I have to perform a query where I can count the number of distinct codes per Id.
|Id | Code
------------
| 1 | C
| 1 | I
| 2 | I
| 2 | C
| 2 | D
| 2 | D
| 3 | C
| 3 | I
| 3 | D
| 4 | I
| 4 | C
| 4 | C
The output should be something like:
|Id | Count | #Code C | #Code I | #Code D
-------------------------------------------
| 1 | 2 | 1 | 1 | 0
| 2 | 3 | 1 | 0 | 2
| 3 | 3 | 1 | 1 | 1
| 4 | 2 | 2 | 1 | 0
Can you give me some advise on this?
This answers the original version of the question.
You are looking for count(distinct):
select id, count(distinct code)
from t
group by id;
If the codes are only to the provided ones, the following query can provide the desired result.
select
pvt.Id,
codes.total As [Count],
COALESCE(C, 0) AS [#Code C],
COALESCE(I, 0) AS [#Code I],
COALESCE(D, 0) AS [#Code D]
from
( select Id, Code, Count(code) cnt
from t
Group by Id, Code) s
PIVOT(MAX(cnt) FOR Code IN ([C], [I], [D])) pvt
join (select Id, count(distinct Code) total from t group by Id) codes on pvt.Id = codes.Id ;
Note: as I can see from sample input data, code 'I' is found in all of Ids. Its count is zero for Id = 3 in the expected output (in the question).
Here is the correct output:
DB Fiddle
Related
I have a table referrals:
id | user_id_owner | firstname | is_active | user_type | referred_at
----+---------------+-----------+-----------+-----------+-------------
3 | 2 | c | t | agent | 3
5 | 3 | e | f | customer | 5
4 | 1 | d | t | agent | 4
2 | 1 | b | f | agent | 2
1 | 1 | a | t | agent | 1
And another table activations
id | user_id_owner | referral_id | amount_earned | activated_at | app_id
----+---------------+-------------+---------------+--------------+--------
2 | 2 | 3 | 3.0 | 3 | a
4 | 1 | 1 | 6.0 | 5 | b
5 | 4 | 4 | 3.0 | 6 | c
1 | 1 | 2 | 2.0 | 2 | b
3 | 1 | 2 | 5.0 | 4 | b
6 | 1 | 2 | 7.0 | 8 | a
I am trying to generate another table from the two tables that has only unique values for referrals.id and returns as one of the columns the count for each apps as best_selling_app_count.
Here is the query I ran:
with agents
as
(select
referrals.id,
referral_id,
amount_earned,
referred_at,
activated_at,
activations.app_id
from referrals
left outer join activations
on (referrals.id = activations.referral_id)
where referrals.user_id_owner = 1),
distinct_referrals_by_id
as
(select
id,
count(referral_id) as activations_count,
sum(coalesce(amount_earned, 0)) as amount_earned,
referred_at,
max(activated_at) as last_activated_at
from
agents
group by id, referred_at),
distinct_referrals_by_app_id
as
(select id, app_id as best_selling_app,
count(app_id) as best_selling_app_count
from agents
group by id, app_id )
select *, dense_rank() over (order by best_selling_app_count desc) best_selling_app_rank
from distinct_referrals_by_id
inner join distinct_referrals_by_app_id
on (distinct_referrals_by_id.id = distinct_referrals_by_app_id.id);
Here is the result I got:
id | activations_count | amount_earned | referred_at | last_activated_at | id | best_selling_app | best_selling_app_count | best_selling_app_rank
----+-------------------+---------------+-------------+-------------------+----+------------------+------------------------+-----------------------
2 | 3 | 14.0 | 2 | 8 | 2 | b | 2 | 1
1 | 1 | 6.0 | 1 | 5 | 1 | b | 1 | 2
2 | 3 | 14.0 | 2 | 8 | 2 | a | 1 | 2
4 | 1 | 3.0 | 4 | 6 | 4 | c | 1 | 2
The problem with this result is that the table has a duplicate id of 2. I only need unique values for the id column.
I tried a workaround by harnessing distinct that gave desired result but I fear the query results may not be reliable and consistent.
Here is the workaround query:
with agents
as
(select
referrals.id,
referral_id,
amount_earned,
referred_at,
activated_at,
activations.app_id
from referrals
left outer join activations
on (referrals.id = activations.referral_id)
where referrals.user_id_owner = 1),
distinct_referrals_by_id
as
(select
id,
count(referral_id) as activations_count,
sum(coalesce(amount_earned, 0)) as amount_earned,
referred_at,
max(activated_at) as last_activated_at
from
agents
group by id, referred_at),
distinct_referrals_by_app_id
as
(select
distinct on(id), app_id as best_selling_app,
count(app_id) as best_selling_app_count
from agents
group by id, app_id
order by id, best_selling_app_count desc)
select *, dense_rank() over (order by best_selling_app_count desc) best_selling_app_rank
from distinct_referrals_by_id
inner join distinct_referrals_by_app_id
on (distinct_referrals_by_id.id = distinct_referrals_by_app_id.id);
I need a recommendation on how best to achieve this.
I am trying to generate another table from the two tables that has only unique values for referrals.id and returns as one of the columns the count for each apps as best_selling_app_count.
