MYSQL Count of duplicate records with condition - mysql

Question

I'm trying to get count of duplicate data, but my query doesn't work correctly.
Every user should have one dev_id but when other user will have this same dev_id I want to know this
Table for example:
dev_id user_id
------------------
111 1
111 1
222 2
111 2
333 3
Should result:
user_id qu
------------------
1 1
2 1
3 0
This is my query
SELECT t1.user_id,
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = t1.dev_id
) AS qu
FROM reports t1
GROUP BY t1.user_id

Answer 1

You can get results by doing:
select r.user_id, count(*) - 1
from reports r
group by r.user_id;
Is this the calculation that you want?

Answer 2

Okay. Let start from simple.
First you need get unique user_id/dev id combinations
select distinct dev_id,user_id from reports
Result will be
dev_id user_id
------------------
111 1
222 2
111 2
333 3
After that you should get number of different user_id per dev_id
select dev_id,c from (
SELECT
dev_id,
count(*)-1 AS c
FROM
(select distinct user_id,dev_id from reports) as fixed_reports
GROUP BY dev_id
) as counts
Result of such query will be
dev_id c
-----------------
111 1
222 0
333 0
Now you should show users which have such dev_id. For that you should join such dev_id list with table from step1(which show which one user_id,dev_id pairs exist)
select distinct fixed_reports2.user_id,counts.c from (
SELECT
dev_id,
count(*)-1 AS c
FROM
(select distinct user_id,dev_id from reports) as fixed_reports
GROUP BY dev_id
) as counts
join
(select distinct user_id,dev_id from reports) as fixed_reports2
on fixed_reports2.dev_id=counts.dev_id
where counts.c>0 and counts.c is not null
"Distinct" here need to skip same rows.
Result should be for internal query
dev_id c
-----------------
111 1
For all query
user_id c
------------------
1 1
2 1
If you sure you need also rows with c=0, then you need do "left join" of fixed_reports2 and large query,that way you will get all rows and rows with c=null will be rows with 0(can be changed by case/when statement)

Answer 3

I think following sql query should solve you problem:
SELECT t1.user_id, t1.dev_id, count(t2.user_id) as qu
FROM (Select Distinct * from reports) t1
Left Join (Select Distinct * from reports) t2
on t1.user_id != t2.user_id and t2.dev_id = t1.dev_id
group by t1.user_Id, t1.dev_id
SQL Fiddle Link

Answer 4

SELECT user_id, (COUNT(user_id) -1) as qu
FROM reports
GROUP BY user_id
This would give desired result in your case, however you can improve it a lot more.
Cheers,,

Answer 5

Your query is broken and would not run on many systems. The problem is that the group with user_id of 2 has two different dev_ids. If you run the "broken query" below you can see that the min() and max() are distinct but the subquery only sees one of those values which is randomly chosen. The last query is corrected by adding dev_id to the groupings which shows you where the "missing" row went in the counts.
SELECT -- broken query
t1.user_id, min(t1.dev_id), max(t1.dev_id),
(select distinct t1.dev_id from reports) as should_have_errored,
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = t1.dev_id
) AS qu
FROM reports t1
GROUP BY t1.user_id;
-- On SQL Server that query returns an error
-- Msg 8120, Level 16, State 1, Line 7
-- Column 'reports.dev_id' is invalid in the select list because it is
-- not contained in either an aggregate function or the GROUP BY clause.
SELECT -- query that duplicates your original query
t1.user_id,
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = max(t1.dev_id) /* <-- see here */
) AS qu
FROM reports t1
GROUP BY t1.user_id;
SELECT t1.user_id, t1.dev_id, -- fixed query
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = t1.dev_id
) AS qu
FROM reports t1
GROUP BY t1.user_id, t1.dev_id
http://sqlfiddle.com/#!9/6576e3/20
Here are some queries that might be useful:
Which dev_ids have multiple user_ids associated with them?
select dev_id
from reports
group by dev_id
having count(distinct user_id) > 1
Which other user_ids share a dev_id with this user_id?
select user_id
from reports r1
where exists (
select 1
from reports r2
where r2.dev_id = r1.dev_id and r2.user_id <> ?
)
Or really that's just equivalent to an inner join which also makes it easy to list everybody at once. Note that each pair will be listed twice:
select r1.user_id, r1.dev_id, r2.user_id as common_user_id
from
reports r1 inner join reports r2
on r2.dev_id = r1.dev_id
where
r1.user_id <> r2.user_id
order by
r1.user_id, r1.dev_id, r2.user_id
And since you've got duplicate rows in your table you'd need to make it select distinct to get unique rows.

Answer 6

Try
SELECT
user_id,
SUM(qu) AS qu
FROM (
SELECT
user_id,
count(*)-1 AS qu
FROM
reports
GROUP BY user_id, dev_id
) AS r
GROUP BY user_id
No need to do a join if all the data you need is in one table.
Edit: changed the group by to dev_id instead of user_id
Edit2: I think you need both dev_id and user_id in the group by clause.
Edit3: Added a subquery to get the desired result. This might be a little cumbersome, perhaps someone has a way to improve this?