a 1
2 a 2
2 a 3
2 a 2
2 a 1
3 b 2
3 b 1
3 b 2
3 b 1
and the resultant i want is:
id name total count count(status3) count(status2) count(status1)
2 a 5 1 2 2
3 b 4 0 2 2
please help me get this result somehow, i can just get id, name or one of them at a time, don't know how to put a clause to get this table at once.
Here's a simple solution using group by and case when.
select id
,count(*) as 'total count'
,count(case status when 3 then 1 end) as 'count(status1)'
,count(case status when 2 then 1 end) as 'count(status3)'
,count(case status when 1 then 1 end) as 'count(status2)'
from t
group by id
id
total count
count(status3)
count(status2)
count(status1)
2
5
1
2
2
3
4
0
2
2
Fiddle
Here's a way to solve it using pivot.
select *
from (select status,id, count(*) over (partition by id) as "total count" from t) tmp
pivot (count(status) for status in ([1],[2],[3])) pvt
d
total count
1
2
3
3
4
2
2
0
2
5
2
2
1
Fiddle
There is a table with values as below,
Id Value
1 1
2 1
3 2
4 2
5 3
6 4
7 4
now need to write a query to retrieve value from the table and output should look as
ID Value
1 1
3 2
5 3
6 4
any suggestion ?
The query you want is nothing to do with being distinct, it's a simple aggregation of value with the minimum ID for each:
select Min(id) Id, value
from table
group by value
I want to create a rank column using existing rank and binary columns. Suppose for example a table with ID, RISK, CONTACT, DATE. The existing rank is RISK, say 1,2,3,NULL, with 3 being the highest. The binary-valued is CONTACT with 0,1 or FAILURE/SUCESS. I want to create a new RANK that will order by RISK once a certain number of successful contacts has been exceeded.
For example, suppose the constraint is a minimum of 2 successful contacts. Then the rank should be created as follows in the two instances below:
Instance 1. Three ID, all have a min of two successful contacts. In that case the rank mirrors the risk:
ID risk contact date rank
1 3 S 1 3
1 3 S 2 3
1 3 F 3 3
1 3 F 4 3
2 2 S 1 2
2 2 S 2 2
2 2 F 3 2
2 2 F 4 2
3 1 S 1 1
3 1 S 2 1
3 1 S 3 1
Instance 2. Suppose ID=1 has only one successful contact. In that case it is relegated to the lowest rank, rank=1, while ID=2 gets the highest value, rank=3, and ID=3 maps to rank=2 because it satisfies the constraint but has a lower risk value than ID=2:
ID risk contact date rank
1 3 S 1 1
1 3 F 2 1
1 3 F 3 1
1 3 F 4 1
2 2 S 1 3
2 2 S 2 3
2 2 F 3 3
2 2 F 4 3
3 1 S 1 2
3 1 S 2 2
3 1 S 3 2
This is SQL, specifically Hive. Thanks in advance.
Edit - I think Gordon Linoff's code does it correctly. In the end, I used three interim tables. The code looks like that:
First,
--numerize risk, contact
select A.* ,
case when A.risk = 'H' then 3
when A.risk = 'M' then 2
when A.risk = 'L' then 1
when A.risk is NULL then NULL
when A.risk = 'NULL' then NULL
else -999 end as RISK_RANK,
case when A.contact = 'Successful' then 1
else NULL end as success
Second,
-- sum_successes_by_risk
select A.* ,
B.sum_successes_by_risk
from T as A
inner join
(select A.person, A.program, A.risk, sum(a.success) as sum_successes_by_risk
from T as A
group by A.person, A.program, A.risk
) as B
on A.program = B.program
and A.person = B.person
and A.risk = B.risk
Third,
--Create table that contains only max risk category
select A.* ,
B.max_risk_rank
from T as A
inner join
(select A.person, max(A.risk_rank) as max_risk_rank
from T as A
group by A.person
) as B
on A.person = B.person
and A.risk_rank = B.max_risk_rank
This is hard to follow, but I think you just want window functions:
select t.*,
(case when sum(case when contact = 'S' then 1 else 0 end) over (partition by id) >= 2
then risk
else 1
end) as new_risk
from t;
Is it possible a query with Group By to show null rows?
let's say that my table has [PlaceID] and [Times].
PlaceID I Times
--------I-------
1 I 2
3 I 1
1 I 1
3 I 2
3 I 4
1 I 2
If I make the following SQL, [PlaceID] will not be visible because there is no data.
SELECT PlaceID, Sum(Times) As SumTimes
FROM tblOrder
GROUP BY PlaceID;
PlaceID I SumTimes
--------I-------
1 I 5
3 I 7
Is it possible to force it and have this output
PlaceID I SumTimes
--------I-------
1 I 5
2 I 0
3 I 7
You need a list of places . . . which I would guess is in the places table.
Then:
select p.placeId, nz(sum(times), 0)
from places as p left join
tblOrder as o
on p.placeId = o.placeId
group by p.placeId;
If there are more than three places, you can add where p.placeId in (1, 2, 3).
My SQL query returns results with 4 columns "A", "B", "C", "D".
Suppose the results are:
A B C D
1 1 1 1
1 1 1 2
2 2 2 1
Is it possible to get the count of duplicate rows with columns "A", "B", "C" in each row.
e.g. the expected result is:
A B C D cnt
1 1 1 1 2
1 1 1 2 2
2 2 2 1 1
I tried using count(*) over. But it returns me the total number of rows returned by the query.
Another information is that in example I have mentioned only 3 columns based on which I need to check the count. But my actual query has such 8 columns. And number of rows in database are huge. So I think group by will not be a feasible option here.
Any hint is appreciable.
Thanks.
Maybe too late, but probably the count over as analytic function (aka window function) within oracle helps you. When I understand your request correctly, this should solve your problem :
create table sne_test(a number(1)
,b number(1)
,c number(1)
,d number(1)
,e number(1)
,f number(1));
insert into sne_test values(1,1,1,1,1,1);
insert into sne_test values(1,1,2,1,1,1);
insert into sne_test values(1,1,2,4,1,1);
insert into sne_test values(1,1,2,5,1,1);
insert into sne_test values(1,2,1,1,3,1);
insert into sne_test values(1,2,1,2,1,2);
insert into sne_test values(2,1,1,1,1,1);
commit;
SELECT a,b,c,d,e,f,
count(*) over (PARTITION BY a,b,c)
FROM sne_test;
A B C D E F AMOUNT
-- -- -- -- -- -- ------
1 1 1 1 1 1 1
1 1 2 4 1 1 3
1 1 2 1 1 1 3
1 1 2 5 1 1 3
1 2 1 1 3 1 2
1 2 1 2 1 2 2
2 1 1 1 1 1 1
To find duplicates you must group the data based on key column
select
count(*)
,empno
from
emp
group by
empno
having
count(*) > 1;
This allows you to aggregate by empno even when multiple records exist for each category (more than one).
You have to use a subquery where you get the count of rows, grouped by A, B and C. And then you join this subquery again with your table (or with your query), like this:
select your_table.A, your_table.B, your_table.C, your_table.D, cnt
from
your_table inner join
(SELECT A, B, C, count(*) as cnt
FROM your_table
GROUP BY A, B, C) t
on t.A = your_table.A
and t.B = your_table.B
and t.C = your_table.C