finding the percentage using case statement - mysql

Question

The following SQL Query :
SELECT
SUM(aol_int) AS AOL,
SUM(android_phone_int) AS Android_Phone,
SUM(androidTablet_int) AS Android_Tablet
FROM mytable;
Is giving me the total values under AOL, Android_Phone and Android_Tablet columns
However, I am trying to get the percentage of AOL, Android_Phone and Android_Tablet and hence I decided to write the above query as follows:
SELECT
SUM(CASE WHEN 'aol_int' THEN 100 END) / Count(aol_int) AS AOL,
SUM(CASE WHEN ‘android_phone_int’ THEN 100 END) /Count(android_phone_int) AS Android_Phone,
SUM(CASE WHEN ‘androidTablet_int’ THEN 100 END)/Count(androidTablet_int) AS Android_Tablet
FROM mytable;
And I am getting NULL in each column. What am I doing wrong here?

Answer 1

If those 3 add up to 100 percent, then you can sumply divide each sum by the total sum to get the percentages. I get the idea that you want to see 20% as "20.00" as opposed to "0.2", so in my example I'll multiple the ration by 100 and round it off at a decimal precision of 2, but it's easy for you to change those as you see fit.
select
round((sum(aol_int) /
(sum(aol_int + android_phone_int + androidtablet_int))) *
100,2) as aol_percentage,
round((sum(android_phone_int) /
(sum(aol_int + android_phone_int + androidtablet_int))) *
100,2) as android_phone_percentage,
round((sum(androidtablet_int) /
(sum(aol_int + android_phone_int + androidtablet_int))) *
100,2) as android_tablet_percentage
from mytable