Hie i had the same issue on Monday but i used a view to kill it see if it might help

1) Create a View (my_table1 - 5 ) which pulls all the data you want in the World. Yote Yote

2) Use Below to Sum and Arrange  from the View

 select `viewalltables`.`Products` AS `Products`,
 sum((case when (`viewalltables`.`priority` = 'Immediate') then 1 else 0 end)) AS `A`,
 sum((case when (`viewalltables`.`priority` = 'Urgent') then 1 else 0 end)) AS `B`,
 sum((case when (`viewalltables`.`priority` = 'High') then 1 else 0 end)) AS `C`,
 sum((case when (`viewalltables`.`priority` = 'Normal') then 1 else 0 end)) AS `D`,
 sum((case when (`viewalltables`.`priority` = 'Low') then 1 else 0 end)) AS `E`,
 sum((case when (`viewalltables`.`name` = 'New') then 1 else 1 end)) AS `F`,
 sum((case when (`viewalltables`.`name` = 'Closed') then 1 else 0 end)) AS `G`,
 sum((case when (`viewalltables`.`name` = 'Resolved') then 1 else 0 end)) AS `H`
 from `viewalltables`
 group by `viewalltables`.`Products`;

3) i think it can cook well






On Thu, Aug 1, 2013 at 9:31 AM, Peter Karunyu <pkarunyu@gmail.com> wrote:
I have a table with 10 columns, and about 20,000 rows.

I want to get the count of the distinct values in 5 of the columns, so normally, I would probably write 5 queries thus:

SELECT column_a, count(column_a) FROM my_table GROUP BY column_a;
SELECT column_b, count(column_b) FROM my_table GROUP BY column_b;
SELECT column_c, count(column_c) FROM my_table GROUP BY column_c;
SELECT column_d, count(column_d) FROM my_table GROUP BY column_d;
SELECT column_e, count(column_e) FROM my_table GROUP BY column_e;

My question is, is there a way I can do this in a more generic/future proof way? 

Maybe a stored procedure which I just pass the table name to? Or some PHP function which I just give the name of the table to? 

Ultimate goal is to write just one query to get the above data.

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