Your question is really complicated with a very complicated SQL query. However, the above is what looks like the actual question. If so, you can use:
select r.*,
a.app_id as most_common_app_id,
a.cnt as most_common_app_id_count
from referrals r left join
(select distinct on (a.referral_id) a.referral_id, a.app_id, count(*) as cnt
from activations a
group by a.referral_id, a.app_id
order by a.referral_id, count(*) desc
) a
on a.referral_id = r.id;
You have not explained the other columns that are in your result set.
I'm trying to find an efficient way to derive the column Expected below from only Id and State. What I want is for the number Expected to increase each time State is 0 (ordered by Id).
+----+-------+----------+
| Id | State | Expected |
+----+-------+----------+
| 1 | 0 | 1 |
| 2 | 1 | 1 |
| 3 | 0 | 2 |
| 4 | 1 | 2 |
| 5 | 4 | 2 |
| 6 | 2 | 2 |
| 7 | 3 | 2 |
| 8 | 0 | 3 |
| 9 | 5 | 3 |
| 10 | 3 | 3 |
| 11 | 1 | 3 |
+----+-------+----------+
I have managed to accomplish this with the following SQL, but the execution time is very poor when the data set is large:
WITH Groups AS
(
SELECT Id, ROW_NUMBER() OVER (ORDER BY Id) AS GroupId FROM tblState WHERE State=0
)
SELECT S.Id, S.[State], S.Expected, G.GroupId FROM tblState S
OUTER APPLY (SELECT TOP 1 GroupId FROM Groups WHERE Groups.Id <= S.Id ORDER BY Id DESC) G
Is there a simpler and more efficient way to produce this result? (In SQL Server 2012 or later)
Just use a cumulative sum:
select s.*,
sum(case when state = 0 then 1 else 0 end) over (order by id) as expected
from tblState s;
Other method uses subquery :
select *,
(select count(*)
from table t1
where t1.id < t.id and state = 0
) as expected
from table t;
I need to select cid, project, and owner from rows in the table below where one or more rows for a cid/project combination has an owner of 1.
cid | project | phase | task | owner
-----------------------------------
1 | 1 | 1 | 1 | 1
1 | 1 | 1 | 2 | 2
1 | 1 | 1 | 3 | 2
2 | 1 | 1 | 1 | 1
2 | 1 | 1 | 2 | 1
3 | 1 | 1 | 3 | 2
My output table should look like the this:
cid | project | phase | task | owner
-----------------------------------
1 | 1 | 1 | 1 | 1
1 | 1 | 1 | 2 | 2
1 | 1 | 1 | 3 | 2
2 | 1 | 1 | 1 | 1
2 | 1 | 1 | 2 | 1
The below query is what I came up with. It does seem to test okay, but my confidence is low. Is the query an effective way to solve the problem?
select task1.cid, task1.project, task1.owner
from
(select cid, project, owner from table) task1
right join
(select distinct cid, project, owner from table where owner = 1) task2
on task1.cid = task2.cid and task1.project = task2.project
(I did not remove the phase and task columns from the sample output so that it would be easier to compare.)
You can simply use a IN clause
select cid, project, owner
from table
where cid in (select distinct id from table where owner = 1)
or a inner join with a subquery
select a.cid, a.project, a.owner
from table a
INNER JOIN ( select distinct cid , project
from table where owner = 1
) t on t.cid = a.cid and t.project = a.project
Given a cross-reference table t relating table a with b:
| id | a_id | b_id |
--------------------
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 3 | 1 | 3 |
| 4 | 2 | 7 |
| 5 | 2 | 3 |
| 6 | 3 | 2 |
| 7 | 3 | 3 |
What would be the conventional way of selecting all a_id whose b_id is a superset of a given set?
For example, for the set (2,3), I would expect the result:
| a_id |
--------
| 1 |
| 3 |
Since a_id 1 and 3 are the only set of b_id that is a superset of (2,3).
The best solution I've found so far (thanks to this answer):
select id
from a
where 2 = (select count(*)
from t
where t.a_id = a.id and t.b_id in (2,3)
);
But I'd prefer to avoid calculating stuff like cardinality before running the query.
You can simply adapt the query as:
select id
from a cross join
(select count(*) as cnt
from t
where . . .
) x
where x.cnt = (select count(*)
from t
where t.a_id = a.id and t.b_id in (2,3)
);
i want to start by saying that i know there are a couple of questions regarding similar problems but they either dont answer my question fully or seem to be incompatible with SQLite.
I want to query all rows with value -1 and the first rows with values other than -1.
And by "first rows" i mean the group of rows that are first with a certain value. the first row is the row that is first stumbled upon depending on the SORT BY clause
An example of the data and outcome:
Data:
a b -1
c d 1
e f 2
g h 2
i j 2
k l -1
Result:
a b -1
c d 1
e f 2
k l -1
And as said above, i am using a SQLite database
Do this as two separate queries, one of them containing an inline view; UNION the two
select blah, blah from T where ...
UNION
select * from
(
select blah, blah from T where something else order by somecolumn limit 1
)
With simple example:
> select * from ex1;
+------+----------+
| id | name |
+------+----------+
| 1 | Pirate |
| 2 | Monkey |
| 3 | Ninja |
| 4 | Spagheti |
| 5 | kumar |
| 6 | siva |
+------+----------+
> select * from ex1 union select "1", "sing" order by case name when 'sing' then 1 else 2 end, name;
+------+----------+
| id | name |
+------+----------+
| 1 | sing |
| 5 | kumar |
| 2 | Monkey |
| 3 | Ninja |
| 1 | Pirate |
| 6 | siva |
| 4 | Spagheti |
+------+----------+