my question is:
If i have table student with student_id and other fields(not matter)
and table exam with three fields: student_id, subject_id(id from another table subject) and exam_mark.
I need to find count of students who have the same exam marks for each subject.
I have stuck on that task, you can get me idea how can i do or explain full solution.
Simple data:
For this scenario query must return count(student_id) = 4
You need nested aggregation:
select count(*)
from
(
select student_id
from tab
group by student_id
-- same min and max means only one value exists
having min(exam_mark) = max(exam_mark)
) as dt
The min/max is usually more efficient than COUNT(DISTINCT exam_mark) = 1.
If you want only students with multiple marks enhance HAVING with
AND count(*) > 1
Related
So I need to select all the students, having the minimum grade for each prof. For example, if Augustinus had two students with grade 1.0, then I would like to see both in the result.
Table of my data
What the result could look like, if the LIMIT was set to 10
So what I basically want is to see the best students that each prof has.
What I have tried is the following:
SELECT professor, student, min(note)
FROM temp
GROUP BY professor
ORDER BY note
The problem of course being that I only get one minimum value for each prof and not all minimum values.
*temp is just the table name
One way to solve these types of problems is to use a subquery to rank the grades for each class in a descending order. This involves a window function. With a second query you can limit the results based on your criteria of 10.
SELECT professor, student, note
FROM
(
SELECT professor,student,note,
row_number() over(partition by professor order by note desc) as downwardrank
) as rankings
WHERE
downwardrank <= 10
Just found a solution myself:
SELECT professor, student, note
FROM temp
WHERE (professor, note) IN
(SELECT professor, min(note)
FROM temp
GROUP BY professor
ORDER BY note)
ORDER BY note, professor, student
LIMIT 10
Say I have a Table called "MARKS" with the columns: Value, subject_id and student_id.
Now I want to write a query to display the names of all students who have secured more than 50 in ALL subjects that they have appeared in.
How can that be achieved?
Example:
Lets say the subjects are maths, english, history.
Even if a student scores 100 in maths, 100 in english but 40 in history, he should be considered as failed and not be displayed.
There are several ways to get what you expect, but in the simplest case the HAVING clause may help. In the following query grouping is done by student_id, so the min function gets minimal value over all subjects for each student_id:
SELECT student_id
FROM marks_table
GROUP BY student_id
HAVING min(marks) > 50;
Then join student names by student_id.
I would say:
select student_id
from table
where student_id not in (
select student_id
from table
where value < 50
)
Beware, if you have nulls in student_id you'll receive incorrect results. Geta round this with a coalesce() in the sub-select
Returns all students with subjects appeared
select student_id,subject_id, value from marks
where
(case when value < 50 then 'failed' else 'pass' end ) = 'pass'
select *
from
(
select student_id, MIN(Value) as sum_value
from MARKS
group by student_id
) summed
where
sum_value > 50
I have two tables that I was going to join, but I understand it's more efficient to use CREATE VIEW. This is what I have:
CREATE OR REPLACE VIEW view0_joinedTablesGrouped
AS
Select table1.*,table2.*
FROM table1
inner join table2 on table1.col =
table2.matchingcol
group by table2.matchingcol;
which causes the following error:
ERROR: column "table1.col" must appear in the GROUP BY clause or be
used in an aggregate function
LINE 3: Select table.*,table2.*
Group By cannot do what you are trying to do.
Consider a simple table:
Name Age
-------
Ann 10
Bill 10
Chris 11
If you try to group by age with:
Select * from Table group by Age
What, exactly, do you expect to appear in the Name column for Age=10? Ann, or Bill or both or neither or ....? There is no good answer.
So, when you group by, every column in the output has to be an aggregate – that means a function of every row in the group.
So these are valid:
Select Age, Count(*) from Table group by Age
Select Age, Max( Length(Name)) from Table group by Age
Select Age, Max(Name) from Table group by Age
But this is impossible to do, and isn't valid:
Select Age,Name from Table group by Age
So your select * is the problem -- you can't just select column values because when you group by there's a whole group of column values for every output row, and you can't stuff all those values into one column of one row.
As for using a view, #systemjack's comment is correct.
I have a table Student in SQL Server with these columns:
[ID], [Age], [Level]
I want the query that returns each age value that appears in Students, and finds the level value that appears most often. For example, if there are more 'a' level students aged 18 than 'b' or 'c' it should print the pair (18, a).
I am new to SQL Server and I want a simple answer with nested query.
