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How to find the average marks of each student separated by comma | SQL Interview question

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27 авг 2024

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Комментарии : 8

@sujanthapa2856 2 месяца назад

with cte as( select name,sum(mk)/count(*) as average_marks from (select name, cast(unnest(string_to_array(marks,',' )) as integer) as mk from scores) group by 1 ) select s.name,string_agg(s.marks,',') as marks , c.average_marks from scores s join cte c on c.name=s.name group by 1,3

@mathman298 24 дня назад

select name , AVG(CAST(value AS FLOAT)) AS avg_value , rank() over(order by AVG(CAST(value AS FLOAT)) desc) rank from ( select * from scores cross apply string_split(Marks,',') ) as a group by name

@AnilBabu19666 4 месяца назад

Nice bro

@datasculptor2895 4 месяца назад

Thank you. Please like share and subscribe to my channel 🙏

@user-gq6cg3ls7f Месяц назад

SELECT name, marks, avg(cast(value as int)) avg_marks, ROW_NUMBER() over (order by (select null)) RN FROM Scores cross apply string_split(marks, ',') group by name, marks order by avg_marks desc

@vijaygupta7059 4 месяца назад

my solution in MSSQL DB: with cte as ( SELECT Name, value as marks FROM Scores cross apply string_split(Marks,',') ), cte2 as ( Select NAME, cast(marks as int) as marks from cte ) Select NAME , avg(marks) as AVG_MARKS_BY_EACH_STUDENT from cte2 group by Name

@datasculptor2895 4 месяца назад

Nice. You can still do without ctes as explained

@vijay.s-ll1yq 3 месяца назад

SELECT *,ROW_NUMBER() OVER(ORDER BY CNT DESC) RNK FROM (select name,avg(CAST(value AS INT)) AS CNT from (select s.name,value from scores s cross apply string_split (s.marks,','))dfd GROUP BY NAME)FD