Count for each record within table - sql

I have a table (stu_grades) that stores student data and their grades at the centers they attended
I want to find out how many times for e.g. each student in that table got 'A' and then 'B' etc at any center
stu_grades
stu_ID|grade1|grade2|Grade3|center
1 A A C 1
2 B B B 2
3 C C A 1
1 C A C 2
the same student could occur more than once in the table with the same grades or even a different grade, same or different center
I especially want to check where the grade has appeared more than 3 or more times and how many centeres they exist in
So the final output should be like:
Stu_ID|Grade|Count|centercount
1 A 3 2 (As they accquired 'A' from 2 centres)
1 C 3 2
2 B 3 1 (As they only exist in 1 centre)
3 C 2 1
3 A 1 1

This is one approach using union all to unpivot the different grades into one column and then doing an aggregation.
select stu_id,grade,count(*) cnt
from (
select stu_id,grade_1 grade from stu_grades
union all
select stu_id,grade_2 grade from stu_grades
union all
select stu_id,grade_3 grade from stu_grades
) t
group by stu_id,grade

This should satisfy your requirement:
SELECT [Stud_ID], Grade, count(*) as GradeCount FROM
(SELECT [Stud_ID],Grade1 as Grade from [Stud_Details]
UNION ALL
SELECT [Stud_ID],Grade2 as Grade from [Stud_Details]
UNION ALL
SELECT [Stud_ID],Grade3 as Grade from [Stud_Details]) AS T
GROUP BY T.[Stud_ID],T.Grade
ORDER BY T.[Stud_ID]
UPDATE:
You can use HAVING after GROUP BY to get gradeCount which are greater than 3.
SELECT [Stud_ID], Grade, count(*) as GrdCountStud FROM
(SELECT [Stud_ID],Grade1 as Grade from [Stud_Details]
UNION ALL
SELECT [Stud_ID],Grade2 as Grade from [Stud_Details]
UNION ALL
SELECT [Stud_ID],Grade3 as Grade from [Stud_Details]) AS T
GROUP BY T.[Stud_ID],T.Grade
HAVING COUNT(*) > 3
ORDER BY T.[Stud_ID]

Related

Find records on group level which are connected to all other record within the group

I have a scenario where I have to find IDs within each group which are connected to all other IDs in the same group. So basically we have to treat each group separately.
In the table below, the group A has 3 IDs 1, 2 and 3. 1 is connected to both 2 and 3, 2 is connected to both 1 and 3, but 3 is not connected to 1 and 2. So 1 and 2 should be output from group A. Similarly in group B only 5 is connected to all other IDs namely 4 and 6 within group B, so 5 should be output. Similarly from group C, that should be 8, and from group D no records should be output.
So the output of the select statement should be 1, 2, 5, 8.
GRP
ID
CONNECTED_TO
A
1
2
A
1
3
A
2
3
A
2
1
A
3
5
B
4
5
B
5
4
B
5
6
B
6
4
C
7
21
C
7
25
C
8
7
D
9
31
D
10
35
D
11
37
I was able to do this if group level was not required, by below SQL:
SELECT ID FROM <table>
where CONNECTED_TO in (select ID from <table>)
group by ID
having count(*) = <number of records - 1>
But not able to find correct SQL for my scenario. Any help is appreciated.
You may use count and count(distinct) functions as the following:
select id
from tbl T
where connected_to in
(
select id from tbl T2
where T2.grp = T.grp
)
group by grp, id
having count(connected_to) =
(
select count(distinct D.id) - 1
from tbl D
where T.grp = D.grp
)
When count(connected_to) group by grp, id equals to the count(distinct id) - 1 with the same grp, this means that the ID is connected to all other IDs.

Is there a way to find active users in SQL?

I'm trying to find the total count of active users in a database. "Active" users here as defined as those who have registered an event on the selected day or later than the selected day. So if a user registered an event on days 1, 2 and 5, they are counted as "active" throughout days 1, 2, 3, 4 and 5.
My original dataset looks like this (note that this is a sample - the real dataset will run to up to 365 days, and has around 1000 users).
Day ID
0 1
0 2
0 3
0 4
0 5
1 1
1 2
2 1
3 1
4 1
4 2
As you can see, all 5 IDs are active on Day 0, and 2 IDs (1 and 2) are active until Day 4, so I'd like the finished table to look like this:
Day Count
0 5
1 2
2 2
3 2
4 2
I've tried using the following query:
select Day as days, sum(case when Day <= days then 1 else 0 end)
from df
But it gives incorrect output (only counts users who were active on each specific days).
I'm at a loss as to what I could try next. Does anyone have any ideas? Many thanks in advance!
I think I would just use generate_series():
select gs.d, count(*)
from (select id, min(day) as min_day, max(day) as max_day
from t
group by id
) t cross join lateral
generate_series(t.min_day, .max_day, 1) gs(d)
group by gs.d
order by gs.d;
If you want to count everyone as active from day 1 -- but not all have a value on day 1 -- then use 1 instead of min_day.
Here is a db<>fiddle.
A bit verbose, but this should do:
with dt as (
select 0 d, 1 id
union all
select 0 d, 2 id
union all
select 0 d, 3 id
union all
select 0 d, 4 id
union all
select 0 d, 5 id
union all
select 1 d, 1 id
union all
select 1 d, 2 id
union all
select 2 d, 1 id
union all
select 3 d, 1 id
union all
select 4 d, 1 id
union all
select 4 d, 2 id
)
, active_periods as (
select id
, min(d) min_d
, max(d) max_d
from dt
group by id
)
, days as (
select distinct d
from dt
)
select d.d
, count(ap.id)
from days d
join active_periods ap on d.d between ap.min_d and ap.max_d
group by 1
order by 1 asc
You need count by day.
select
id,
count(*)
from df
GROUP BY
id