You can do this using window functions:
select t.*
from (select age, level, count(*) as cnt,
row_number() over (partition by age order by count(*) desc) as seqnum
from student s
group by age, level
) t
where seqnum = 1;
The inner query aggregates the data to count the number of levels for each age. The row_number() enumerates these for each age (the partition by with the largest first). The where clause then chooses the highest values.
In the case of ties, this returns just one of the values. If you want all of them, use rank() instead of row_number().
One more option with ROW_NUMBER ranking function in the ORDER BY clause. WITH TIES used when you want to return two or more rows that tie for last place in the limited results set.
SELECT TOP 1 WITH TIES age, level
FROM dbo.Student
GROUP BY age, level
ORDER BY ROW_NUMBER() OVER(PARTITION BY age ORDER BY COUNT(*) DESC)
Or the second version of the query using amount each pair of age and level, and max values of count pair age and level per age.
SELECT *
FROM (
SELECT age, level, COUNT(*) AS cnt,
MAX(COUNT(*)) OVER(PARTITION BY age) AS mCnt
FROM dbo.Student
GROUP BY age, level
)x
WHERE x.cnt = x.mCnt
Demo on SQLFiddle
Another option but will require later version of sql-server:
;WITH x AS
(
SELECT age,
level,
occurrences = COUNT(*)
FROM Student
GROUP BY age,
level
)
SELECT *
FROM x x
WHERE EXISTS (
SELECT *
FROM x y
WHERE x.occurrences > y.occurrences
)
I realise it doesn't quite answer the question as it only returns the age/level combinations where there are more than one level for the age.
Maybe someone can help to amend it so it includes the single level ages aswell in the result set: http://sqlfiddle.com/#!3/d597b/9
with combinations as (
select age, level, count(*) occurrences
from Student
group by age, level
)
select age, level
from combinations c
where occurrences = (select max(occurrences)
from combinations
where age = c.age)
This finds every age and level combination in the Students table and counts the number of occurrences of each level.
Then, for each age/level combination, find the one whose occurrences are the highest for that age/level combination. Return the age and level for that row.
This has the advantage of not being tied to SQL Server - it's vanilla SQL. However, a window function like Gordon pointed out may perform better on SQL Server.
Say I have a list of student names and their marks. I want to find out the highest mark and the student, how can I write one select statement to do that?
Assuming you mean marks rather than remarks, use:
select name, mark
from students
where mark = (
select max(mark)
from students
)
This will generally result in a fairly efficient query. The subquery should be executed once only (unless your DBMS is brain-dead) and the result fed into the second query. You may want to ensure that you have an index on the mark column.
If you don't want to use a subquery:
SELECT name, remark
FROM students
ORDER BY remark DESC
LIMIT 1
select name, remarks
from student
where remarks =(select max(remarks) from student)
If you are using a database that supports windowing,
SELECT name, mark FROM
(SELECT name, mark, rank() AS rk
FROM student_marks OVER (ORDER BY mark DESC)
) AS subqry
WHERE subqry.rk=1;
This probably does not run as fast as the mark=(SELECT MAX(mark)... style query, but it would be worth checking out.
In SQL Server:
SELECT TOP 1 WITH TIES *
FROM Students
ORDER BY Mark DESC
This will return all the students that have the highest mark, whether there is just one of them or more than one. If you want only one row, drop the WITH TIES specifier. (But the actual row is not guaranteed to be always the same then.)
You can create view and join it with original table:
V1
select id , Max(columName)
from t1
group by id
select * from t1
where t1.id = V1.id and t1.columName = V1.columName
this is right if you need Max Values with related info
I recently had a need for something "kind of similar" to this post and wanted to share a technique. Say you have an Order and OrderDetail table, and you want to return info from the Order table along with the product name associated with the highest priced detail row. Here's a way to pull that off without subtables, RANK, etc.. The key is to create and aggregate that combined the key and value from the detailed table and then just max on that and substring out the value you want.
create table CustOrder(ID int)
create table CustOrderDetail(OrderID int, Price money, ProdName varchar(20))
insert into CustOrder(ID) values(1)
insert into CustOrderDetail(OrderID,Price,ProdName) values(1,10,'AAA')
insert into CustOrderDetail(OrderID,Price,ProdName) values(1,50,'BBB')
insert into CustOrderDetail(OrderID,Price,ProdName) values(1,10,'CCC')
select
o.ID,
JoinAggregate=max(convert(varchar,od.price)+'*'+od.prodName),
maxProd=
SUBSTRING(
max(convert(varchar,od.price)+'*'+od.prodName)
,CHARINDEX('*',max(convert(varchar,od.price)+'*'+convert(varchar,od.prodName))
)+1,9999)
from
CustOrder o
inner join CustOrderDetail od on od.orderID = o.ID
group by
o.ID