In SQL ,How do I retrieve name FROM track and ORDER BY bit rate (bit/s) then name . columns : name, bytes and milliseconds

Answer 1

the name of tracks where the bit rate are greater than 300 kilobits per second is:

select name from track where name = name and (bytes *8*1000)/(milliseconds *1000) > 300 order by name asc;

Answer 2

select name from track order by ( ([bytes]*8) / ([milliseconds]*1000) ) * 125, name

Explanation :

(1) calculate bit rate ie. (bits/seconds)
= bytes*8 / milliseconds*1000

(2) To counter external multipliers that we added in bit rate equation, multiply step 1 equation with (1000/8) ie. 125.

(3) So, bit rate in bit/s will be ( ([bytes]*8) / ([milliseconds]*1000) ) * 125.

(4) Use this formula with Order By Clause.

Answer 3

Hi there thanks it was accepted. my appreciations to you🐸

Answer 4

Get the name of the tracks where the bit rate is greater than 300 kilobits per second (Kb/s). How to do it? previous answer didn't helped me :(

select name from track where ((bytes*8000)/(milliseconds*1000))>300

Answer 5

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Answer 6

Get the name of the tracks where the bit rate is greater than 300 kilobits per second (Kb/s). How to do it? previous answer didn't helped me :(

Answer 7

you need to multiply by 8 and divided milliseconds by 1000 then divide the whole thing by 1000

Answer 8

Select name, bytes*8 / milliseconds*1000 from track