Smarter GROUP BY

Consider Table like this.
I will call it Test
Id A B C D
1 1 1 8 25
2 1 2 5 35
3 1 3 2 75
4 2 2 2 45
5 3 2 5 26
Now I want rows with max 'Id' Grouped by 'A'
Id A B C D
3 1 3 2 75
4 2 2 2 45
5 3 2 5 26
-
--Work, but I do not want
SELECT MAX(Id), A FROM Test GROUP BY A
--I want but do not work
SELECT MAX(Id), A, B, C, D FROM Test GROUP BY A
--Work but I do not want
SELECT MAX(Id), A, B, C, D FROM Test GROUP BY A, B, C, D
--Work and I want
SELECT old.Id, old.A, new.B, new.C, new.D
FROM(
SELECT
MAX(Id) AS Id, A
FROM
Test GROUP BY A
)old
JOIN Test new
ON old.Id = new.Id
Is there a better way to write last query without join
Most databases support window functions:
select *
from (
select *, row_number() over (partition by a order by id desc) rn
from test
) t
where rn = 1
Most DBMS now support Common Table Expressions (CTE). You can use one.
;with maxa as (
select row_number() over(partition by a order by id desc) rn,
id,a,b,c,d from test
)
select id,a,b,c,d
from maxa
where rn=1

SQL union same number of columns, same data types, different data

I have two result sets that look approximately like this:
Id Name Count
1 Asd 1
2 Sdf 4
3 Dfg 567
4 Fgh 23
But the Count column data is different for the second one and I would like both to be displayed, about like this:
Id Name Count from set 1 Count from set two
1 Asd 1 15
2 Sdf 4 840
3 Dfg 567 81
4 Fgh 23 9
How can I do this in SQL (with union if possible)?
My current SQL, hope this will better explain what I want to do:
(SELECT Id, Name, COUNT(*) FROM Customers where X)
union
(SELECT Id, Name, COUNT(*) FROM Customers where Y)
select *
from
(
SELECT 'S1' as dataset, Id, Name, COUNT(*) as resultcount FROM Customers where X
union
SELECT 'S2',Id, Name, COUNT(*) FROM Customers where Y
) s
pivot
(sum(resultcount) for dataset in (s1,s2)) p
You can do something like:
;WITH Unioned
AS
(
SELECT 'Set1' FromWhat, Id, Name FROM Table1
UNION ALL
SELECT 'Set2', Id, Name FROM Table2
)
SELECT
Id,
Name,
SUM(CASE FromWhat WHEN 'Set1' THEN 1 ELSE 0 END) 'Count from set 1',
SUM(CASE FromWhat WHEN 'Set2' THEN 1 ELSE 0 END) 'Count from set 2'
FROM Unioned
GROUP BY Id, Name;
SQL Fiddle Demo

Exclude value of a record in a group if another is present

In the example table below, I'm trying to figure out a way to sum amount over id for all marks where mark 'C' doesn't exist within an id. When mark 'C' does exist in an id, I want the sum of amounts over that id, excluding the amount against mark 'A'. As illustration, my desired output is at the bottom. I've considered using partitions and the EXISTS command, but I'm having trouble conceptualizing the solution. If any of you could take a look and point me in the right direction, it would be greatly appreciated :)
sample table:
id mark amount
------------------
1 A 1
2 A 3
2 B 2
3 A 2
4 A 1
4 B 3
5 A 1
5 C 3
6 A 2
6 C 2
desired output:
id sum(amount)
-----------------
1 1
2 5
3 2
4 4
5 3
6 2
select
id,
case
when count(case mark when 'C' then 1 else null end) = 0
then
sum(amount)
else
sum(case when mark <> 'A' then amount else 0 end)
end
from sampletable
group by id
Here is my effort:
select id, sum(amount) from table t where not t.id = 'A' group by id
having id in (select id from table t where mark = 'C')
union
select id, sum(amount) from table t where t.id group by id
having id not in (select id from table t where mark = 'C')
SELECT
id,
sum(amount) AS sum_amount
FROM atable t
WHERE mark <> 'A'
OR NOT EXISTS (
SELECT *
FROM atable
WHERE id = t.id
AND mark = 'C'
)
GROUP BY
id
;